行列A(2x2)に対してA, A+E, A-Eのいずれかは逆行列を持つ証明

行列$A(2x2)$に対して$A, A+E, A-E$のいずれかは逆行列$A^{-1}$を持つ証明(Eは単位行列$\left[\begin{array}{cc}1&0\\0&1\\\end{array}\right]$)

$$\begin{array}{rcl} A&=& \displaystyle \left[ \begin{array}{cc} a & b \\ c & d \\ \end{array} \right]\\ A^{-1}&=& \href{https://shikitenkai.blogspot.com/2021/05/blog-post_96.html}{\displaystyle \frac{1}{ |A| } \displaystyle \left[ \begin{array}{cc} d & -b \\ -c & a \\ \end{array} \right]}\\ A+E &=& \displaystyle \left[ \begin{array}{cc} a & b \\ c & d \\ \end{array} \right] + \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right]\\ &=& \displaystyle \left[ \begin{array}{cc} a+1 & b \\ c & d+1 \\ \end{array} \right]\\ (A+E)^{-1}&=& \displaystyle \frac{1}{ |A+E| } \displaystyle \left[ \begin{array}{cc} d+1 & -b \\ -c & a+1 \\ \end{array} \right]\\ A-E &=& \displaystyle \left[ \begin{array}{cc} a & b \\ c & d \\ \end{array} \right] - \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right]\\ &=& \displaystyle \left[ \begin{array}{cc} a-1 & b \\ c & d-1 \\ \end{array} \right]\\ (A+E)^{-1}&=& \displaystyle \frac{1}{ |A-E| } \displaystyle \left[ \begin{array}{cc} d-1 & -b \\ -c & a-1 \\ \end{array} \right]\\ \end{array}$$ $A, A+E, A-E$の全てが逆行列を持たないとすると $|A|, |A+E|, |A-E|$の全てが同時にゼロ$0$になるのでこれを仮定として始める． $$\begin{align} |A|&=&ad-bc&=&0 \tag{1}\label{a1}\\ |A+E|&=&(a+1)(d+1)-bc&=&ad-bc+a+d+1=0 \tag{2}\label{a2}\\ |A-E|&=&(a-1)(d-1)-bc&=&ad-bc-a-d+1=0 \tag{3}\label{a3}\\ \end{align}$$ $\eqref{a1}$を$\eqref{a2}$に代入すると$a+d=-1$． $$\begin{align} ad-bc+a+d+1&=&0 \\ 0+a+d+1&=&0\\ a+d&=&-1\tag{4}\label{a4}\\ \end{align}$$ $\eqref{a1}$を$\eqref{a3}$に代入すると$a+d=1$． $$\begin{align} ad-bc-a-d+1&=&0 \\ 0-a-d+1&=&0\\ -a-d&=&-1\\ a+d&=&1\tag{5}\label{a5}\\ \end{align}$$ $a+d=-1\eqref{a4}$と$a+d=1\eqref{a5}$で矛盾するので， $|A|, |A+E|, |A-E|$の全てが同時にゼロ$0$になるという前提が間違いということになり， $|A|, |A+E|, |A-E|$のいずれかはゼロ$0$でない．よって$A, A+E, A-E$のいずれかは逆行列を持つ．

標本平均の分布の歪度

標本平均($\overline{X}$)の分布の歪度

$$\begin{array}{rcl} \displaystyle \beta_1(\overline{X}) &=&\displaystyle \frac{E\left[\left(\overline{X}-\mu\right)^3\right]}{\left(V\left[\overline{X}\right]^{\frac{1}{2}}\right)^3}\\ &=&\displaystyle \frac{\frac{\mu_3}{n^2}}{\left\{ \left( \frac{\sigma^2}{n} \right) ^{\frac{1}{2}} \right\}^3} &&\displaystyle\,\dotso\,\href{https://shikitenkai.blogspot.com/2019/07/overlinexmu3.html}{E\left[\left(\overline{X}-\mu\right)^3\right]=\mu_3\left(\overline{X}\right)=\frac{\mu_3}{n^2}} ,\,\href{https://shikitenkai.blogspot.com/2019/06/specimen-random-variable_3.html}{V\left[\overline{X}\right]=\frac{\sigma^2}{n}}\\ &=&\displaystyle \frac{\frac{\mu_3}{n^2}}{\left( \frac{\sigma}{\sqrt{n}} \right)^3}\\ &=&\displaystyle \frac{\frac{\mu_3}{n^2}}{ \frac{\sigma^3}{n\sqrt{n}} }\\ &=&\displaystyle \frac{\mu_3}{n^2} \frac{n\sqrt{n}}{\sigma^3}\\ &=&\displaystyle \frac{\mu_3}{\sigma^3} \frac{1}{\sqrt{n}}\\ &=&\displaystyle \frac{\beta_1}{\sqrt{n}} &&\displaystyle\,\dotso\,\href{https://shikitenkai.blogspot.com/2019/07/mu-sigma2beta1.html}{\frac{\mu_3}{\sigma^3}=\beta_1}\\ \end{array}$$ よって標本数$n$を増やすことで$\beta_1\left(\overline{X}\right)$は$0$に近づいていく．

標本の二乗和

標本分散の式展開

$$\begin{array}{rcl} \displaystyle \sum_{k=1}^{n}(X_k-\overline{X})^2 &=&\displaystyle \sum_{k=1}^{n}(X_k^2-2\overline{X} X_k+\overline{X}^2)\\ &=&\displaystyle \sum_{k=1}^{n}X_k^2-2\overline{X} \sum_{k=1}^{n}X_k+\overline{X}^2\sum_{k=1}^{n}1\\ &=&\displaystyle \sum_{k=1}^{n}X_k^2-2n\overline{X}^2+n\overline{X}^2\\ &&\displaystyle\,\dotso\,\sum_{k=1}^{n}X_k=n\overline{X}\\ &=&\displaystyle \sum_{k=1}^{n}X_k^2-n\overline{X}^2\\ \end{array}$$

標本の二乗和

$$\begin{array}{rcl} \displaystyle \sum_{k=1}^{n}X_k^2 &=&\displaystyle \sum_{k=1}^{n}(X_k-\overline{X})^2+n\overline{X}^2\\ \end{array}$$

標本平均の母平均まわりの3次モーメント

標本平均$\overline{X}$の母平均$\mu$まわりの3次モーメント(=標本平均$\overline{X}$の3次の中心(化)モーメント)

$$\begin{eqnarray} \mathrm{E}\left[(\overline{X}-\mu)^3\right] &=&\mathrm{E}\left[\left\{\left(\frac{1}{n}\sum_{k=1}^{n}X_k\right) - \mu\right\}^3\right] \;\cdots\;\overline{X}=\frac{1}{n}\sum_{k=1}^{n}X_k \\&=&\mathrm{E}\left[\left\{\left(\frac{1}{n}\sum_{k=1}^{n}X_k\right) - \left(\frac{1}{n}\sum_{k=1}^{n}\mu\right)\right\}^3\right] \;\cdots\;C=\frac{n}{n}C=\frac{1}{n}C\sum_{k=1}^{n}1=\frac{1}{n}\sum_{k=1}^{n}C\;(C:kによらない数，\sumにとって定数) \\&=&\mathrm{E}\left[\left[\frac{1}{n}\left\{\left(\sum_{k=1}^{n}X_k\right)-\left(\sum_{k=1}^{n}\mu\right)\right\}\right]^3\right] \\&=&\mathrm{E}\left[\frac{1}{n^3}\left\{\left(\sum_{k=1}^{n}X_k\right)-\left(\sum_{k=1}^{n}\mu\right)\right\}^3\right] \;\cdots\;(AB)^C=A^CB^C \\&=&\mathrm{E}\left[\frac{1}{n^3}\left\{\sum_{k=1}^{n}\left(X_k-\mu\right)\right\}^3\right] \;\cdots\;\sum_{k=1}^{n}X_k-\sum_{k=1}^{n}Y_k=\sum_{k=1}^{n}\left(X_k-Y_k\right) \end{eqnarray}$$ 総和の指数計算において掛け合わせる添え字の組合せについて考える． $$\begin{eqnarray} \left(\sum_{k=1}^{n}A_k\right)^3 &=&\left(\sum_{k=1}^{n} A_k \right)\left(\sum_{l=1}^{n}A_l\right)\left(\sum_{m=1}^{n}A_m\right) \\&=&(A_1+A_2+\cdots+A_k+\cdots+A_n)(A_1+A_2+\cdots+A_l+\cdots+A_n)(A_1+A_2+\cdots+A_m+\cdots+A_n) \\&=&_3\mathrm{P}_0\times\left(\sum_{k=1}^{n} A_k^3 \right)\;\cdots\;3つとも同じ添え字(どれか0個の添え字が異なるケース) \\&&+_3\mathrm{P}_1\times\left(\sum_{k \neq l} A_k^2 A_l\right)\;\cdots\;いずれか2つが同じ添え字(どれか1個の添え字が異なるのケース) \\&&+_3\mathrm{P}_2\times\left(\sum_{k \lt l \lt m} A_k A_l A_m\right)\;\cdots\;すべての添え字が異なるケース(どれか2個の添え字が異なるのケース) \\&&\;\cdots\;各ケースでの(重複する数 \times 組合せで総和)の和 \\&=&\frac{3!}{(3-0)!}\left(\sum_{k=1}^{n} A_k^3 \right) +\frac{3!}{(3-1)!}\left(\sum_{k \neq l} A_k^2 A_l\right) +\frac{3!}{(3-2)!}\left(\sum_{k \lt l \lt m} A_k A_l A_m\right) \\&=&\frac{3\times2\times1}{3\times2\times1}\left(\sum_{k=1}^{n} A_k^3 \right) +\frac{3\times2\times1}{2\times1}\left(\sum_{k \neq l} A_k^2 A_l\right) +\frac{3\times2\times1}{1}\left(\sum_{k \lt l \lt m} A_k A_l A_m\right) \\&=&1\cdot\left(\sum_{k=1}^{n} A_k^3 \right) +3\cdot\left(\sum_{k \neq l} A_k^2 A_l\right) +6\cdot\left(\sum_{k \lt l \lt m} A_k A_l A_m\right) \end{eqnarray}$$ よって， $$\begin{eqnarray} \mathrm{E}\left[(\overline{X}-\mu)^3\right] &=&\mathrm{E}\left[\frac{1}{n^3}\left\{\sum_{k=1}^{n}\left(X_k-\mu\right)\right\}^3\right] \\&=&\mathrm{E}\left[\frac{1}{n^3}\left\{ \sum_{k=1}^{n} \left(X_k-\mu\right)^3 +3\sum_{k \neq l} \left(X_k-\mu\right)^2\left(X_l-\mu\right) +6\sum_{k \lt l \lt m} \left(X_k-\mu\right)\left(X_l-\mu\right)\left(X_m-\mu\right) \right\}\right] \\&=&\frac{1}{n^3}\mathrm{E}\left[ \sum_{k=1}^{n} \left(X_k-\mu\right)^3 +3\sum_{k \neq l} \left(X_k-\mu\right)^2\left(X_l-\mu\right) +6\sum_{k \lt l \lt m} \left(X_k-\mu\right)\left(X_l-\mu\right)\left(X_m-\mu\right) \right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-expected-value.html}{\mathrm{E}[cX]=c\mathrm{E}[X]} \\&=&\frac{1}{n^3}\left[ \mathrm{E}\left[\sum_{k=1}^{n} \left(X_k-\mu\right)^3\right] +\mathrm{E}\left[3\sum_{k \neq l} \left(X_k-\mu\right)^2\left(X_l-\mu\right)\right] +\mathrm{E}\left[6\sum_{k \lt l \lt m} \left(X_k-\mu\right)\left(X_l-\mu\right)\left(X_m-\mu\right)\right] \right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-expected-value.html}{\mathrm{E}[X+Y]=\mathrm{E}[X]+\mathrm{E}[Y]} \\&=&\frac{1}{n^3}\left[ \mathrm{E}\left[\sum_{k=1}^{n} \left(X_k-\mu\right)^3\right] +3\mathrm{E}\left[\sum_{k \neq l} \left(X_k-\mu\right)^2\left(X_l-\mu\right)\right] +6\mathrm{E}\left[\sum_{k \lt l \lt m} \left(X_k-\mu\right)\left(X_l-\mu\right)\left(X_m-\mu\right)\right] \right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-expected-value.html}{\mathrm{E}[cX]=c\mathrm{E}[X]} \\&=&\frac{1}{n^3}\left[ \sum_{k=1}^{n} \mathrm{E}\left[ \left(X_k-\mu\right)^3\right] +3\sum_{k \neq l} \mathrm{E}\left[ \left(X_k-\mu\right)^2\left(X_l-\mu\right)\right] +6\sum_{k \lt l \lt m} \mathrm{E}\left[ \left(X_k-\mu\right)\left(X_l-\mu\right)\left(X_m-\mu\right)\right] \right] \\&&\;\cdots\;\mathrm{E}\left[\sum_{k=1}^n A_k\right]=\mathrm{E}\left[A_1+A_2+\;\cdots\;+A_n\right]=\mathrm{E}\left[A_1\right]+\mathrm{E}\left[A_2\right]+\cdots+\mathrm{E}\left[A_n\right]=\sum_{k=1}^n\mathrm{E}\left[A_k\right] ,\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-expected-value.html}{\mathrm{E}[X+Y]=\mathrm{E}[X]+\mathrm{E}[Y]} \\&=&\frac{1}{n^3}\left[ \sum_{k=1}^{n} \mathrm{E}\left[\left(X_k-\mu\right)^3\right] +3\sum_{k \neq l} \mathrm{E}\left[\left(X_k-\mu\right)^2\right]\mathrm{E}\left[X_l-\mu\right] +6\sum_{k \lt l \lt m} \mathrm{E}\left[X_k-\mu\right]\mathrm{E}\left[X_l-\mu\right]\mathrm{E}\left[X_m-\mu\right] \right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-expected-value.html}{X,Yが独立の場合\,\,\mathrm{E}[XY]=\mathrm{E}[X]\mathrm{E}[Y]} \end{eqnarray}$$ ここで1次の中心(化)モーメントについて考える． $$\begin{eqnarray} \mathrm{E}\left[X_i-\mu\right] &=& \mathrm{E}\left[X_i\right]-\mathrm{E}\left[\mu\right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-expected-value.html}{\mathrm{E}[X-Y]=\mathrm{E}[X]-\mathrm{E}[Y]} \\&=& \mu-\mu \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/specimen-random-variable.html}{\mathrm{E}[X_i]=\mathrm{E}[X]=\mu},\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-expected-value.html}{\mathrm{E}[C]=C\;(C定数)} \\&=& 0 \end{eqnarray}$$ これを用いて $$\begin{eqnarray} \mathrm{E}\left[(\overline{X}-\mu)^3\right] \\&=&\frac{1}{n^3}\left[ \sum_{k=1}^{n} \mathrm{E}\left[\left(X_k-\mu\right)^3\right] +3\sum_{k \neq l} \mathrm{E}\left[\left(X_k-\mu\right)^2\right]\mathrm{E}\left[X_l-\mu\right] +6\sum_{k \lt l \lt m} \mathrm{E}\left[X_k-\mu\right]\mathrm{E}\left[X_l-\mu\right]\mathrm{E}\left[X_m-\mu\right] \right] \\&=&\frac{1}{n^3}\left[ \sum_{k=1}^{n} \mathrm{E}\left[\left(X_k-\mu\right)^3\right] +3\sum_{k \neq l} \left(\mathrm{E}\left[\left(X_k-\mu\right)^2\right]\cdot0\right) +6\sum_{k \lt l \lt m} \left(0\cdot0\cdot0\right) \right] \\&=&\frac{1}{n^3}\left[ \sum_{k=1}^{n} \mathrm{E}\left[\left(X_k-\mu\right)^3\right] +0+0 \right] \\&=&\frac{1}{n^3}\sum_{k=1}^{n} \mathrm{E}\left[\left(X_k-\mu\right)^3\right] \\&=&\frac{1}{n^3}\sum_{k=1}^{n} \mu_3 \,\cdots\,\href{https://shikitenkai.blogspot.com/2019/07/mu-sigma2beta1.html}{\mathrm{E}\left[\left(X-\mu\right)^3\right]=\mu_3\;:3次の中心(化)モーメント} \\&=&\frac{1}{n^3}n\mu_3 \\&=&\frac{\mu_3}{n^2} \\&=&\mu_3\left(\overline{X}\right)\;\cdots\;\mu_3\left(\overline{X}\right):標本平均\overline{X}の母平均\muまわりの3次モーメント(3次の中心(化)モーメント) \end{eqnarray}$$

標本平均$\overline{X}$の母平均$\mu$まわりの3次モーメント(標本平均$\overline{X}$の3次の中心(化)モーメント)を歪度$\beta_1$で表す

$$\begin{eqnarray} \mathrm{E}\left[(\overline{X}-\mu)^3\right] &=&\mu_3\left(\overline{X}\right) \\&=&\frac{\mu_3}{n^2} \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/07/mu-sigma2beta1.html}{\mu_3=\mu_3\left(X\right)=\beta_1\sigma^3\;:3次の中心(化)モーメント} \\&=&\frac{\beta_1 \sigma^3}{n^2} \end{eqnarray}$$

標本平均$\overline{X}$の歪度$\beta_1\left(\overline{X}\right)$

$$\begin{eqnarray} \href{https://shikitenkai.blogspot.com/2019/07/blog-post_22.html}{\beta_1\left(\overline{X}\right)=\frac{\beta_1}{\sqrt{n}}} \end{eqnarray}$$

母集団の歪度

母集団(平均 $\mu$, 分散 $\sigma^2$)の歪度$\beta_1$

$$\begin{array}{rcl} \displaystyle \beta_1 &=&\displaystyle \frac{\mathrm{E}\left[\left(X-\mu\right)^3\right]}{\mathrm{V}\left[X\right]^\frac{3}{2}}\,\dots\,歪度（わいど, skewness）\\ &=&\displaystyle \frac{\mu_3}{\sigma^3}\\ && \displaystyle \,\dotso\, \mu_3 = \mathrm{E}\left[\left(X-\mu\right)^3\right] \,\,\, \left(確率変数Xの母平均\muまわりの3次モーメント=確率変数Xの3次の中心化モーメント\right)\\ \end{array}$$

確率変数$X$の母平均$\mu$まわりの3次のモーメント$\mu_3$を歪度$\beta_1$で表す

$$\begin{array}{rcl} \displaystyle \mu_3 &=&\displaystyle \beta_1\sigma^3\\ \end{array}$$

標本平均まわりの3次モーメントの和

“標本平均$\overline{X}$まわりの3次モーメントの和”を“母平均$\mu$まわりの3次モーメント$\mu_3$”で表す

$$\begin{array}{rcl} \displaystyle \sum_{k=1}^{n}E\left[(X_k-\overline{X})^3\right] &=&\displaystyle E\left[\sum_{k=1}^{n}(X_k-\overline{X})^3\right]\\ &&\displaystyle\,\dotso\,E\left[X\right]+E\left[Y\right]=E\left[X+Y\right]\\ &=&\displaystyle E\left[\sum_{k=1}^{n}\left\{ \displaystyle \left( X_k-\mu \right) \displaystyle -\left( \overline{X}-\mu \right) \displaystyle \right\}^3\right]\\ &&\displaystyle\,\dotso\,(A-B)=(A-C)-(B-C)\\ &=&\displaystyle E\left[\sum_{k=1}^{n}\left\{ \displaystyle \left( X_k-\mu \right)^3 \displaystyle -3\left( X_k-\mu \right)^2\left( \overline{X}-\mu \right) \displaystyle +3\left( X_k-\mu \right)\left( \overline{X}-\mu \right)^2 \displaystyle - \left( \overline{X}-\mu \right)^3 \displaystyle \right\}\right]\\ &&\displaystyle\,\dotso\,(A-B)^3=A^3-3A^2B+3AB^2-B^3\\ &=&\displaystyle E\left[ \displaystyle \sum_{k=1}^{n}\left( X_k-\mu \right)^3 \displaystyle -3\left\{ \left( \overline{X}-\mu \right) \sum_{k=1}^{n} \left( X_k-\mu \right)^2 \right\} \displaystyle +3\left\{ \left( \overline{X}-\mu \right)^2 \sum_{k=1}^{n} \left( X_k-\mu \right) \right\} \displaystyle - \left( \overline{X}-\mu \right)^3 \sum_{k=1}^{n}1 \displaystyle \right]\\ &&\displaystyle\,\dotso\,\sum_{k=1}^{n} (X+Y)=\sum_{k=1}^{n} X+\sum_{k=1}^{n} Y\\ &=&\displaystyle E\left[\sum_{k=1}^{n}\left( X_k-\mu \right)^3\right] \displaystyle -3E\left[ \left( \overline{X}-\mu \right) \sum_{k=1}^{n} \left( X_k-\mu \right)^2 \right] \displaystyle +3E\left[ \left( \overline{X}-\mu \right)^2 \sum_{k=1}^{n} \left( X_k-\mu \right) \right] \displaystyle - E\left[\left( \overline{X}-\mu \right)^3 \sum_{k=1}^{n}1\right] \displaystyle \\ &&\displaystyle\,\dotso\,E\left[X+Y\right]=E\left[X\right]+E\left[Y\right]\\ &=&\displaystyle \sum_{k=1}^{n}E\left[\left( X_k-\mu \right)^3\right] \displaystyle -3E\left[ \left( \overline{X}-\mu \right) \sum_{k=1}^{n} \left( X_k-\mu \right)^2 \right] \displaystyle +3E\left[ \left( \overline{X}-\mu \right)^2 \left( \sum_{k=1}^{n}X_k - \sum_{k=1}^{n}\mu \right) \right] \displaystyle - E\left[n\left( \overline{X}-\mu \right)^3\right] \displaystyle \\ &&\displaystyle\,\dotso\, \displaystyle E\left[X+Y\right]=E\left[X\right]+E\left[Y\right] ,\quad \sum_{k=1}^{n} (X+Y)=\sum_{k=1}^{n} X+\sum_{k=1}^{n} Y\\ &=&\displaystyle \sum_{k=1}^{n}E\left[\left( X_k-\mu \right)^3\right] \displaystyle -3E\left[ \left( \overline{X}-\mu \right) \sum_{k=1}^{n} \left( X_k-\mu \right)^2 \right] \displaystyle +3E\left[ \left( \overline{X}-\mu \right)^2 \left( n\overline{X} - n\mu \right) \right] \displaystyle - E\left[n\left( \overline{X}-\mu \right)^3\right] \displaystyle \\ &&\displaystyle\,\dotso\, \sum_{k=1}^{n}X_k=n\overline{X} ,\quad \sum_{k=1}^{n}\mu=n\mu\\ &=&\displaystyle n\mu_3 \displaystyle -3E\left[ \left( \overline{X}-\mu \right) \sum_{k=1}^{n} \left( X_k-\mu \right)^2\right] \displaystyle +3E\left[ n\left( \overline{X}-\mu \right)^3\right] \displaystyle -nE\left[\left( \overline{X}-\mu \right)^3\right] \displaystyle \\ &&\displaystyle\,\dotso\, \href{https://shikitenkai.blogspot.com/2019/07/mu-sigma2beta1.html}{E\left[\left( X_k-\mu \right)^3\right]=\mu_3}\\ &=&\displaystyle n\mu_3 \displaystyle -3E\left[ \left( \overline{X}-\mu \right) \sum_{k=1}^{n} \left( X_k-\mu \right)^2\right] \displaystyle +3nE\left[ \left( \overline{X}-\mu \right)^3\right] \displaystyle - nE\left[\left( \overline{X}-\mu \right)^3\right] \displaystyle \\ &=&\displaystyle n\mu_3 \displaystyle -3E\left[ \left( \overline{X}-\mu \right) \sum_{k=1}^{n} \left( X_k-\mu \right)^2\right] \displaystyle +2nE\left[ \left( \overline{X}-\mu \right)^3\right] \displaystyle \\ &=&\displaystyle n\mu_3 \displaystyle -3E\left[ \left( \overline{X}-\mu \right) \sum_{k=1}^{n} \left( X_k-\mu \right)^2\right] \displaystyle +2n\frac{\mu_3}{n^2} \displaystyle \\ &&\displaystyle\,\dotso\, \href{https://shikitenkai.blogspot.com/2019/07/overlinexmu3.html}{E\left[\left( \overline{X}-\mu \right)^3\right]=\frac{\mu_3}{n^2}}\\ &=&\displaystyle n\mu_3 \displaystyle -3E\left[ \left( \overline{X}-\mu \right) \sum_{k=1}^{n} \left( X_k-\mu \right)^2\right] \displaystyle +2\frac{\mu_3}{n} \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \left( \overline{X}-\mu \right) \sum_{k=1}^{n} \left( X_k-\mu \right)^2\right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \left( \overline{X}-\mu \right) \sum_{k=1}^{n} \left( X_k^2-2\mu X_k+\mu^2 \right) \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \left( \overline{X}-\mu \right) \left( \sum_{k=1}^{n}X_k^2-2\mu \sum_{k=1}^{n}X_k+\mu^2\sum_{k=1}^{n}1 \right) \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \left( \overline{X}-\mu \right) \left( \sum_{k=1}^{n}X_k^2-2\mu n\overline{X}+n\mu^2 \right) \right] \displaystyle \\ &&\displaystyle\,\dotso\, \sum_{k=1}^{n}X_k=n\overline{X} ,\quad \sum_{k=1}^{n}1=n\\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \displaystyle \overline{X}\left( \sum_{k=1}^{n}X_k^2- 2\mu n\overline{X}+ n\mu^2 \right) \displaystyle -\mu\left( \sum_{k=1}^{n}X_k^2-2\mu n\overline{X}+ n\mu^2 \right) \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \displaystyle \left( \overline{X}\sum_{k=1}^{n}X_k^2-2n\mu \overline{X}^2+ n\mu^2\overline{X} \right) \displaystyle -\left( \mu\sum_{k=1}^{n}X_k^2-2\mu^2 n\overline{X}+n\mu^3 \right) \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \displaystyle \overline{X}\sum_{k=1}^{n}X_k^2-2n\mu \overline{X}^2+ n\mu^2\overline{X} \displaystyle -\mu\sum_{k=1}^{n}X_k^2+2\mu^2 n\overline{X}-n\mu^3 \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \displaystyle \overline{X}\sum_{k=1}^{n}X_k^2 \displaystyle -2n\mu \overline{X}^2- \mu \sum_{k=1}^{n}X_k^2 \displaystyle + n\mu^2\overline{X} +2\mu^2 n\overline{X} \displaystyle -n\mu^3 \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \displaystyle \overline{X}\sum_{k=1}^{n}X_k^2 \displaystyle - \mu \left(2n \overline{X}^2+\sum_{k=1}^{n}X_k^2\right) \displaystyle + n\mu^2\left( \overline{X} +2 \overline{X}\right) \displaystyle -n\mu^3 \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \displaystyle \overline{X}\sum_{k=1}^{n}X_k^2 \displaystyle - \mu\left(2n \overline{X}^2+\sum_{k=1}^{n}X_k^2\right) \displaystyle + 3n\mu^2\overline{X} \displaystyle -n\mu^3 \displaystyle \right] \displaystyle \\ &&\displaystyle\,\dotso\, \displaystyle \href{https://shikitenkai.blogspot.com/2019/07/blog-post_41.html}{\sum_{k=1}^{n}X_k^2=\sum_{k=1}^{n}(X_k-\overline{X})^2+n\overline{X}^2}\\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \displaystyle \overline{X}\,\left(\sum_{k=1}^{n}\left(X_k-\overline{X}\right)^2+n\overline{X}^2\right) \displaystyle - \mu \left(2n \overline{X}^2+\left(\sum_{k=1}^{n}\left(X_k-\overline{X}\right)^2+n\overline{X}^2\right)\right) \displaystyle +3n\mu^2\overline{X} \displaystyle -n\mu^3 \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \displaystyle \overline{X}\sum_{k=1}^{n}\left(X_k-\overline{X}\right)^2 \displaystyle +n\overline{X}^3 \displaystyle -2n\mu \overline{X}^2 \displaystyle -\mu\sum_{k=1}^{n}\left(X_k-\overline{X}\right)^2 \displaystyle - n\mu \overline{X}^2 \displaystyle +3n\mu^2\overline{X} \displaystyle -n\mu^3 \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \displaystyle \left( \overline{X}-\mu \right) \sum_{k=1}^{n}\left(X_k-\overline{X}\right)^2 \displaystyle +n\overline{X}^3 \displaystyle -3n\mu \overline{X}^2 \displaystyle +3n\mu^2\overline{X} \displaystyle -n\mu^3 \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \displaystyle \left( \overline{X}-\mu \right) \sum_{k=1}^{n}\left(X_k-\overline{X}\right)^2 \displaystyle \right] \displaystyle -3E\left[ \displaystyle n\overline{X}^3 \displaystyle -3n\mu \overline{X}^2 \displaystyle +3n\mu^2\overline{X} \displaystyle -n\mu^3 \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[\overline{X}-\mu \right] \displaystyle \sum_{k=1}^{n}\left(X_k-\overline{X}\right)^2 \displaystyle -3E\left[ \displaystyle n\overline{X}^3 \displaystyle -3n\mu \overline{X}^2 \displaystyle +3n\mu^2\overline{X} \displaystyle -n\mu^3 \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3\left(E\left[\overline{X} \right]-\mu\right) \displaystyle \sum_{k=1}^{n}\left(X_k-\overline{X}\right)^2 \displaystyle -3E\left[ \displaystyle n\overline{X}^3 \displaystyle -3n\mu \overline{X}^2 \displaystyle +3n\mu^2\overline{X} \displaystyle -n\mu^3 \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3\left(\mu-\mu\right) \displaystyle \sum_{k=1}^{n}\left(X_k-\overline{X}\right)^2 \displaystyle -3E\left[ \displaystyle n\overline{X}^3 \displaystyle -3n\mu \overline{X}^2 \displaystyle +3n\mu^2\overline{X} \displaystyle -n\mu^3 \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \displaystyle n\overline{X}^3 \displaystyle -3n\mu \overline{X}^2 \displaystyle +3n\mu^2\overline{X} \displaystyle -n\mu^3 \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[n \displaystyle \left(\overline{X}^3 \displaystyle -3\mu \overline{X}^2 \displaystyle +3\mu^2\overline{X} \displaystyle -\mu^3\right) \displaystyle \right] \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3E\left[ \displaystyle n\left(\overline{X}-\mu\right)^3 \displaystyle \right] \displaystyle \\ &&\displaystyle\,\dotso\,A^3-3A^2B+3AB^2-B^3=(A-B)^3\\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3nE\left[ \displaystyle \left(\overline{X}-\mu\right)^3 \displaystyle \right] \displaystyle \\ &&\displaystyle\,\dotso\,E[cX]=cE[X]\\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3n\frac{\mu_3}{n^2} \displaystyle \\ &&\displaystyle\,\dotso\, \href{https://shikitenkai.blogspot.com/2019/07/overlinexmu3.html}{E\left[\left( \overline{X}-\mu \right)^3\right]=\frac{\mu_3}{n^2}}\\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} \right) \displaystyle -3\frac{\mu_3}{n} \displaystyle \\ &=&\displaystyle \mu_3\left( \frac{n^2+2}{n} -3\frac{n}{n}\right) \displaystyle \\ &=&\displaystyle \mu_3\frac{n^2-3n+2}{n}\\ &=&\displaystyle \mu_3\frac{(n-1)(n-2)}{n}\\ \end{array}$$

“標本平均$\overline{X}$まわりの3次モーメントの和”から“母平均$\mu$まわりの3次モーメント$\mu_3$”を推定する$\hat{\mu}_3$

$$\begin{array}{rcl} \displaystyle \hat{\mu}_3 &=&\displaystyle \frac{n}{(n-1)(n-2)}E\left[\sum_{k=1}^{n}(X_k-\overline{X})^3\right]\\ \end{array}$$

不偏分散の期待値

不偏分散の期待値

$$\begin{array}{rclcl} \hat{\sigma}^2 &=& \href{https://shikitenkai.blogspot.com/2019/07/specimen-random-variable.html}{\displaystyle \frac{1}{n-1}\sum_{k=1}^{n}\left( \displaystyle X_k - \overline{X} \displaystyle \right)^2}\,\dotso\,不偏分散(unbiased \, variance)\\ E\left[\hat{\sigma}^2\right] &=&E\left[ \displaystyle\frac{1}{n-1}\sum_{k=1}^{n} \left(X_k -\overline{X}\right)^2 \right]\\ &=&\displaystyle\frac{1}{n-1}E\left[\sum_{k=1}^{n} \left(X_k -\overline{X}\right)^2 \right]\\ &=&\displaystyle\frac{1}{n-1}\left(n-1\right)\sigma^2 \,\dotso\,\displaystyle \href{https://shikitenkai.blogspot.com/2019/07/overlinex2.html}{E\left[\sum_{k=1}^{n} \left(X_k -\overline{X}\right)^2 \right]=\left(n-1\right)\sigma^2}\\ &=&\displaystyle\sigma^2\\ \end{array}$$

標本分散の期待値

$$\begin{array}{rcl} s^2&=&\displaystyle \frac{1}{n}\sum_{k=1}^{n}\left( \displaystyle X_k - \overline{X} \displaystyle \right)^2\,\dotso\,標本分散(sample \, variance)\\ E\left[s^2\right]&=&E\left[ \displaystyle\frac{1}{n}\sum_{k=1}^{n} \left(X_k -\overline{X}\right)^2 \right]\\ &=&\displaystyle \frac{1}{n} E\left[ \sum_{k=1}^{n} \left(X_k -\overline{X}\right)^2 \right]\\ &=&\displaystyle \frac{1}{n} \left(n-1\right)\sigma^2 \,\dotso\,\displaystyle \href{https://shikitenkai.blogspot.com/2019/07/overlinex2.html}{E\left[\sum_{k=1}^{n} \left(X_k -\overline{X}\right)^2 \right]=\left(n-1\right)\sigma^2}\\ &=&\displaystyle \frac{n-1}{n}\sigma^2 \\ \end{array}$$

標本平均まわりの2次モーメントの和

標本平均$\overline{X}$まわりの2次モーメント

$$\begin{array}{rclcl} \displaystyle E\left[ \left(X_k -\overline{X}\right)^2 \right] \end{array}$$

標本平均$\overline{X}$まわりの2次モーメントの和

$$\begin{array}{rclcl} \displaystyle \sum_{k=1}^{n}E\left[ \left(X_k -\overline{X}\right)^2 \right] &=&\displaystyle E\left[ \sum_{k=1}^{n} \left(X_k -\overline{X}\right)^2 \right]\\ &=&\displaystyle E\left[ \sum_{k=1}^{n} \left(\left(X_k - \mu\right) - \left(\overline{X} - \mu\right)\right)^2 \right]\\ &=&\displaystyle E\left[ \sum_{k=1}^{n} \left(\left(X_k - \mu\right)^2 -2\left(X_k - \mu\right)\left(\overline{X} - \mu\right) + \left(\overline{X} - \mu\right)^2\right) \right]\\ &=&\displaystyle E\left[ \sum_{k=1}^{n} \left(X_k - \mu\right)^2 -2\left(\overline{X} - \mu\right)\sum_{k=1}^{n}\left(X_k - \mu\right) + \left(\overline{X} - \mu\right)^2\sum_{k=1}^{n}1 \right]\\ &=&\displaystyle E\left[ \sum_{k=1}^{n} \left(X_k - \mu\right)^2 -2\left(\overline{X} - \mu\right)\sum_{k=1}^{n}\left(X_k - \mu\right) + n\left(\overline{X} - \mu\right)^2 \right]\\ &=&\displaystyle E\left[ \sum_{k=1}^{n} \left(X_k - \mu\right)^2\right] +E\left[-2\left(\overline{X} - \mu\right)\sum_{k=1}^{n}\left(X_k - \mu\right) + n\left(\overline{X} - \mu\right)^2\right] \\ &=&\displaystyle \sum_{k=1}^{n}E\left[ \left(X_k - \mu\right)^2\right] + E\left[-2\left(\overline{X} - \mu\right)\left(\sum_{k=1}^{n}X_k - \sum_{k=1}^{n}\mu\right) + n\left(\overline{X} - \mu\right)^2\right] \\ &=&\displaystyle \sum_{k=1}^{n}V\left[X_k\right] + E\left[-2\left(\overline{X} - \mu\right)\left(n\overline{X} - n\mu\right) + n\left(\overline{X} - \mu\right)^2\right] \,\dots\,\displaystyle\sum_{k=1}^{n}X_k=n\overline{X}\\ &=&\displaystyle \sum_{k=1}^{n}\sigma^2 + E\left[-2\left(\overline{X} - \mu\right)n\left(\overline{X} - \mu\right) + n\left(\overline{X} - \mu\right)^2\right] \\ &=&\displaystyle n\sigma^2 + E\left[-n\left(\overline{X} - \mu\right)^2\right] \\ &=&\displaystyle n\sigma^2 - nE\left[\left(\overline{X} - \mu\right)^2\right] \\ &=&\displaystyle n\sigma^2 - nV\left[\overline{X}\right] \\ &=&\displaystyle n\sigma^2 - n\frac{\sigma^2}{n} \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/specimen-random-variable_3.html}{V\left[\overline{X}\right]=\frac{\sigma^2}{n}} \\&=&\displaystyle \left(n-1\right)\sigma^2 \\ \end{array}$$

“標本平均$\overline{X}$まわりの2次モーメントの和”から母分散$\sigma^2$を推定する$\hat{\sigma}^2$

$$\begin{array}{rclcl} \displaystyle \hat{\sigma}^2 &=&\displaystyle \frac{1}{\left(n-1\right)}\sum_{k=1}^{n}E\left[ \left(X_k -\overline{X}\right)^2 \right]\\ &=&\displaystyle \frac{1}{\left(n-1\right)}E\left[\sum_{k=1}^{n} \left(X_k -\overline{X}\right)^2 \right]\\ &=&\displaystyle \frac{1}{\left(n-1\right)}E\left[\sum_{k=1}^{n} \left(X_k^2 \right)-\overline{X}^2 \right] \end{array}$$

X軸と二次凾数で囲まれる面積

$y=0$(X軸)と$(x-\alpha)(x-\beta)=0$(二次凾数)とで囲まれる領域の面積

$x^2-(\alpha+\beta)x+\alpha\beta=0\,\,\,(\alpha \lt \beta)$であり$x^2$の係数が正で，X軸の下側に面積ができる場合(下に凸)
$$\begin{array}{rcl} \displaystyle \int_{\alpha}^{\beta} (x-\alpha)(x-\beta) \mathrm{d}x &=& \displaystyle \int_{\alpha}^{\beta} (x-\alpha)(x-\beta-\alpha+\alpha) \mathrm{d}x\\ &=& \displaystyle \int_{\alpha}^{\beta} (x-\alpha)((x-\alpha)-(\beta-\alpha)) \mathrm{d}x\\ &=& \displaystyle \int_{\alpha}^{\beta} (x-\alpha)^2-(\beta-\alpha)(x-\alpha) \mathrm{d}x\\ &=& \displaystyle \int_{\alpha}^{\beta} (x-\alpha)^2\mathrm{d}x - \int_{\alpha}^{\beta}(\beta-\alpha)(x-\alpha) \mathrm{d}x\\ &=& \displaystyle \left[ \frac{(x-\alpha)^3}{3} \right]_\alpha^\beta - \left[ \frac{(\beta-\alpha)(x-\alpha)^2}{2} \right]_\alpha^\beta\\ &=& \displaystyle \left[ \frac{(\beta-\alpha)^3}{3} - 0 \right] - \left[ \frac{(\beta-\alpha)(\beta-\alpha)^2}{2} - 0 \right]\\ &=& \displaystyle \frac{(\beta-\alpha)^3}{3} - \frac{(\beta-\alpha)^3}{2}\\ &=& \displaystyle \frac{2(\beta-\alpha)^3 - 3(\beta-\alpha)^3}{6}\\ &=& \displaystyle \frac{-(\beta-\alpha)^3}{6}\\ \end{array}$$ これは第一種オイラー積分の一つの形であり，$\frac{1}{6}$公式と呼ばることがある式である．

連続型確率変数(continuous random variable) の一様分布(uniform distribution)の分散(variance)

$$\begin{array}{rcl} \displaystyle M_X^{(m)}(0)&\equiv&\frac{ \mathrm{d}^m }{ \mathrm{d}^m t } M_X(t)|_{t=0}\\ &=&\displaystyle E[X^m\mathrm{e}^{tX}]|_{t=0}\\ &=&\displaystyle E[X^m]\\ \end{array}$$

積率母凾数の二階微分

$$\begin{array}{rcl} \displaystyle M_X^{(2)} &=& \displaystyle \frac{\mathrm{d}^2}{\mathrm{d}t^2}\left\{ \displaystyle \href{https://shikitenkai.blogspot.com/2019/07/continuous-random-variable-uniform.html}{\frac{\mathrm{e}^{tb}-\mathrm{e}^{ta}}{t(b-a)}} \displaystyle \right\}\\ &=& \displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \displaystyle \href{https://shikitenkai.blogspot.com/2019/07/continuous-random-variable-uniform_8.html}{\frac{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}}{t^2(b-a)}} \displaystyle \right\}\\ &=& \displaystyle \frac{1}{b-a}\left[ \displaystyle (t^{-2})'\left\{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}\right\} \displaystyle +(t^{-2})\left\{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}\right\}' \displaystyle \right]\\ &=& \displaystyle \frac{1}{b-a}\left[ \displaystyle (-2t^{-3})\left\{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}\right\} \displaystyle +(t^{-2})\left[\left\{(tb-1)\mathrm{e}^{tb}\right\}'-\left\{(ta-1)\mathrm{e}^{ta}\right\}'\right] \displaystyle \right]\\ &=& \displaystyle \frac{1}{b-a}\left[ \displaystyle (-2t^{-3})\left\{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}\right\} \displaystyle +(t^{-2})\left[ \left\{ (tb-1)'\mathrm{e}^{tb}+(tb-1)(\mathrm{e}^{tb})' \right\} -\left\{ (ta-1)'\mathrm{e}^{ta}+(ta-1)(\mathrm{e}^{ta})' \right\} \right] \displaystyle \right]\\ &=& \displaystyle \frac{1}{b-a}\left[ \displaystyle (-2t^{-3})\left\{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}\right\} \displaystyle +(t^{-2})\left[ \left\{ b\mathrm{e}^{tb}+(tb-1)b\mathrm{e}^{tb} \right\} -\left\{ a\mathrm{e}^{ta}+(ta-1)a\mathrm{e}^{ta} \right\} \right] \displaystyle \right]\\ &=& \displaystyle \frac{1}{b-a}\left[\frac{ \displaystyle -2\left\{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}\right\} \displaystyle +t\left[ \left\{ b\mathrm{e}^{tb}+(tb-1)b\mathrm{e}^{tb} \right\} -\left\{ a\mathrm{e}^{ta}+(ta-1)a\mathrm{e}^{ta} \right\} \right]}{t^3} \displaystyle \right]\\ &=& \displaystyle \frac{1}{t^3(b-a)}\left[ \displaystyle -2(tb-1)\mathrm{e}^{tb}+2(ta-1)\mathrm{e}^{ta} \displaystyle +tb\mathrm{e}^{tb}+tb(tb-1)\mathrm{e}^{tb} \displaystyle -ta\mathrm{e}^{ta}-ta(ta-1)\mathrm{e}^{ta} \displaystyle \right]\\ &=& \displaystyle \frac{1}{t^3(b-a)}\left[ \displaystyle \left\{-2(tb-1)+tb+tb(tb-1)\right\} \mathrm{e}^{tb} \displaystyle +\left\{2(ta-1)-ta-ta(ta-1)\right\} \mathrm{e}^{ta} \displaystyle \right]\\ &=& \displaystyle \frac{1}{t^3(b-a)}\left\{ \displaystyle ((tb-1)^2+1) \mathrm{e}^{tb} \displaystyle -((ta-1)^2+1) \mathrm{e}^{ta} \displaystyle \right\}\\ &=& \displaystyle \frac{((tb-1)^2+1) \mathrm{e}^{tb}-((ta-1)^2+1) \mathrm{e}^{ta}}{t^3(b-a)}\\ \end{array}$$

原点周りの二次モーメント

$$\begin{array}{rcl} \displaystyle E[X^2]&=&\displaystyle M_X^{(2)}(0)\\ &=&\displaystyle \lim_{t \to 0}\left\{ \displaystyle \frac{((tb-1)^2+1) \mathrm{e}^{tb}-((ta-1)^2+1) \mathrm{e}^{ta}}{t^3(b-a)} \displaystyle \right\}\,\dotso\,0を代入すると分母が0になってしまうので極限で考える．\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left\{ \displaystyle ((tb-1)^2+1) \mathrm{e}^{tb} - ((ta-1)^2+1) \mathrm{e}^{ta} \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left\{ \displaystyle ((tb-1)^2+1) \left(\frac{(tb)^0}{0!}+\frac{(tb)^1}{1!}+\frac{(tb)^2}{2!}+\frac{(tb)^3}{3!}\right) \displaystyle -((ta-1)^2+1) \left(\frac{(ta)^0}{0!}+\frac{(ta)^1}{1!}+\frac{(ta)^2}{2!}+\frac{(ta)^3}{3!}\right) \displaystyle \right\} \displaystyle \right]\\ && \,\dotso\,\href{https://shikitenkai.blogspot.com/2019/07/blog-post.html}{\mathrm{e}^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dotsb} (マクローリン展開), t^3が分母にあるのでt^4の項以上は分子にtが残ることになるのでt^3の項までで計算を進める．\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left\{ \displaystyle (t^2b^2-2tb+2) \left(1+tb+t^2\frac{b^2}{2}+t^3\frac{b^3}{6}\right) \displaystyle -(t^2a^2-2ta+2) \left(1+ta+t^2\frac{a^2}{2}+t^3\frac{a^3}{6}\right) \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left[ \left\{ \left( t^2b^2 +t^3b^3 +t^4\frac{b^4}{2} +t^5\frac{b^5}{6} \right) +\left( -2tb -2t^2b^2 -t^3b^3 -t^4\frac{b^4}{3} \right) +\left(2+2tb + t^2b^2 +t^3\frac{b^3}{3} \right) \right\} -\left\{ \left( t^2a^2 +t^3a^3 +t^4\frac{a^4}{2} +t^5\frac{a^5}{6} \right) +\left( -2ta -2t^2a^2 -t^3a^3 -t^4\frac{a^4}{3} \right) +\left(2+2ta + t^2a^2 +t^3\frac{a^3}{3} \right) \right\} \displaystyle \right] \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left\{ \left( 2+t^3\frac{b^3}{3}+t^4\frac{b^4}{6}+t^5\frac{b^5}{6} \right) -\left( 2+t^3\frac{a^3}{3}+t^4\frac{a^4}{6}+t^5\frac{a^5}{6} \right) \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left\{ t^3\frac{b^3-a^3}{3} +t^4\frac{b^4-a^4}{6} +t^5\frac{b^5-a^5}{6} \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^3(b-a)} \displaystyle \left\{ t^3\frac{(b-a)(b^2+ab+a^2)}{3} +t^4\frac{(b-a)(b+a)(a^2+b^2)}{6} +t^5\frac{(b-a)\frac{b^5-a^5}{b-a}}{6} \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left\{ \frac{b^2+ab+a^2}{3} +t \frac{(b+a)(a^2+b^2)}{6} +t^2\frac{\left(\frac{b^5-a^5}{b-a}\right)}{6} \displaystyle \right\}\\ &=&\displaystyle \frac{b^2+ab+a^2}{3}=\frac{a^2+ab+b^2}{3} \,\dotso\,tが分子にある(掛けられている)項は全て0．\\ \end{array}$$

分散(二次の中心モーメント)

$$\begin{array}{rcl} \displaystyle V[X] &=&\displaystyle E[X^2]-E[X]^2\\ &=&\displaystyle \frac{b^2+ab+a^2}{3}-\left(\href{https://shikitenkai.blogspot.com/2019/07/continuous-random-variable-uniform_8.html}{\frac{b+a}{2}}\right)^2\\ &=&\displaystyle \frac{b^2+ab+a^2}{3}-\frac{b^2+2ab+a^2}{4}\\ &=&\displaystyle \frac{4(b^2+ab+a^2)-3(b^2+2ab+a^2)}{12}\\ &=&\displaystyle \frac{4b^2+4ab+4a^2-3b^2-6ab-3a^2}{12}\\ &=&\displaystyle \frac{b^2-2ab+a^2}{12}\\ &=&\displaystyle \frac{(b-a)^2}{12}\\ \end{array}$$

連続型確率変数(continuous random variable) の一様分布(uniform distribution)の期待値(expected value)

$$\begin{array}{rcl} \displaystyle M_X^{(m)}(0)&\equiv&\frac{ \mathrm{d}^m }{ \mathrm{d}^m t } M_X(t)|_{t=0}\\ &=&\displaystyle E[X^m\mathrm{e}^{tX}]|_{t=0}\\ &=&\displaystyle E[X^m]\\ \end{array}$$

積率母凾数の一階微分

$$\begin{array}{rcl} \displaystyle M_X^{(1)} &=& \displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \displaystyle \href{https://shikitenkai.blogspot.com/2019/07/continuous-random-variable-uniform.html}{\frac{\mathrm{e}^{tb}-\mathrm{e}^{ta}}{t(b-a)}} \displaystyle \right\}\\ &=& \displaystyle \frac{1}{b-a} \displaystyle \left\{ \displaystyle \left(t^{-1}\right)'\left(\mathrm{e}^{tb}-\mathrm{e}^{ta}\right) \displaystyle +\left(t^{-1}\right)\left(\mathrm{e}^{tb}-\mathrm{e}^{ta}\right)' \displaystyle \right\}\\ &=& \displaystyle \frac{1}{b-a} \displaystyle \left\{ \displaystyle \left(-t^{-2}\right)\left(\mathrm{e}^{tb}-\mathrm{e}^{ta}\right) \displaystyle +\left(t^{-1}\right)\left(b\mathrm{e}^{tb}-a\mathrm{e}^{ta}\right) \displaystyle \right\}\\ &=& \displaystyle \frac{1}{b-a} \displaystyle \left( \displaystyle -\frac{\mathrm{e}^{tb}-\mathrm{e}^{ta}}{t^2} \displaystyle +\frac{b\mathrm{e}^{tb}-a\mathrm{e}^{ta}}{t} \displaystyle \right)\\ &=& \displaystyle \frac{1}{b-a} \displaystyle \left\{ \displaystyle -\frac{\mathrm{e}^{tb}-\mathrm{e}^{ta}}{t^2} \displaystyle +\frac{(b\mathrm{e}^{tb}-a\mathrm{e}^{ta})t}{t^2} \displaystyle \right\}\\ &=& \displaystyle \frac{1}{b-a} \displaystyle \left( \displaystyle \frac{tb\mathrm{e}^{tb}-\mathrm{e}^{tb}-ta\mathrm{e}^{ta}+\mathrm{e}^{ta}}{t^2} \displaystyle \right)\\ &=& \displaystyle \frac{1}{t^2(b-a)} \displaystyle \left\{ \displaystyle (tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta} \displaystyle \right\}\\ &=& \displaystyle \frac{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}}{t^2(b-a)}\\ \end{array}$$

原点周りの一次モーメント=期待値

$$\begin{array}{rcl} \displaystyle E[X]&=&\displaystyle M_X^{(1)}(0)\\ &=&\displaystyle \lim_{t \to 0}\left\{ \displaystyle \frac{(tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta}}{t^2(b-a)} \displaystyle \right\}\,\dotso\,0を代入すると分母が0になってしまうので極限で考える．\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^2(b-a)}\left\{ \displaystyle (tb-1)\mathrm{e}^{tb}-(ta-1)\mathrm{e}^{ta} \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^2(b-a)}\left\{ \displaystyle (tb-1)\left(\frac{(tb)^0}{0!}+\frac{(tb)^1}{1!}+\frac{(tb)^2}{2!}\right) \displaystyle -(ta-1)\left(\frac{(ta)^0}{0!}+\frac{(ta)^1}{1!}+\frac{(ta)^2}{2!}\right) \displaystyle \right\} \displaystyle \right]\\ && \,\dotso\,\href{https://shikitenkai.blogspot.com/2019/07/blog-post.html}{\mathrm{e}^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\dotsb} (マクローリン展開), t^2が分母にあるのでt^3の項以上は分子にtが残ることになるのでt^2の項までで計算を進める．\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^2(b-a)}\left\{ \displaystyle (tb-1)\left(1+tb+t^2\frac{b^2}{2}\right) \displaystyle -(ta-1)\left(1+ta+t^2\frac{a^2}{2}\right) \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^2(b-a)}\left[ \displaystyle \left\{\left(tb+t^2b^2+t^3\frac{b^3}{2}\right)-\left(1+tb+t^2\frac{b^2}{2}\right)\right\} \displaystyle -\left\{\left(ta+t^2a^2+t^3\frac{a^3}{2}\right)-\left(1+ta+t^2\frac{a^2}{2}\right)\right\} \displaystyle \right] \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^2(b-a)}\left\{ \displaystyle \left(-1+t(b-b)+t^2(b^2-\frac{b^2}{2})+t^3\frac{b^3}{2}\right) \displaystyle -\left(-1+t(a-a)+t^2(a^2-\frac{a^2}{2})+t^3\frac{a^3}{2}\right) \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \displaystyle \frac{1}{t^2(b-a)}\left\{ \displaystyle \left(-1+t^2\frac{b^2}{2}+t^3\frac{b^3}{2}\right) \displaystyle -\left(-1+t^2\frac{a^2}{2}+t^3\frac{a^3}{2}\right) \displaystyle \right\} \displaystyle \right]\\ &=&\displaystyle \lim_{t \to 0}\left\{ \displaystyle \frac{1}{t^2(b-a)}\left( \displaystyle t^2\frac{b^2-a^2}{2} \displaystyle +t^3\frac{b^3-a^3}{2} \displaystyle \right) \displaystyle \right\}\\ &=&\displaystyle \lim_{t \to 0} \left[\frac{1}{t^2(b-a)}\left\{ \displaystyle t^2 \frac{ (b-a)(b+a) }{2} \displaystyle +t^3 \frac{ (b-a)(b^2+ab+a^2) }{2} \displaystyle \right\}\right] \,\dotso\,b^2-a^2=(b-a)(b+a),\,b^3-a^3=(b-a)(b^2+ab+a^2)\\ &=&\displaystyle \lim_{t \to 0} \displaystyle \left\{ \displaystyle \frac{ (b+a) }{2} \displaystyle +t \frac{ (b^2+ab+a^2) }{2} \displaystyle \right\}\\ &=&\displaystyle \frac{b+a}{2}=\frac{a+b}{2} \,\dotso\,tが分子にある(掛けられている)項は全て0．\\ \end{array}$$

連続型確率変数(continuous random variable) の一様分布(uniform distribution)の積率母凾数(moment-generating function)

$$f_X(x) = \begin{cases} \displaystyle \frac{1}{b-a} & \quad \left\{ a\leq x \leq b\right\}\\ \displaystyle 0 & \quad \left\{ x\lt a,\,b\lt x\right\}\\ \end{cases}$$ $$\begin{array}{rcl} \displaystyle M_X(t)&\equiv&\displaystyle E[\mathrm{e}^{tX}]\\ &=&\displaystyle \int_{-\infty}^{\infty}\mathrm{e}^{tx}f_X(x)\mathrm{d}x\\ &=&\displaystyle \int_{a}^{b}\mathrm{e}^{tx}\left(\frac{1}{b-a}\right)\mathrm{d}x\\ &=&\displaystyle \frac{1}{b-a}\int_{a}^{b}\mathrm{e}^{tx}\mathrm{d}x\\ &=&\displaystyle \frac{1}{b-a}\int_{ta}^{tb}\mathrm{e}^{s}\frac{1}{t}\mathrm{d}s\,\dotso\,s=tx,\,\frac{\mathrm{d}s}{\mathrm{d}x}=t,\, \mathrm{d}x=\frac{1}{t}\mathrm{d}s,\,a\to ta,\,b\to tb\\ &=&\displaystyle \frac{1}{b-a}\frac{1}{t}\int_{ta}^{tb}\mathrm{e}^{s}\mathrm{d}s\\ &=&\displaystyle \frac{1}{t(b-a)}\left[\mathrm{e}^{s}\right]_{ta}^{tb}\\ &=&\displaystyle \frac{1}{t(b-a)}\left[\mathrm{e}^{tb}-\mathrm{e}^{ta}\right]\\ &=&\displaystyle \frac{\mathrm{e}^{tb}-\mathrm{e}^{ta}}{t(b-a)}\\ \end{array}$$

離散型確率変数(discrete random variable) の一様分布(uniform distribution)の分散(variance)

$$\begin{array}{rcl} \displaystyle M_X^{(m)}(0)&\equiv&\frac{ \mathrm{d}^m }{ \mathrm{d}^m t } M_X(t)|_{t=0}\\ &=&\displaystyle E[X^m\mathrm{e}^{tX}]|_{t=0}\\ &=&\displaystyle E[X^m]\\ \end{array}$$

積率母凾数の二階微分

$$\begin{array}{rcl} \displaystyle M_X^{(2)} &=&\displaystyle \frac{\mathrm{d^2}}{\mathrm{d}t^2}\left\{ \displaystyle \href{https://shikitenkai.blogspot.com/2019/07/uniform-distribution.html}{\frac{1}{n}\frac{\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{(\mathrm{e}^{t}-1)}} \displaystyle \right\}\\ &=&\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \displaystyle \href{https://shikitenkai.blogspot.com/2019/07/discrete-random-variable-uniform.html}{\frac{\mathrm{e}^{t}}{n}\frac{n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1}{(\mathrm{e}^{t}-1)^2}} \displaystyle \right\}\\ &=&\displaystyle \frac{1}{n} \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \displaystyle \mathrm{e}^{t} \frac{n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1}{(\mathrm{e}^{t}-1)^2} \displaystyle \right\}\\ &=&\displaystyle \frac{1}{n} \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \displaystyle \mathrm{e}^{t} (n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)(\mathrm{e}^{t}-1)^{-2} \displaystyle \right\}\\ &=&\displaystyle \frac{1}{n} \left\{ (\mathrm{e}^{t})'(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)(\mathrm{e}^{t}-1)^{-2} +\mathrm{e}^{t}(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)'(\mathrm{e}^{t}-1)^{-2} +\mathrm{e}^{t}(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)\left((\mathrm{e}^{t}-1)^{-2}\right)' \right\}\\ &=&\displaystyle \frac{1}{n} \left\{ \mathrm{e}^{t}(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)(\mathrm{e}^{t}-1)^{-2} +\mathrm{e}^{t}((n\mathrm{e}^{(n+1)t})'-((n+1)\mathrm{e}^{nt})'+(1)')(\mathrm{e}^{t}-1)^{-2} +\mathrm{e}^{t}(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)\left(-2\mathrm{e}^{t}(\mathrm{e}^{t}-1)^{-3}\right) \right\}\\ &=&\displaystyle \frac{1}{n} \left\{ \mathrm{e}^{t}(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)(\mathrm{e}^{t}-1)^{-2} +\mathrm{e}^{t}((n(n+1)\mathrm{e}^{(n+1)t})-(n(n+1)\mathrm{e}^{nt})+0)(\mathrm{e}^{t}-1)^{-2} +\mathrm{e}^{t}(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)\left(-2\mathrm{e}^{t}(\mathrm{e}^{t}-1)^{-3}\right) \right\}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \left\{ (n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)(\mathrm{e}^{t}-1)^{-2} +((n(n+1)\mathrm{e}^{(n+1)t})-(n(n+1)\mathrm{e}^{nt}))(\mathrm{e}^{t}-1)^{-2} +(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)\left(-2\mathrm{e}^{t}(\mathrm{e}^{t}-1)^{-3}\right) \right\}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \left\{\frac{ (n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1)(\mathrm{e}^{t}-1) +((n(n+1)\mathrm{e}^{(n+1)t})-(n(n+1)\mathrm{e}^{nt}))(\mathrm{e}^{t}-1) -2\mathrm{e}^{t}(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1) }{(\mathrm{e}^{t}-1)^{3}} \right\}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \left\{\frac{ (n\mathrm{e}^{(n+2)t}-(n+1)\mathrm{e}^{(n+1)t}+\mathrm{e}^{t})-(n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1) +((n(n+1)\mathrm{e}^{(n+2)t})-(n(n+1)\mathrm{e}^{(n+1)t}))-((n(n+1)\mathrm{e}^{(n+1)t})-(n(n+1)\mathrm{e}^{nt})) -(2n\mathrm{e}^{(n+2)t}-2(n+1)\mathrm{e}^{(n+1)t}+2\mathrm{e}^{t}) }{(\mathrm{e}^{t}-1)^{3}} \right\}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \left\{\frac{ n\mathrm{e}^{(n+2)t}-(n+1)\mathrm{e}^{(n+1)t}+\mathrm{e}^{t}-n\mathrm{e}^{(n+1)t}+(n+1)\mathrm{e}^{nt}-1 +n(n+1)\mathrm{e}^{(n+2)t}-n(n+1)\mathrm{e}^{(n+1)t}-n(n+1)\mathrm{e}^{(n+1)t}+n(n+1)\mathrm{e}^{nt} -2n\mathrm{e}^{(n+2)t}+2(n+1)\mathrm{e}^{(n+1)t}-2\mathrm{e}^{t} }{(\mathrm{e}^{t}-1)^{3}} \right\}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \left\{\frac{ (n+n(n+1)-2n)\mathrm{e}^{(n+2)t} +(-(n+1)-n-n(n+1)-n(n+1)+2(n+1))\mathrm{e}^{(n+1)t} +((n+1)+n(n+1))\mathrm{e}^{nt} +(1-2)\mathrm{e}^{t} -1}{(\mathrm{e}^{t}-1)^{3}} \right\}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \left\{\frac{ n^2\mathrm{e}^{(n+2)t} -(2n^2+2n-1)\mathrm{e}^{(n+1)t} +(n^2+2n+1)\mathrm{e}^{nt} -\mathrm{e}^{t} -1}{(\mathrm{e}^{t}-1)^{3}} \right\}\\ \end{array}$$

原点周りの二次モーメント

$$\begin{array}{rcl} \displaystyle E[X^2]&=&\displaystyle M_X^{(2)}(0)\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\mathrm{e}^{t}}{n} \left\{\frac{ n^2\mathrm{e}^{(n+2)t} -(2n^2+2n-1)\mathrm{e}^{(n+1)t} +(n^2+2n+1)\mathrm{e}^{nt} -\mathrm{e}^{t} -1}{(\mathrm{e}^{t}-1)^{3}} \right\} \right]\,\dotso\,0を代入すると分母が0になってしまうので極限で考える．\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\mathrm{e}^{t}}{n} \left\{\frac{ n^2\mathrm{e}^{(n+2)t} -(2n^2+2n-1)\mathrm{e}^{(n+1)t} +(n^2+2n+1)\mathrm{e}^{nt} -\mathrm{e}^{t} -1}{ \mathrm{e}^{3t} -3\mathrm{e}^{2t} +3\mathrm{e}^{t} -1} \right\} \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(\frac{t^0}{0!}+\frac{t^1}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}\right)}{n} \left\{\frac{ n^2\left( \frac{((n+2)t)^0}{0!}+\frac{((n+2)t)^1}{1!}+\frac{((n+2)t)^2}{2!}+\frac{((n+2)t)^3}{3!}+\frac{((n+2)t)^4}{4!} \right) -(2n^2+2n-1)\left( \frac{((n+1)t)^0}{0!}+\frac{((n+1)t)^1}{1!}+\frac{((n+1)t)^2}{2!}+\frac{((n+1)t)^3}{3!}+\frac{((n+1)t)^4}{4!} \right) +(n^2+2n+1)\left( \frac{(nt)^0}{0!}+\frac{(nt)^1}{1!}+\frac{(nt)^2}{2!}+\frac{(nt)^3}{3!}+\frac{(nt)^4}{4!} \right) -\left( \frac{t^0}{0!}+\frac{t^1}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!} \right) -1}{ \left( \frac{(3t)^0}{0!}+\frac{(3t)^1}{1!}+\frac{(3t)^2}{2!}+\frac{(3t)^3}{3!}+\frac{(3t)^4}{4!} \right) -3\left( \frac{(2t)^0}{0!}+\frac{(2t)^1}{1!}+\frac{(2t)^2}{2!}+\frac{(2t)^3}{3!}+\frac{(2t)^4}{4!} \right) +3\left( \frac{t^0}{0!}+\frac{t^1}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!} \right) -1} \right\} \right]\\ && \displaystyle \,\dotso\,\href{https://shikitenkai.blogspot.com/2019/07/blog-post.html}{\mathrm{e}^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\dotsb} (マクローリン展開), ひとまず4乗の項までで計算を進める．\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}\right)}{n} \left\{\frac{ n^2\left( 1 +(n+2)t +\frac{(n+2)^2}{2}t^2 +\frac{(n+2)^3}{6}t^3 +\frac{(n+2)^4}{24}t^4 \right) -(2n^2+2n-1)\left( 1 +(n+1)t +\frac{(n+1)^2}{2}t^2 +\frac{(n+1)^3}{6}t^3 +\frac{(n+1)^4}{24}t^4 \right) +(n^2+2n+1)\left( 1 +nt +\frac{n^2}{2}t^2 +\frac{n^3}{6}t^3 +\frac{n^4}{24}t^4 \right) -\left( 1 +t +\frac{1}{2}t^2 +\frac{1}{6}t^3 +\frac{1}{24}t^4 \right) -1}{ 1 +3t +\frac{9}{2}t^2 +\frac{27}{6}t^3 +\frac{81}{24}t^4 -3 -6t -\frac{12}{2}t^2 -\frac{24}{6}t^3 -\frac{48}{24}t^4 +3 +3t +\frac{3}{2}t^2 +\frac{3}{6}t^3 +\frac{3}{24}t^4 -1} \right\}\right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}\right)}{n} \left\{\frac{ n^2 \left(1+(n+2)t +\frac{(n+2)^2}{2}t^2 +\frac{(n+2)^3}{6}t^3 +\frac{(n+2)^4}{24}t^4 \right) -(2n^2+2n-1)\left(1+(n+1)t +\frac{(n+1)^2}{2}t^2 +\frac{(n+1)^3}{6}t^3 +\frac{(n+1)^4}{24}t^4 \right) +(n^2+2n+1) \left(1+nt +\frac{n^2}{2}t^2 +\frac{n^3}{6}t^3 +\frac{n^4}{24}t^4 \right) - \left(1+t +\frac{1}{2}t^2 +\frac{1}{6}t^3 +\frac{1}{24}t^4 \right) -1}{t^3+\frac{3}{2}t^4} \right\}\right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}\right)}{nt^3(1+\frac{3}{2}t)} \left\{ n^2 \left(1+(n+2)t +\frac{(n+2)^2}{2}t^2 +\frac{(n+2)^3}{6}t^3 +\frac{(n+2)^4}{24}t^4 \right) -(2n^2+2n-1)\left(1+(n+1)t +\frac{(n+1)^2}{2}t^2 +\frac{(n+1)^3}{6}t^3 +\frac{(n+1)^4}{24}t^4 \right) +(n^2+2n+1) \left(1+nt +\frac{n^2}{2}t^2 +\frac{n^3}{6}t^3 +\frac{n^4}{24}t^4 \right) - \left(1+t +\frac{1}{2}t^2 +\frac{1}{6}t^3 +\frac{1}{24}t^4 \right) -1 \right\}\right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}\right)}{nt^3(1+\frac{3}{2}t)} \left\{ n^2 +n^2(n+2)t +\frac{n^2(n+2)^2}{2}t^2 +\frac{n^2(n+2)^3}{6}t^3 +\frac{n^2(n+2)^4}{24}t^4 -(2n^2+2n-1) -(2n^2+2n-1)(n+1)t -\frac{(2n^2+2n-1)(n+1)^2}{2}t^2 -\frac{(2n^2+2n-1)(n+1)^3}{6}t^3 +\frac{(2n^2+2n-1)(n+1)^4}{24}t^4 +(n^2+2n+1) +(n^2+2n+1)nt +\frac{(n^2+2n+1)n^2}{2}t^2 +\frac{(n^2+2n+1)n^3}{6}t^3 +\frac{(n^2+2n+1)n^4}{24}t^4 -1 -t -\frac{1}{2}t^2 -\frac{1}{6}t^3 +\frac{1}{24}t^4 -1 \right\}\right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}\right)}{nt^3(1+\frac{3}{2}t)} \left\{ (n^2 -(2n^2+2n-1) +(n^2+2n+1) -1 -1) +(n^2(n+2) -(2n^2+2n-1)(n+1) +(n^2+2n+1)n -1 )t +(\frac{n^2(n+2)^2}{2} -\frac{(2n^2+2n-1)(n+1)^2}{2} +\frac{(n^2+2n+1)n^2}{2} -\frac{1}{2} )t^2 +(\frac{n^2(n+2)^3}{6} -\frac{(2n^2+2n-1)(n+1)^3}{6} +\frac{(n^2+2n+1)n^3}{6} -\frac{1}{6} )t^3 +(\frac{n^2(n+2)^4}{24} -\frac{(2n^2+2n-1)(n+1)^4}{24} +\frac{(n^2+2n+1)n^4}{24} -\frac{1}{24} )t^4 \right\}\right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}\right)}{nt^3(1+\frac{3}{2}t)} \left\{ (0) +(0)t +(0)t^2 +\frac{n(n+1)(2n+1)}{6}t^3 +\frac{n(n+1)(3n^2+5n+1)}{12}t^4 \right\}\right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}\right)}{nt^3(1+\frac{3}{2}t)} nt^3 \left\{ \frac{(n+1)(2n+1)}{6} +\frac{(n+1)(3n^2+5n+1)}{12}t \right\}\right]\\ &&\,\dotso\,分母はt^3の項からが残っている．t^4以上の項はtが残るのでマクローリン展開はt^4で十分となる\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{t^2}{2}+\frac{t^3}{6}+\frac{t^4}{24}\right)}{1+\frac{3}{2}t} \left\{ \frac{(n+1)(2n+1)}{6} +\frac{(n+1)(3n^2+5n+1)}{12}t \right\}\right]\\ &=&\displaystyle \frac{(n+1)(2n+1)}{6}\\ &&\,\dotso\,tが分子にある(掛けられている)項は全て0．\\ \end{array}$$

分散

$$\begin{array}{rcl} \displaystyle V[X]&=&\displaystyle E[X^2]-E[X]^2\\ &=&\displaystyle \frac{(n+1)(2n+1)}{6}-\left(\href{https://shikitenkai.blogspot.com/2019/07/discrete-random-variable-uniform.html}{\frac{n+1}{2}}\right)^2\\ &=&\displaystyle \frac{(n+1)(2n+1)}{6}-\frac{(n+1)^2}{4}\\ &=&\displaystyle \frac{2(n+1)(2n+1)-3(n+1)^2}{12}\\ &=&\displaystyle \frac{(n+1)(2(2n+1)-3(n+1))}{12}\\ &=&\displaystyle \frac{(n+1)(4n+2-3n-3)}{12}\\ &=&\displaystyle \frac{(n+1)(n-1)}{12}\\ &=&\displaystyle \frac{n^2-1}{12}\\ \end{array}$$

離散型確率変数(discrete random variable) の一様分布(uniform distribution)の期待値(expected value)

$$\begin{array}{rcl} \displaystyle M_X^{(m)}(0)&\equiv&\frac{ \mathrm{d}^m }{ \mathrm{d}^m t } M_X(t)|_{t=0}\\ &=&\displaystyle E[X^m\mathrm{e}^{tX}]|_{t=0}\\ &=&\displaystyle E[X^m]\\ \end{array}$$

積率母凾数の一階微分

$$\begin{array}{rcl} \displaystyle M_X^{(1)} &=& \displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \displaystyle \href{https://shikitenkai.blogspot.com/2019/07/uniform-distribution.html}{\frac{1}{n}\frac{\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{(\mathrm{e}^{t}-1)}} \displaystyle \right\}\\ &=& \displaystyle \frac{1}{n} \displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \displaystyle \frac{\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{(\mathrm{e}^{t}-1)} \displaystyle \right\}\\ &=& \displaystyle \frac{1}{n} \displaystyle \frac{\mathrm{d}}{\mathrm{d}t}\left\{ \displaystyle \mathrm{e}^{t}(\mathrm{e}^{nt}-1)(\mathrm{e}^{t}-1)^{-1} \displaystyle \right\}\\ &=&\displaystyle \frac{1}{n} \left\{ (\mathrm{e}^{t})'(\mathrm{e}^{nt}-1)(\mathrm{e}^{t}-1)^{-1} + \mathrm{e}^{t}(\mathrm{e}^{nt}-1)'(\mathrm{e}^{t}-1)^{-1} + \mathrm{e}^{t}(\mathrm{e}^{nt}-1)((\mathrm{e}^{t}-1)^{-1})' \right\}\\ &&\,\dotso\,(uvw)'=u'(vw)+u(vw)'=u'(vw)+u(v'w+vw')=u'vw+uv'w+uvw'\\ &=&\displaystyle \frac{1}{n} \left\{ \mathrm{e}^{t}(\mathrm{e}^{nt}-1)(\mathrm{e}^{t}-1)^{-1} + \mathrm{e}^{t}(n\mathrm{e}^{nt})(\mathrm{e}^{t}-1)^{-1} + \mathrm{e}^{t}(\mathrm{e}^{nt}-1)(-\mathrm{e}^{t}(\mathrm{e}^{t}-1)^{-2}) \right\}\\ &=&\displaystyle \frac{1}{n} \left\{ \frac{\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{(\mathrm{e}^{t}-1)} + \frac{\mathrm{e}^{t}(n\mathrm{e}^{nt})}{(\mathrm{e}^{t}-1)} + \frac{\mathrm{e}^{t}\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{-(\mathrm{e}^{t}-1)^2} \right\}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \left\{ \frac{(\mathrm{e}^{nt}-1)}{(\mathrm{e}^{t}-1)} + \frac{(n\mathrm{e}^{nt})}{(\mathrm{e}^{t}-1)} + \frac{\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{-(\mathrm{e}^{t}-1)^2} \right\}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \frac{(\mathrm{e}^{nt}-1)(\mathrm{e}^{t}-1) + (n\mathrm{e}^{nt})(\mathrm{e}^{t}-1) - \mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{(\mathrm{e}^{t}-1)^2}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \frac{(\mathrm{e}^{nt}\mathrm{e}^{t}-\mathrm{e}^{nt}-\mathrm{e}^{t}+1) + (n\mathrm{e}^{nt}\mathrm{e}^{t}-n\mathrm{e}^{nt}) - (\mathrm{e}^{nt}\mathrm{e}^{t}-\mathrm{e}^{t})}{(\mathrm{e}^{t}-1)^2}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \frac{\mathrm{e}^{nt}\mathrm{e}^{t}-\mathrm{e}^{nt}-\mathrm{e}^{t}+1 + n\mathrm{e}^{nt}\mathrm{e}^{t}-n\mathrm{e}^{nt} - \mathrm{e}^{nt}\mathrm{e}^{t}+\mathrm{e}^{t}}{(\mathrm{e}^{t}-1)^2}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \frac{\mathrm{e}^{nt}(\mathrm{e}^{t} - 1 +n\mathrm{e}^{t} -n-\mathrm{e}^{t}) +1}{(\mathrm{e}^{t}-1)^2}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \frac{\mathrm{e}^{nt}(n\mathrm{e}^{t}-(n+1))+1}{(\mathrm{e}^{t}-1)^2}\\ &=&\displaystyle \frac{\mathrm{e}^{t}}{n} \frac{n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1}{(\mathrm{e}^{t}-1)^2}\\ \end{array}$$

原点周りの一次モーメント=期待値

$$\begin{array}{rcl} \displaystyle E[X]&=&\displaystyle M_X^{(1)}(0)\\ &=&\displaystyle \lim_{t \to 0}\left\{ \frac{\mathrm{e}^{t}}{n} \displaystyle \frac{n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1}{(\mathrm{e}^{t}-1)^2} \right\}\,\dotso\,0を代入すると分母が0になってしまうので極限で考える．\\ &=&\displaystyle \lim_{t \to 0}\left\{ \frac{\mathrm{e}^{t}}{n} \displaystyle \frac{n\mathrm{e}^{(n+1)t}-(n+1)\mathrm{e}^{nt}+1}{\mathrm{e}^{2t}-2\mathrm{e}^{t}+1} \right\}\\ &=&\displaystyle \lim_{t \to 0} \left\{\frac{\left(\frac{t^0}{0!}+\frac{t^1}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}\right)}{n} \frac{n\left(\frac{((n+1)t)^0}{0!}+\frac{((n+1)t)^1}{1!}+\frac{((n+1)t)^2}{2!}+\frac{((n+1)t)^3}{3!}\right) -(n+1)\left(\frac{(nt)^0}{0!}+\frac{(nt)^1}{1!}+\frac{(nt)^2}{2!}+\frac{(nt)^3}{3!}\right) +1}{ \left(\frac{(2t)^0}{0!}+\frac{(2t)^1}{1!}+\frac{(2t)^2}{2!}+\frac{(2t)^3}{3!}\right) -2 \left(\frac{t^0}{0!}+\frac{t^1}{1!}+\frac{t^2}{2!}+\frac{t^3}{3!}\right) +1} \right\}\\ && \displaystyle \,\dotso\,\href{https://shikitenkai.blogspot.com/2019/07/blog-post.html}{\mathrm{e}^x=\sum_{k=0}^{\infty}\frac{x^k}{k!}=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dotsb} (マクローリン展開), ひとまずt^3の項までで計算を進める．\\ &=&\displaystyle \lim_{t \to 0}\left\{ \frac{\left(1+t+\frac{1}{2}t^2+\frac{1}{6}t^3\right)}{n} \frac{n\left(1+(n+1)t+\frac{(n+1)^2}{2}t^2+\frac{(n+1)^3}{6}t^3\right) -(n+1)\left(1+nt+\frac{n^2}{2}t^2+\frac{n^3}{6}t^3\right) +1}{ \left(1+(2t)+\frac{(2t)^2}{2}+\frac{(2t)^3}{6}\right) -2 \left(1+(t)+\frac{t^2}{2}+\frac{t^3}{6}\right) +1} \right\}\\ &=&\displaystyle \lim_{t \to 0}\left\{ \frac{\left(1+t+\frac{1}{2}t^2+\frac{1}{6}t^3\right)}{n} \frac{n\left(1+(n+1)t+\frac{(n+1)^2}{2}t^2+\frac{(n+1)^3}{6}t^3\right) -(n+1)\left(1+nt+\frac{n^2}{2}t^2+\frac{n^3}{6}t^3\right) +1}{ 1+2t+2t^2+\frac{4}{3}t^3 -2-2t-t^2-\frac{1}{3}t^3 +1} \right\}\\ &=&\displaystyle \lim_{t \to 0}\left\{ \frac{\left(1+t+\frac{1}{2}t^2+\frac{1}{6}t^3\right)}{n} \frac{n\left(1+(n+1)t+\frac{(n+1)^2}{2}t^2+\frac{(n+1)^3}{6}t^3\right) -(n+1)\left(1+nt+\frac{n^2}{2}t^2+\frac{n^3}{6}t^3\right) +1}{t^2(1+t)} \right\}\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{1}{2}t^2+\frac{1}{6}t^3\right)}{nt^2(1+t)} \left\{n+n(n+1)t+\frac{n(n+1)^2}{2}t^2+\frac{n(n+1)^3}{6}t^3 -(n+1)-(n+1)nt-\frac{(n+1)n^2}{2}t^2-\frac{(n+1)n^3}{6}t^3 +1\right\} \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{1}{2}t^2+\frac{1}{6}t^3\right)}{nt^2(1+t)} \left\{(n-(n+1)+1) +(n(n+1)-(n+1)n)t +(\frac{n(n+1)^2}{2}-\frac{(n+1)n^2}{2})t^2 +(\frac{n(n+1)^3}{6}-\frac{(n+1)n^3}{6})t^3 \right\} \right]\\ &=&\displaystyle \lim_{t \to 0}\left[ \frac{\left(1+t+\frac{1}{2}t^2+\frac{1}{6}t^3\right)}{nt^2(1+t)} \left\{(0) +(0)t +(\frac{n+1}{2})nt^2 +(\frac{(n+1)(2n+1)}{6})nt^3 \right\} \right]\\ &=&\displaystyle \lim_{t \to 0}\left\{ \frac{\left(1+t+\frac{1}{2}t^2+\frac{1}{6}t^3\right)}{nt^2(1+t)} nt^2\left(\frac{n+1}{2}+\frac{(n+1)(2n+1)}{6}t\right) \right\}\\ &&\,\dotso\,分母はt^2の項からが残っている．t^3以上の項はtが残るのでマクローリン展開はt^3で十分となる．\\ &=&\displaystyle \lim_{t \to 0}\left\{ \left(1+t+\frac{1}{2}t^2+\frac{1}{6}t^3\right) \left(\frac{n+1}{2}+\frac{(n+1)(2n+1)}{6}t\right) \right\}\\ &=&\displaystyle \frac{n+1}{2}\\ &&\,\dotso\,tが分子にある(掛けられている)項は全て0．\\ \end{array}$$

離散型確率変数(discrete random variable) の一様分布(uniform distribution)の積率母凾数(moment-generating function)

$$f_X(x) = \begin{cases} \displaystyle \frac{1}{n} & \quad x \in \left\{1,2, \dots ,n\right\}\\ \displaystyle 0 & \quad x \notin \left\{1,2, \dots ,n\right\} \end{cases}$$ $$\begin{array}{rcl} \displaystyle M_X(t)&\equiv&\displaystyle E[\mathrm{e}^{tX}]\\ &=&\displaystyle \sum_{x=1}^{n}\mathrm{e}^{tx}\left(\frac{1}{n}\right)\\ &=&\displaystyle \frac{1}{n} \sum_{x=1}^{n}\mathrm{e}^{tx}\\ &=&\displaystyle \frac{1}{n} \left(\mathrm{e}^t+\mathrm{e}^{2t}+\dotsb+\mathrm{e}^{nt}\right)\\ &=&\displaystyle \frac{1}{n} \frac{\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{\mathrm{e}^{t}-1} \,\dotso\,a+ar+ar^2+\dotsb+ar^{n-1}=\frac{a(1-r^n)}{1-r}=\frac{a(r^n-1)}{r-1}\;(r \neq 1)\\ &=&\displaystyle \frac{\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{n(\mathrm{e}^{t}-1)}\\ \end{array}$$

偶数の二重階乗(double factorial / semifactorial)の逆数(reciprocal)の和(無限級数(infinite series))

$$\begin{array}{rcl} \frac{1}{0!!}+\frac{1}{2!!}+\frac{1}{4!!}+\cdots &=&\displaystyle \sum_{k=0}^{\infty}\frac{1}{2^k k!} \\&&\;\dots\;n!!(二重階乗) \neq (n!)! (=階乗凾数の二回反復),\;偶数nの二重階乗n!!=\prod_{k=0}^{\frac{n}{2}}(2k),偶数2k(k\geq0)の二重階乗(2k)!!=2^kk! \\&=&\displaystyle \sum_{k=0}^{\infty}\frac{\left(\frac{1}{2}\right)^k}{k!} \\&=&\displaystyle \mathrm{e}^{\frac{1}{2}} \,\dotso\, \href{https://shikitenkai.blogspot.com/2019/07/blog-post.html}{\sum_{k=0}^{\infty}\frac{x^k}{k!}= \mathrm{e}^x}\\ &=&\displaystyle \sqrt{\mathrm{e}}\\ \end{array}$$

正規分布(normal distribution)の積率母凾数(moment-generating function)と期待値(expected value)・分散(variance)

正規分布

$$\begin{array}{rcl} N(\mu, \sigma^2)&=&\frac{1}{\sqrt{2\pi \sigma^2}}\mathrm{e}^{\frac{-(x-\mu)^2}{2\sigma^2}} \end{array}$$

積率母凾数

$$\begin{array}{rcl} \displaystyle M_X(t)&\equiv&\displaystyle E[\mathrm{e}^{tX}]\\ &=&\displaystyle \int_{-\infty}^{\infty}(\mathrm{e}^{tx})\frac{1}{\sqrt{2\pi \sigma^2}}\mathrm{e}^{-\frac{(x-\mu)^2}{2\sigma^2}} \mathrm{d}x\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty}(\mathrm{e}^{tx})\mathrm{e}^{-\frac{(x-\mu)^2}{2\sigma^2}} \mathrm{d}x\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty}\mathrm{e}^{-\frac{(x-\mu)^2}{2\sigma^2}+tx} \mathrm{d}x\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty}\mathrm{e}^{-\frac{(x-\mu)^2+(2\sigma^2)(tx)}{2\sigma^2}} \mathrm{d}x\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty}\mathrm{e}^{-\frac{x^2-2x\mu+\mu^2-2x\sigma^2t}{2\sigma^2}} \mathrm{d}x\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty} \displaystyle \mathrm{e}^{-\frac{x^2-2x\mu+\mu^2-2x\sigma^2t}{2\sigma^2}-\frac{2\mu\sigma^2 t+\sigma^4t^2}{2\sigma^2}+\frac{2\mu\sigma^2 t+\sigma^4t^2}{2\sigma^2}} \displaystyle \mathrm{d}x\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty} \displaystyle \mathrm{e}^{-\frac{x^2-2x\mu+\mu^2-2x\sigma^2t+2\mu\sigma^2 t+\sigma^4t^2}{2\sigma^2}+\frac{2\mu\sigma^2 t+\sigma^4t^2}{2\sigma^2}} \displaystyle \mathrm{d}x\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty} \displaystyle \mathrm{e}^{-\frac{(x-\mu-\sigma^2t)^2}{2\sigma^2}+\frac{2\mu\sigma^2 t+\sigma^4t^2}{2\sigma^2}} \displaystyle \mathrm{d}x \,\dotso\,a^2-2ab+b^2-2ac+2bc+c^2=(a-b-c)^2\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty} \displaystyle \mathrm{e}^{-\frac{(x-\mu-\sigma^2t)^2}{2\sigma^2}+(\mu t + \frac{\sigma^2t^2}{2})} \displaystyle \mathrm{d}x\\ &=&\displaystyle \frac{1}{\sqrt{2\pi \sigma^2}} \displaystyle \int_{-\infty}^{\infty} \mathrm{e}^{-\frac{(x-\mu-\sigma^2t)^2}{2\sigma^2}} \displaystyle \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})} \displaystyle \mathrm{d}x\\ &=&\displaystyle \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})} \displaystyle \frac{1}{\sqrt{2\pi \sigma^2}} \displaystyle \int_{-\infty}^{\infty}\mathrm{e}^{-\frac{(x-\mu-\sigma^2t)^2}{2\sigma^2}}\mathrm{d}x \\ &=&\displaystyle \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})}\,\dotso\,\frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^{\infty}\mathrm{e}^{-\frac{(x-\mu-\sigma^2t)^2}{2\sigma^2}}\mathrm{d}x = N(\mu+\sigma^2t,\sigma^2)の総和=1\\ \end{array}$$

期待値・分散

$$\begin{array}{rcl} \displaystyle M_X^{(m)}(0)&\equiv&\frac{ \mathrm{d}^m }{ \mathrm{d}^m t } M_x(t)|_{t=0}\\ &=&\displaystyle E[X^m\mathrm{e}^{tX}]|_{t=0}\\ &=&\displaystyle E[X^m]\\ \end{array}$$ $$\begin{array}{rcl} \displaystyle E[X]&=&\displaystyle M_X^{(1)}(0)\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} t }\left(\mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})}\right) \right\}|_{t=0}\\ &=&\displaystyle \left\{ \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})}(\mu+\sigma^2t) \right\}|_{t=0}\\ &=&\displaystyle \mathrm{e}^{(\mu 0 + \frac{\sigma^20^2}{2})}(\mu+\sigma^20) \\ &=&\displaystyle \mathrm{e}^{0}(\mu+0) \\ &=&\displaystyle \mu \\ \end{array}$$ $$\begin{array}{rcl} \displaystyle E[X^2]&=&\displaystyle M_X^{(2)}(0)\\ &=&\displaystyle \left\{ \frac{ \mathrm{d}^2 }{ \mathrm{d} t^2 }\left(\mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})}\right) \right\}|_{t=0}\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} t }\left( \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})}(\mu+\sigma^2t)\right) \right\}|_{t=0}\\ &=&\displaystyle \left[ \displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} t } \left( \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})} \right)\right\} \left( \mu+\sigma^2t \right) \displaystyle + \left( \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})} \right) \left\{ \frac{ \mathrm{d} }{ \mathrm{d} t } \left( \mu+\sigma^2t \right) \right\} \right]|_{t=0}\\ &=&\displaystyle \left[ \displaystyle \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})} \left( \mu+\sigma^2t \right)^2 \displaystyle + \mathrm{e}^{(\mu t + \frac{\sigma^2t^2}{2})} \sigma^2 \displaystyle \right]|_{t=0}\\ &=&\displaystyle \mathrm{e}^{(\mu 0 + \frac{\sigma^20^2}{2})} \left( \mu+\sigma^20 \right)^2 \displaystyle + \mathrm{e}^{(\mu 0 + \frac{\sigma^20^2}{2})} \sigma^2\\ &=&\displaystyle \mu^2+\sigma^2\\ \end{array}$$ $$\begin{array}{rcl} V[X]&=&E[X^2]-E[X]^2\\ &=&\mu^2+\sigma^2-\mu^2\\ &=&\sigma^2 \end{array}$$

ポアソン分布(Poisson distribution)の積率母凾数(moment-generating function)と期待値(expected value)・分散(variance)

ポアソン分布

$$\begin{array}{rcl} \displaystyle Po(\lambda)&=&\displaystyle \frac{\lambda^{x}}{x!}\mathrm{e}^{-\lambda}\\ \end{array}$$

積率母凾数

$$\begin{array}{rcl} \displaystyle M_X(t)&\equiv&\displaystyle E[\mathrm{e}^{tX}]\\ &=&\displaystyle \sum_{x=0}^{\infty}(\mathrm{e}^{tx})\frac{\lambda^{x}}{x!}\mathrm{e}^{-\lambda}\\ &=&\displaystyle \mathrm{e}^{-\lambda}\sum_{x=0}^{\infty}(\mathrm{e}^{tx})\frac{\lambda^{x}}{x!}\\ &=&\displaystyle \mathrm{e}^{-\lambda}\sum_{x=0}^{\infty}\frac{\mathrm{e}^{tx}\lambda^{x}}{x!}\\ &=&\displaystyle \mathrm{e}^{-\lambda}\sum_{x=0}^{\infty}\frac{(\mathrm{e}^{t}\lambda)^{x}}{x!}\\ &=&\displaystyle \mathrm{e}^{-\lambda}\mathrm{e}^{\mathrm{e}^{t}\lambda} \,\dotso\,\href{https://shikitenkai.blogspot.com/2019/07/blog-post.html}{\sum_{x=0}^{\infty}\frac{a^{x}}{x!}=\mathrm{e}^a}\\ &=&\displaystyle \mathrm{e}^{\mathrm{e}^{t}\lambda-\lambda}=\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)} \end{array}$$

期待値・分散

$$\begin{array}{rcl} \displaystyle M_X^{(m)}(0)&\equiv&\frac{ \mathrm{d}^m }{ \mathrm{d}^m t } M_x(t)|_{t=0}\\ &=&\displaystyle E[X^m\mathrm{e}^{tX}]|_{t=0}\\ &=&\displaystyle E[X^m]\\ \end{array}$$ $$\begin{array}{rcl} \displaystyle E[X]&=&\displaystyle M_X^{(1)}(0)\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} t }(\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)}) \right\}|_{t=0}\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} s }(\mathrm{e}^{s})\frac{ \mathrm{d}s}{ \mathrm{d}t} \right\}|_{t=0} \,\dotso\,s=\lambda(\mathrm{e}^{t}-1),\frac{ \mathrm{d}s}{ \mathrm{d}t}=\lambda\mathrm{e}^{t}\\ &=&\displaystyle \left\{ (\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)})(\lambda\mathrm{e}^{t}) \right\}|_{t=0} \,\dotso\,\frac{ \mathrm{d} }{ \mathrm{d} x }\mathrm{e}^{x}=\mathrm{e}^{x}\\ &=&\displaystyle \left\{ \lambda(\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)+t}) \right\}|_{t=0}\\ &=&\displaystyle \lambda(\mathrm{e}^{\lambda(\mathrm{e}^{0}-1)+0})\\ &=&\displaystyle \lambda\mathrm{e}^0 \,\dotso\,a^0=1\\ &=&\lambda \,\dotso\,a^0=1\\ \end{array}$$ $$\begin{array}{rcl} \displaystyle E[X^2]&=&\displaystyle M_X^{(2)}(0)\\ &=&\displaystyle \left\{ \frac{ \mathrm{d}^2 }{ \mathrm{d} t^2 }(\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)}) \right\}|_{t=0}\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} t } \lambda(\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)+t}) \right\}|_{t=0} \,\dotso\,E[X]の展開から．\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} s }(\lambda\mathrm{e}^{s})\frac{ \mathrm{d}s}{ \mathrm{d}t} \right\}|_{t=0} \,\dotso\,s=\lambda(\mathrm{e}^{t}-1)+t,\frac{ \mathrm{d}s}{ \mathrm{d}t}=\lambda\mathrm{e}^{t}+1\\ &=&\displaystyle \left\{ (\lambda\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)+1})(\lambda\mathrm{e}^{t}+1) \right\}|_{t=0} \,\dotso\,\frac{ \mathrm{d} }{ \mathrm{d} x }C\mathrm{e}^{x}=C\mathrm{e}^{x}\\ &=&\displaystyle (\lambda\mathrm{e}^{\lambda(\mathrm{e}^{0}-1)+1})(\lambda\mathrm{e}^{0}+1)\\ &=&\displaystyle \lambda\mathrm{e}^0(\lambda+1) \,\dotso\,a^0=1\\ &=&\lambda(\lambda+1) \,\dotso\,a^0=1\\ \end{array}$$ $$\begin{array}{rcl} \displaystyle V[X]&=&\displaystyle E[X^2]-E[X]^2\\ &=&\lambda(\lambda+1)-\lambda^2\\ &=&\lambda^2+\lambda-\lambda^2\\ &=&\lambda\\ \end{array}$$

階乗(factorial)の逆数(reciprocal)の和(無限級数(infinite series))

$$\begin{array}{rcl} \displaystyle f(x) &=& \displaystyle \sum_{k=0}^{\infty}\frac{f^{(k)}(a)}{x!}(x-a)^k\,\dotso\,a点まわりのテイラー展開\\ &=& \displaystyle \frac{1}{0!}f^{(0)}(a)(x-a)^0+\frac{1}{1!}f^{(1)}(a)(x-a)^1+\frac{1}{2!}f^{(2)}(a)(x-a)^2+\dotsb\\ \end{array}$$

$e^x$のマクローリン展開

$$\begin{array}{rcl} \displaystyle \mathrm{e}^x　 &=& \displaystyle \sum_{k=0}^{\infty}\frac{1}{k!}\left\{\left(\frac{\mathrm{d}^k}{\mathrm{d}x^k}\mathrm{e}^x\right)|_{x=0}\right\}(x-0)^k \,\dotso\,0点まわりのテイラー展開(Taylor\,series)=マクローリン展開(Maclaurin\,expansion)\\ &=& \displaystyle \frac{1}{0!}\left\{\left(\frac{\mathrm{d}^0}{\mathrm{d}x^0}\mathrm{e}^x\right)|_{x=0}\right\}(x-0)^0 \displaystyle +\frac{1}{1!}\left\{\left(\frac{\mathrm{d}^1}{\mathrm{d}x^1}\mathrm{e}^x\right)|_{x=0}\right\}(x-0)^1 \displaystyle +\frac{1}{2!}\left\{\left(\frac{\mathrm{d}^2}{\mathrm{d}x^2}\mathrm{e}^x\right)|_{x=0}\right\}(x-0)^2 \displaystyle +\dotsb\\ &=& \displaystyle \frac{1}{0!}\left\{\left(\mathrm{e}^x\right)|_{x=0}\right\}(x-0)^0 \displaystyle +\frac{1}{1!}\left\{\left(\mathrm{e}^x\right)|_{x=0}\right\}(x-0)^1 \displaystyle +\frac{1}{2!}\left\{\left(\mathrm{e}^x\right)|_{x=0}\right\}(x-0)^2 \displaystyle +\dotsb\\ &=& \displaystyle \frac{1}{0!}\,1\,x^0 \displaystyle +\frac{1}{1!}\,1\,x^1 \displaystyle +\frac{1}{2!}\,1\,x^2 \displaystyle +\dotsb \,\dotso\,a^0=1\\ &=& \displaystyle \sum_{k=0}^{\infty}\frac{1}{k!}x^k\\ &=& \displaystyle \sum_{k=0}^{\infty}\frac{x^k}{k!}\\ \end{array}$$ $$\begin{array}{rcl} \displaystyle \sum_{k=0}^{\infty}\frac{x^k}{k!}&=&\displaystyle \mathrm{e}^x\\ \displaystyle \sum_{k=0}^{\infty}\frac{1}{k!}=\sum_{k=0}^{\infty}\frac{1^k}{k!}&=&\displaystyle \mathrm{e}^1=\mathrm{e}\,\dotso\,x=1\\ \end{array}$$

$e^{Cx}$のマクローリン展開

$$\begin{array}{rcl} \displaystyle \mathrm{e}^{Cx}　 &=& \displaystyle \sum_{k=0}^{\infty}\frac{1}{k!}\left\{\left(\frac{\mathrm{d}^k}{\mathrm{d}x^k}\mathrm{e}^{Cx}\right)|_{x=0}\right\}(x-0)^k \,\dotso\,0点まわりのテイラー展開(Taylor\,series)=マクローリン展開(Maclaurin\,expansion)\\ &=& \displaystyle \frac{1}{0!}\left\{\left(\frac{\mathrm{d}^0}{\mathrm{d}x^0}\mathrm{e}^{Cx}\right)|_{x=0}\right\}(x-0)^0 \displaystyle +\frac{1}{1!}\left\{\left(\frac{\mathrm{d}^1}{\mathrm{d}x^1}\mathrm{e}^{Cx}\right)|_{x=0}\right\}(x-0)^1 \displaystyle +\frac{1}{2!}\left\{\left(\frac{\mathrm{d}^2}{\mathrm{d}x^2}\mathrm{e}^{Cx}\right)|_{x=0}\right\}(x-0)^2 \displaystyle +\dotsb\\ &=& \displaystyle \frac{1}{0!}\left\{\left(\mathrm{e}^{Cx}\right)|_{x=0}\right\}(x-0)^0 \displaystyle +\frac{1}{1!}\left\{\left(C\mathrm{e}^{Cx}\right)|_{x=0}\right\}(x-0)^1 \displaystyle +\frac{1}{2!}\left\{\left(C^2\mathrm{e}^{Cx}\right)|_{x=0}\right\}(x-0)^2 \displaystyle +\dotsb\\ &=& \displaystyle \frac{1}{0!}\,1\,x^0 \displaystyle +\frac{1}{1!}\,C\,x^1 \displaystyle +\frac{1}{2!}\,C^2\,x^2 \displaystyle +\dotsb \,\dotso\,a^0=1\\ &=& \displaystyle \frac{1}{0!}\,C^0\,x^0 \displaystyle +\frac{1}{1!}\,C^1\,x^1 \displaystyle +\frac{1}{2!}\,C^2\,x^2 \displaystyle +\dotsb \,\dotso\,a^0=1\\ &=& \displaystyle \sum_{k=0}^{\infty}\frac{(Cx)^k}{k!}\\ \end{array}$$ $$\begin{array}{rcl} \displaystyle \sum_{k=0}^{\infty}\frac{(Cx)^k}{k!} &=& \mathrm{e}^{Cx}\\ \end{array}$$

回転行列と三角凾数の加法定理

角度$\alpha$と角度$\beta$の回転行列を掛ける

$$\begin{array}{rcl} \begin{pmatrix} cos(\alpha) & -sin(\alpha) \\ sin(\alpha) & cos(\alpha) \\ \end{pmatrix} \begin{pmatrix} cos(\beta) & -sin(\beta) \\ sin(\beta) & cos(\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix} &=& \begin{pmatrix} cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta) & -cos(\alpha)sin(\beta)-sin(\alpha)cos(\beta) \\ sin(\alpha)cos(\beta)+cos(\alpha)sin(\beta) & -sin(\alpha)sin(\beta)+cos(\alpha)cos(\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\\ &=& \begin{pmatrix} cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta) & -(sin(\alpha)cos(\beta)+cos(\alpha)sin(\beta)) \\ sin(\alpha)cos(\beta)+cos(\alpha)sin(\beta) & -sin(\alpha)sin(\beta)+cos(\alpha)cos(\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\\ &=& \begin{pmatrix} cos(\alpha+\beta) & -sin(\alpha+\beta) \\ sin(\alpha+\beta) & cos(\alpha+\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\,\dotso\,これが角度(\alpha+\beta)の回転行列と等しい．\\ \end{array}$$ 対応する要素同士が加法定理となる． $$\begin{array}{rcl} cos(\alpha+\beta)&=&cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta)\\ sin(\alpha+\beta)&=&sin(\alpha)cos(\beta)+cos(\alpha)sin(\beta)\\ \end{array}$$

角度$\alpha$と角度$-\beta$の回転行列を掛ける

$$\begin{array}{rcl} \begin{pmatrix} cos(\alpha) & -sin(\alpha) \\ sin(\alpha) & cos(\alpha) \\ \end{pmatrix} \begin{pmatrix} cos(-\beta) & -sin(-\beta) \\ sin(-\beta) & cos(-\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix} &=& \begin{pmatrix} cos(\alpha)cos(-\beta)-sin(\alpha)sin(-\beta) & -cos(\alpha)sin(-\beta)-sin(\alpha)cos(-\beta) \\ sin(\alpha)cos(-\beta)+cos(\alpha)sin(-\beta) & -sin(\alpha)sin(-\beta)+cos(\alpha)cos(-\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\\ &=& \begin{pmatrix} cos(\alpha)cos(\beta)+sin(\alpha)sin(\beta) & cos(\alpha)sin(\beta)-sin(\alpha)cos(\beta) \\ sin(\alpha)cos(\beta)-cos(\alpha)sin(\beta) & sin(\alpha)sin(\beta)+cos(\alpha)cos(\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\,\dotso\,cos(-\beta)=cos(\beta), sin(-\beta)=-sin(\beta)を適用．\\ &=& \begin{pmatrix} cos(\alpha)cos(\beta)+sin(\alpha)sin(\beta) & -(sin(\alpha)cos(\beta)-cos(\alpha)sin(\beta)) \\ sin(\alpha)cos(\beta)-cos(\alpha)sin(\beta) & cos(\alpha)cos(\beta)+sin(\alpha)sin(\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\\ &=& \begin{pmatrix} cos(\alpha-\beta) & -sin(\alpha-\beta) \\ sin(\alpha-\beta) & cos(\alpha-\beta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\,\dotso\,これが角度(\alpha-\beta)の回転行列と等しい．\\ \end{array}$$ $$\begin{array}{rcl} cos(\alpha -\beta)&=&cos(\alpha)cos(\beta)+sin(\alpha)sin(\beta)\\ sin(\alpha-\beta)&=&sin(\alpha)cos(\beta)-cos(\alpha)sin(\beta)\\ \end{array}$$

角度$\alpha$と角度$\beta$が一致($\alpha=\beta=\theta$)

$$\begin{array}{rcl} \begin{pmatrix} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \\ \end{pmatrix} \begin{pmatrix} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix} &=& \begin{pmatrix} cos(\theta)cos(\theta)-sin(\theta)sin(\theta) & -cos(\theta)sin(\theta)-sin(\theta)cos(\theta) \\ sin(\theta)cos(\theta)+cos(\theta)sin(\theta) & -sin(\theta)sin(\theta)+cos(\theta)cos(\theta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\\ &=& \begin{pmatrix} cos(\theta)^2-sin(\theta)^2 & -2sin(\theta)cos(\theta) \\ 2sin(\theta)cos(\theta) & cos(\theta)^2-sin(\theta)^2 \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\\ &=& \begin{pmatrix} cos(2\theta) & -sin(2\theta) \\ sin(2\theta) & cos(2\theta) \\ \end{pmatrix} \begin{pmatrix} x\\y \end{pmatrix}\,\dotso\,これが角度(2\theta)の回転行列と等しい．\\ \end{array}$$ 対応する要素同士が倍角公式となる． $$\begin{array}{rcl} cos(2\theta)&=&cos(\theta)^2-sin(\theta)^2\\ sin(2\theta)&=&2sin(\theta)cos(\theta)\\ \end{array}$$

標本確率変数(Specimen random variable) / 標本確率変数の期待値と分散

$$\begin{array}{rcl} 母集団の確率変数&:&X\\ 標本確率変数&:&X_k (k=1,2,\dotsc ,n)\\ \end{array}$$ $$\begin{array}{rcl} \mu&=&E[X]\\ \sigma^2&=&V[X]\\ \end{array}$$ 標本確率変数$X_k$は母集団の確率変数$X$と同じ確率分布(母集団)に従うので，$X_k$の期待値・分散は$X$の期待値・分散と等しい． $$\begin{array}{rclcl} E[X_k]&=&E[X]&=&\mu\\ V[X_k]&=&V[X]&=&\sigma^2\\ \end{array}$$

標本確率変数(Specimen random variable) / 標本平均・標本分散・不偏分散

$n$個の標本(試行して得られたもの)に対して求めたものが，標本平均,標本分散，不偏分散となる

$$\begin{array}{rcl} 標本確率変数&:&X_k (k=1,2,\dotsc ,n)\\ \end{array}$$ $$\begin{array}{rclcl} \overline{X}&=&\displaystyle\frac{1}{n}\sum_{k=1}^{n} X_k &\dots&標本平均(sample \, mean)\\ s^2&=&\displaystyle\frac{1}{n}\sum_{k=1}^{n} (X_k - \overline{X})^2 &\dots&標本分散(sample \, variance)\\ \hat{\sigma}^2&=&\displaystyle\frac{1}{n-1}\sum_{k=1}^{n} (X_k - \overline{X})^2 &\dots&不偏分散(unbiased \, variance)\\ \end{array}$$

全事象をもとに求めたものが母平均,母分散となる

$$\begin{array}{rclcl} \mu&=&\displaystyle\frac{1}{N}\sum_{k=1}^{N} X_i &\dots&母平均(population \, mean)\\ \sigma^2&=&\displaystyle\frac{1}{N}\sum_{i=1}^{N} (X_i - \mu)^2 &\dots&母分散(population \, variance)\\ \end{array}$$

標本確率変数(Specimen random variable) / 標本平均の期待値

$$\begin{array}{rcl} 標本確率変数&:&X_k (k=1,2,\dotsc ,n)\\ 標本平均(sample \, mean)&:&\overline{X}=\displaystyle\frac{1}{n}\sum_{k=1}^{n} X_k\\ \end{array}$$ $$\begin{array}{rclcl} \mu&=&E[X]\\ E[X_k]&=&E[X]&=&\mu\\ 　　　　E[cX]&=&c E[X]\\ \end{array}$$ $$\begin{array}{rclcl} E[\overline{X}]&=&E[\displaystyle\frac{1}{n}\sum_{k=1}^{n} X_k]\\ &=&\displaystyle\frac{1}{n}E[\sum_{k=1}^{n} X_k]\\ &=&\displaystyle\frac{1}{n}\sum_{k=1}^{n} E[X_k]\\ &=&\displaystyle\frac{1}{n}\sum_{k=1}^{n} \mu\\ &=&\displaystyle\frac{1}{n}n\mu\\ &=&\displaystyle\mu\\ \end{array}$$

二項分布(binomial distribution)

$$B(n, p) = f_X(x) = \begin{cases} \displaystyle _nC_x\,p^x(1-p)^{n-x} & \quad x \in \left\{0,1,2, \dotsc ,n\right\}\\ \displaystyle　0 & \quad x \notin \left\{0,1,2, \dotsc ,n\right\} \end{cases}$$

二項分布の期待値

二項分布 $B(n, p)$

$$B(n, p) = f_X(x) = \begin{cases} \displaystyle _nC_x\,p^x(1-p)^{n-x} & \quad x \in \left\{0,1,2, \dotsc ,n\right\}\\ \displaystyle　0 & \quad x \notin \left\{0,1,2, \dotsc ,n\right\} \end{cases}$$

二項分布の期待値

$$\begin{array}{rcl} E[X]&=&\displaystyle \sum_{x=0}^{n}\left(x\right)\left( _nC_x\,p^x(1-p)^{n-x} \right)\\ &=& \displaystyle \sum_{x=0}^{n}\left(x\right)\left(\left( \frac{n!}{x!(n-x)!}\right)p^x(1-p)^{n-x}\right)\\ &=& \displaystyle \sum_{x=0}^{n} \frac{n!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}\dotso \frac{x}{x!}=\frac{1}{(x-1)!}\\ &=& \displaystyle \sum_{x=0}^{n} \frac{n(n-1)!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}\dotso n!=n(n-1)!\\ &=& \displaystyle \sum_{x=0}^{n} \frac{n(n-1)!}{(x-1)!(n-x)!}pp^{x-1}(1-p)^{n-x}\dotso p^x=pp^{x-1}\\ &=& \displaystyle np\sum_{x=0}^{n} \frac{(n-1)!}{(x-1)!(n-x)!}p^{x-1}(1-p)^{n-x}\\ &=& \displaystyle np\sum_{x=0}^{n} \frac{(n-1)!}{(x-1)!(n-1+1-x)!}p^{x-1}(1-p)^{n-1+1-x}\\ &=& \displaystyle np\sum_{x=0}^{n} \frac{(n-1)!}{(x-1)!(n-1-(x-1))!}p^{x-1}(1-p)^{n-1-(x-1)}\dotso 1-x=-(x-1)\\ &=& \displaystyle np\sum_{x'=0}^{n-1} \frac{(n-1)!}{x'!(n-1-x')!}p^{x'}(1-p)^{n-1-x'}\dotso x'=x-1\\ &=& \displaystyle np\,1 \dotso B(n-1, p)の総和は1に等しい．\\ &=& \displaystyle np\\ &=& \mu \dotso 母集団の期待値\\ \end{array}$$

二項分布の分散

二項分布 $B(n, p)$

$$B(n, p) = f_X(x) = \begin{cases} \displaystyle _nC_x\,p^x(1-p)^{n-x} & \quad x \in \left\{0,1,2, \dotsc ,n\right\}\\ \displaystyle　0 & \quad x \notin \left\{0,1,2, \dotsc ,n\right\} \end{cases}$$

二項分布の分散

$$\begin{array}{rcl} E[X(X-1)]&=&\displaystyle \sum_{x=0}^{n}\left(x\left(x-1\right)\right)\left( _nC_x\,p^x(1-p)^{n-x} \right)\\ &=& \displaystyle \sum_{x=0}^{n}\left(x\left(x-1\right)\right)\left(\left( \frac{n!}{x!(n-x)!}\right)p^x(1-p)^{n-x}\right)\\ &=& \displaystyle \sum_{x=0}^{n} \frac{n!}{(x-2)!(n-x)!}p^x(1-p)^{n-x}\dotso \frac{x^2}{x!}=\frac{1}{(x-2)!}\\ &=& \displaystyle \sum_{x=0}^{n} \frac{n(n-1)(n-2)!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}\dotso n!=n(n-1)(n-2)!\\ &=& \displaystyle \sum_{x=0}^{n} \frac{n(n-1)(n-2)!}{(x-1)!(n-x)!}p^2p^{x-2}(1-p)^{n-x}\dotso p^x=p^2p^{x-2}\\ &=& \displaystyle \sum_{x=0}^{n} \frac{n(n-1)(n-2)!}{(x-2)!(n-x))!}p^2 p^{x-2}(1-p)^{n-x}\\ &=& \displaystyle n(n-1)p^2\sum_{x=0}^{n} \frac{(n-2)!}{(x-2)!(n-2+2-x))!} p^{x-2}(1-p)^{n-2+2+x}\\ &=& \displaystyle n(n-1)p^2\sum_{x=0}^{n} \frac{(n-2)!}{(x-2)!(n-2-(x-2))!} p^{x-2}(1-p)^{n-2-(x-2)}\dotso 2-x=-(x-2)\\ &=& \displaystyle n(n-1)p^2 \sum_{x'=0}^{n-2} \frac{(n-2)!}{x'!(n-2-x')!}p^{x'}(1-p)^{n-2-x'}\dotso x'=x-2\\ &=& \displaystyle n(n-1)p^2\,1 \dotso B(n-2, p)の総和は1に等しい．\\ &=& \displaystyle n(n-1)p^2\\ V[X]&=&\displaystyle E[X^2]-E[X]^2\\ &=&\displaystyle E[X^2]-E[X]+E[X]-E[X]^2\\ &=&\displaystyle E[X^2-X]+E[X]-E[X]^2\dotso E[X] \pm E[Y]=E[X \pm Y]\\ &=&\displaystyle E[X(X-1)]+E[X]-E[X]^2\\ &=&\displaystyle n(n-1)p^2+np-(np)^2\\ &=&\displaystyle n^2p^2-np^2+np-n^2p^2\\ &=&\displaystyle -np^2+np\\ &=&\displaystyle np(-p+1)\\ &=&\displaystyle np(1-p)\\ &=& \sigma^2 \dotso 母集団の分散\\ \end{array}$$

ド・モアブルの定理(de Moivre's theorem)と三角凾数(trigonometric function)の3倍角の公式

ド・モアブルの定理

$$\displaystyle (\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta)$$

3倍角の公式(Triple-angle formulae)

$$\begin{array}{rcl} \displaystyle (\cos\theta+i\sin\theta)^3 &=& \displaystyle \cos^3\theta+3\cos^2\theta(i\sin\theta)+3\cos\theta(i\sin\theta)^2+(i\sin\theta)^3\\ &=& \displaystyle \cos^3\theta+i3\cos^2\theta\sin\theta-3\cos\theta\sin^2\theta-i\sin^3\theta\\ &=& \displaystyle \left(\cos^3\theta-3\cos\theta\sin^2\theta\right)+i\left(3\cos^2\theta\sin\theta-\sin^3\theta\right)\\ &=& \displaystyle \cos(3\theta)+i\sin(3\theta)\,\dotso\,これがド・モアブルの定理の結果と等しい． \end{array}$$ $$\begin{array}{rcl} \displaystyle \cos(3\theta)&=& \cos^3\theta-3\cos\theta\sin^2\theta\\ &=& \cos^3\theta-3\cos\theta(1-\cos^2\theta)\\ &=& \cos^3\theta-3\cos\theta+3\cos^3\theta\\ &=& 4\cos^3\theta-3\cos\theta\\ \displaystyle \sin(3\theta)&=& 3\cos^2\theta\sin\theta-\sin^3\theta\\ &=& 3\sin\theta(1-\sin^2\theta)-\sin^3\theta\\ &=& 3\sin\theta-3\sin^3\theta-\sin^3\theta\\ &=& 3\sin\theta-4\sin^3\theta\\ \end{array}$$

4倍角の場合

$$\begin{array}{rcl} \displaystyle (\cos\theta+i\sin\theta)^4 &=& \displaystyle \cos^4\theta+4\cos^3\theta(i\sin\theta)+6\cos^2\theta(i\sin\theta)^2+4\cos\theta(i\sin\theta)^3+(i\sin\theta)^4\\ &=& \displaystyle \cos^4\theta+i4\cos^3\theta\sin\theta-6\cos^2\theta\sin^2\theta-i4\cos\theta\sin^3\theta+\sin^4\theta\\ &=& \displaystyle \left(\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta\right)+i\left(4\cos^3\theta\sin\theta-4\cos\theta\sin^3\theta\right)\\ &=& \displaystyle \cos(4\theta)+i\sin(4\theta)\,\dotso\,これがド・モアブルの定理の結果と等しい． \end{array}$$ $$\begin{array}{rcl} \displaystyle \cos(4\theta)&=&\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta\\ &=&\cos^4\theta-6\cos^2\theta(1-\cos^2\theta)+(1-\cos^2\theta)^2\\ &=&\cos^4\theta-6\cos^2\theta+6\cos^4\theta+1-2\cos^2\theta+\cos^4\theta\\ &=&\cos^4\theta+6\cos^4\theta+\cos^4\theta-6\cos^2\theta-2\cos^2\theta+1\\ &=&8\cos^4\theta-8\cos^2\theta+1\\ \displaystyle \sin(4\theta)&=&4\cos^3\theta\sin\theta-4\cos\theta\sin^3\theta\\ &=&\cos\theta(4\cos^2\theta\sin\theta-4\sin^3\theta)\\ &=&\cos\theta(4\sin\theta(1-\sin^2\theta)-4\sin^3\theta)\\ &=&\cos\theta(4\sin\theta-4\sin^3\theta-4\sin^3\theta)\\ &=&\cos\theta(4\sin\theta-8\sin^2\theta)\\ \end{array}$$

階乗(factorial)の対数(logarithm)の微分(differential calculus)

スターリングの近似(Stirling's approximation)

$$\begin{array}{rcl} \displaystyle n! & \simeq & \sqrt{2\pi n}\left(\dfrac{n}{e}\right)^n\\ \end{array}$$

階乗の対数

$$\begin{array}{rcl} \displaystyle \log(n!)&\simeq& \log\left( \sqrt{2\pi n}\left(\dfrac{n}{e}\right)^n \right)\\ &\simeq& \log\left( \sqrt{2\pi n}\,n^n\,e^{-n} \right)\\ \displaystyle &\simeq& \log\left( \sqrt{2\pi} \right) \displaystyle + \log\left( \sqrt{n} \right) \displaystyle + \log\left(n^n \right) \displaystyle + \log\left(e^{-n} \right)\\ \displaystyle &\simeq& \log\left( \sqrt{2\pi} \right) \displaystyle + \frac{1}{2}\log\left( n \right) \displaystyle + n\log\left(n \right) \displaystyle - n \log\left(e \right)\\ \displaystyle &\simeq& \log\left( \sqrt{2\pi} \right) \displaystyle + \frac{1}{2}\log\left( n \right) \displaystyle + n\log\left(n \right) \displaystyle - n\\ \end{array}$$

階乗の対数の微分

$$\begin{array}{rcl} \displaystyle \left(\log(n!)\right)' \displaystyle &\simeq& \left(\log\left( \sqrt{2\pi} \right)\right)' \displaystyle + \left(\frac{1}{2}\log\left( n \right)\right)' \displaystyle + \left(n\log\left(n \right)\right)' \displaystyle - \left(n\right)'\\ \displaystyle &\simeq& \left(\log\left( \sqrt{2\pi} \right)\right)' \displaystyle + \left(\frac{1}{2}\log\left( n \right)\right)' \displaystyle + \left(n(\log\left(n \right))'+(n)'\log\left(n \right)\right) \displaystyle - \left(n\right)'\\ \displaystyle &\simeq& 0 \displaystyle + \frac{1}{2}\left(\frac{1}{n}\right) \displaystyle + \left(n\left(\frac{1}{n}\right)+1\,\log\left(n \right)\right) \displaystyle - 1\\ \displaystyle &\simeq& \frac{1}{2n}+1+ \log\left(n \right)-1\\ \displaystyle &\simeq& \frac{1}{2n}+\log\left(n \right) \end{array}$$

$n$が十分に大きい場合

$$\begin{array}{rcl} \displaystyle \left(\log(n!)\right)' \displaystyle &\simeq& \frac{1}{2n}+\log\left(n \right)\\ \displaystyle &\simeq& 0+\log\left(n \right)\,\dotso\,nが十分い大きい \frac{1}{2n} \to 0\\ \displaystyle &\simeq& \log\left(n \right) \end{array}$$

二項分布(binomial distribution)からポアソン分布(Poisson distribution)を導く

二項分布 $B(n, p)$

$$B(n, p) = f_X(x) = \begin{cases} \displaystyle _nC_x\,p^x(1-p)^{n-x} & \quad x \in \left\{0,1,2, \dotsc ,n\right\}\\ \displaystyle　0 & \quad x \notin \left\{0,1,2, \dotsc ,n\right\} \end{cases}$$

$_nC_x$の展開と変形

$$\begin{array}{rcl} \displaystyle _nC_x&=&\displaystyle \frac{n!}{x!(n-x)!}\\ &=&\displaystyle \frac{1}{x!}\frac{n!}{(n-x)!}\\ &=&\displaystyle \frac{1}{x!}n(n-1)(n-2)\dotsm(n-x+1)\\ &=&\displaystyle \frac{1}{x!}\{\frac{n\,n}{n}\frac{n\,(n-1)}{n}\frac{n\,(n-2)}{n}\dotsm\frac{n\,(n-x+1)}{n}\}\\ &=&\displaystyle \frac{1}{x!}n^x\{\frac{n}{n}\frac{(n-1)}{n}\frac{(n-2)}{n}\dotsm\frac{(n-x+1)}{n}\}\\ &=&\displaystyle \frac{n^x}{x!}\{\frac{n}{n}(\frac{n}{n}-\frac{1}{n})(\frac{n}{n}-\frac{2}{n})\dotsm(\frac{n}{n}-\frac{x-1}{n})\}\\ &=&\displaystyle \frac{n^x}{x!}\{1(1-\frac{1}{n})(1-\frac{2}{n})\dotsm(1-\frac{x-1}{n})\}\\ &=&\displaystyle \frac{n^x}{x!}\{(1-\frac{1}{n})(1-\frac{2}{n})\dotsm(1-\frac{x-1}{n})\}\\ \end{array}$$

$(1-p)^{n-x}$の展開と変形

$$\begin{array}{rcl} \displaystyle (1-p)^{n-x}&=&\displaystyle (1-p)^n(1-p)^{-x}\\ &=&\displaystyle \left[\left[\{1+(-p)\}^{\frac{1}{-p}}\right]^{-p}\right]^n(1-p)^{-x} \,\dotso\,a=(a^\frac{1}{x})^{x}\\ &=&\displaystyle \{\left(\mathrm{e}\right)^{-p}\}^n(1-p)^{-x} \,\dotso\,\left(1+x\right)^{\frac{1}{x}}=\left(1+\frac{1}{y}\right)^{y}=\mathrm{e}\\ &=&\displaystyle \mathrm{e}^{-np}(1-p)^{-x}\,\dotso\,\left(a^x\right)^y=a^{xy}\\ \end{array}$$

展開と変形の結果を当てはめる

$$\begin{array}{rcl} \displaystyle _nC_x\,p^x(1-p)^{n-x}&=&\displaystyle \frac{n^x}{x!}\{(1-\frac{1}{n})(1-\frac{2}{n})\dotsm(1-\frac{k-1}{n})\}p^x\mathrm{e}^{-np}(1-p)^{-x}\\ &=&\displaystyle \frac{1}{x!}\{(1-\frac{1}{n})(1-\frac{2}{n})\dotsm(1-\frac{k-1}{n})\}n^{x}p^x\mathrm{e}^{-np}(1-p)^{-x}\\ &=&\displaystyle \frac{1}{x!}\{(1-\frac{1}{n})(1-\frac{2}{n})\dotsm(1-\frac{k-1}{n})\}(np)^{x}\mathrm{e}^{-np}(1-p)^{-x}\\ \end{array}$$

$\lambda=np$，$\lambda$を一定に$n$を十分大きくとると, $p$は極めて小さくなる

$$\begin{array}{rcl} \displaystyle f_X(x)&=&\displaystyle \frac{1}{x!}\{(1-\frac{1}{n})(1-\frac{2}{n})\dotsm(1-\frac{k-1}{n})\}(np)^{x}\mathrm{e}^{-np}(1-p)^{-x}\\ &=&\displaystyle \frac{1}{x!}\{(1-\frac{1}{n})(1-\frac{2}{n})\dotsm(1-\frac{k-1}{n})\}(\lambda)^{x}\mathrm{e}^{-\lambda}(1-p)^{-x}\,\dotso\,\lambda=np\\ &=&\displaystyle \frac{1}{x!}\{(1)(1)\dotsm(1)\}(\lambda)^{x}\mathrm{e}^{-\lambda}(1)^{-x}\,\dotso\,\frac{C}{n}\to 0, p\to 0\\ &=&\displaystyle \frac{1}{x!}\lambda^{x}\mathrm{e}^{-\lambda}\\ &=&\displaystyle \frac{\lambda^{x}}{x!}\mathrm{e}^{-\lambda}\\ &=&\displaystyle Po(\lambda)\\ \end{array}$$

積率母凾数(moment-generating function)を考える

$$\begin{array}{rcl} \displaystyle M_X(t)&\equiv&\displaystyle E[\mathrm{e}^{tX}]\\ &=&\displaystyle \sum_{x=0}^{\infty}(\mathrm{e}^{tx})\frac{\lambda^{x}}{x!}\mathrm{e}^{-\lambda}\\ &=&\displaystyle \mathrm{e}^{-\lambda}\sum_{x=0}^{\infty}(\mathrm{e}^{tx})\frac{\lambda^{x}}{x!}\\ &=&\displaystyle \mathrm{e}^{-\lambda}\sum_{x=0}^{\infty}\frac{\mathrm{e}^{tx}\lambda^{x}}{x!}\\ &=&\displaystyle \mathrm{e}^{-\lambda}\sum_{x=0}^{\infty}\frac{(\mathrm{e}^{t}\lambda)^{x}}{x!}\\ &=&\displaystyle \mathrm{e}^{-\lambda}\mathrm{e}^{\mathrm{e}^{t}\lambda} \,\dotso\,\href{https://shikitenkai.blogspot.com/2019/07/blog-post.html}{\sum_{x=0}^{\infty}\frac{a^{x}}{x!}=\mathrm{e}^a}\\ &=&\displaystyle \mathrm{e}^{\mathrm{e}^{t}\lambda-\lambda}=\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)} \end{array}$$

期待値，分散

$$\begin{array}{rcl} \displaystyle M_X^{(m)}(0)&\equiv&\frac{ \mathrm{d}^m }{ \mathrm{d}^m t } M_x(t)|_{t=0}\\ &=&\displaystyle E[X^m\mathrm{e}^{tX}]|_{t=0}\\ &=&\displaystyle E[X^m]\\ \end{array}$$ $$\begin{array}{rcl} \displaystyle E[X]&=&\displaystyle M_X^1(0)\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} t }(\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)}) \right\}|_{t=0}\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} s }(\mathrm{e}^{s})\frac{ \mathrm{d}s}{ \mathrm{d}t} \right\}|_{t=0} \,\dotso\,s=\lambda(\mathrm{e}^{t}-1),\frac{ \mathrm{d}s}{ \mathrm{d}t}=\lambda\mathrm{e}^{t}\\ &=&\displaystyle \left\{ (\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)})(\lambda\mathrm{e}^{t}) \right\}|_{t=0} \,\dotso\,\frac{ \mathrm{d} }{ \mathrm{d} x }\mathrm{e}^{x}=\mathrm{e}^{x}\\ &=&\displaystyle \left\{ \lambda(\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)+t}) \right\}|_{t=0}\\ &=&\displaystyle \lambda(\mathrm{e}^{\lambda(\mathrm{e}^{0}-1)+0})\\ &=&\displaystyle \lambda\mathrm{e}^0 \,\dotso\,a^0=1\\ &=&\lambda \,\dotso\,a^0=1\\ \end{array}$$ $$\begin{array}{rcl} \displaystyle E[X^2]&=&\displaystyle M_X^2(0)\\ &=&\displaystyle \left\{ \frac{ \mathrm{d}^2 }{ \mathrm{d} t^2 }(\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)}) \right\}|_{t=0}\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} t } \lambda(\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)+t}) \right\}|_{t=0} \,\dotso\,E[X]の展開から．\\ &=&\displaystyle \left\{ \frac{ \mathrm{d} }{ \mathrm{d} s }(\lambda\mathrm{e}^{s})\frac{ \mathrm{d}s}{ \mathrm{d}t} \right\}|_{t=0} \,\dotso\,s=\lambda(\mathrm{e}^{t}-1)+t,\frac{ \mathrm{d}s}{ \mathrm{d}t}=\lambda\mathrm{e}^{t}+1\\ &=&\displaystyle \left\{ (\lambda\mathrm{e}^{\lambda(\mathrm{e}^{t}-1)+1})(\lambda\mathrm{e}^{t}+1) \right\}|_{t=0} \,\dotso\,\frac{ \mathrm{d} }{ \mathrm{d} x }C\mathrm{e}^{x}=C\mathrm{e}^{x}\\ &=&\displaystyle (\lambda\mathrm{e}^{\lambda(\mathrm{e}^{0}-1)+1})(\lambda\mathrm{e}^{0}+1)\\ &=&\displaystyle \lambda\mathrm{e}^0(\lambda+1) \,\dotso\,a^0=1\\ &=&\lambda(\lambda+1) \,\dotso\,a^0=1\\ \end{array}$$ $$\begin{array}{rcl} \displaystyle V[X]&=&\displaystyle E[X^2]-E[X]^2\\ &=&\lambda(\lambda+1)-\lambda^2\\ &=&\lambda^2+\lambda-\lambda^2\\ &=&\lambda\\ \end{array}$$

標本確率変数(Specimen random variable) / 標本平均の分散

$$\begin{array}{rcl} 母集団の確率変数&:&X\\ 標本確率変数&:&X_k (k=1,2,\dotso ,n)\\ 標本平均(sample \, mean)&:&\overline{X}=\displaystyle\frac{1}{n}\sum_{k=1}^{n} X_k\\ \end{array}$$ $$\begin{array}{rcl} \sigma^2&=&V[X]\,\dotso\,母集団の分散\\ \end{array}$$ $$\begin{array}{rclcl} V[\overline{X}]&=&V[\displaystyle\frac{1}{n}\sum_{k=1}^{n} X_k]\\ &=&\displaystyle\left( \frac{1}{n} \right)^2 V[\sum_{k=1}^{n} X_k] \,\dotso\,\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-variance.html}{V[cX]=c^2V[X]}\\ &=&\displaystyle\frac{1}{n^2}\sum_{k=1}^{n} V[X_k] \;\cdots\;X_k同士が互いに独立\left(\mathrm{Con}\left(X_i, X_j\right)=0\right)であるならV[\sum_{k=1}^{n} X_k]=\sum_{k=1}^{n} V[X_k] \\&=&\displaystyle\frac{1}{n^2}\sum_{k=1}^{n} \sigma^2 \,\dotso\,\href{https://shikitenkai.blogspot.com/2019/06/specimen-random-variable.html}{V[X_k]=V[X]=\sigma^2}\\ &=&\displaystyle\frac{1}{n^2}n\sigma^2\\ &=&\displaystyle\frac{\sigma^2}{n}\\ \end{array}$$ よって標本数$n$を増やすことで$\mathrm{V}[\overline{X}]$は$0$に近づいていく．

連続型確率変数(continuous random variable) / 期待値(expected value)

$$\begin{array}{rcl} 母集団(population)の確率変数&:&X,Y\\ 確率密度凾数(probability \, density \, function, \, PDF)&:&\displaystyle\int_{a}^{b}f_X(x)\mathrm{d}x=P(a \leq x \leq b)\\ &&\displaystyle\int_{-\infty}^{\infty}f_X(x)\mathrm{d}x=1\\ 累積分布凾数(cumulative \, distribution \, function, \, CDF)&:&\displaystyle F_X(x)=\int_{-\infty}^{x}f_X(t)\mathrm{d}t\\ &&\displaystyle f_X(x)=\frac{\mathrm{d}}{\mathrm{d}x}F_X(x)\\ 同時確率密度凾数(joint\,probability\,density\,function)&:&\displaystyle\int_{a}^{b}\int_{c}^{d}f_{XY}(x, y)\mathrm{d}x\mathrm{d}y=P(a \leq x \leq b,\,c \leq y \leq d)\\ &&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f_{XY}(x, y)\mathrm{d}x\mathrm{d}y=1\\ 周辺確率密度凾数(marginal\, probability\, density\, function)&:&\displaystyle f_X(x)=\int_{-\infty}^{\infty}f_{XY}(x,y)\mathrm{d}y \end{array}$$ $$\begin{array}{rcl} E[g(X)]&\equiv&\displaystyle\int_{-\infty}^{\infty} g(x) f_X(x) \mathrm{d}x\\ E[X]&=&\displaystyle\int_{-\infty}^{\infty} (x) f_X(x) \mathrm{d}x\\ E[cX] &=&\displaystyle\int_{-\infty}^{\infty} (c x) f_X(x) \mathrm{d}x\,\dotso\,cは定数\\ &=&c\displaystyle\int_{-\infty}^{\infty} x f_X(x) \mathrm{d}x\\ &=&c E[X] \\ E[X \pm t]&=&\displaystyle\int_{-\infty}^{\infty} (x \pm t) f_X(x) \mathrm{d}x\,\dotso\,tは定数\\ &=&\displaystyle\int_{-\infty}^{\infty} (x f_X(x) \pm t f_X(x)) \mathrm{d}x\\ &=&\displaystyle\int_{-\infty}^{\infty} x f_X(x) \pm \displaystyle\int_{-\infty}^{\infty} t f_X(x) \mathrm{d}x\\ &=&\displaystyle\int_{-\infty}^{\infty} x f_X(x) \pm t \displaystyle\int_{-\infty}^{\infty} f_X(x) \mathrm{d}x\\ &=&E[X] \pm t\\ E[X \pm Y]&=&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} (x \pm y) f_{XY}(x, y) \mathrm{d}x\mathrm{d}y\\ &=&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} (x f_{XY}(x, y) \pm y f_{XY}(x, y)) \mathrm{d}x\mathrm{d}y\\ &=&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} x f_{XY}(x, y) \mathrm{d}x\mathrm{d}y \pm \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} y f_{XY}(x, y)\mathrm{d}x\mathrm{d}y\\ &=&\displaystyle\int_{-\infty}^{\infty} x \int_{-\infty}^{\infty} f_{XY}(x, y) \mathrm{d}y\,\mathrm{d}x \pm \displaystyle\int_{-\infty}^{\infty} y \int_{-\infty}^{\infty} f_{XY}(x, y)\mathrm{d}x\,\mathrm{d}y\\ &=&\displaystyle\int_{-\infty}^{\infty} x f_{X}(x) \mathrm{d}x \pm \displaystyle\int_{-\infty}^{\infty} y f_{Y}(y) \mathrm{d}y\,\dotso\,周辺確率密度凾数を適用\\ &=& E[X] \pm E[Y] \\ \end{array}$$

X,Yが独立の場合

$$\begin{array}{rcl} E[XY]&=&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(x y)f_{XY}(x, y)\mathrm{d}x\mathrm{d}y\\ &=&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}xyf_{X}(x)f_{Y}(y)\mathrm{d}x\mathrm{d}y \,\dotso\,X,Yが独立\,f_{XY}(x, y)=f_{X}(x)f_{Y}(y)\\ &=&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}xf_{X}(x)yf_{Y}(y)\mathrm{d}x\mathrm{d}y\\ &=&\displaystyle\left(\int_{-\infty}^{\infty}xf_{X}(x)\mathrm{d}x\right)\left(\int_{-\infty}^{\infty}yf_{Y}(y)\mathrm{d}y\right) \,\dotso\,\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x)g(y)\mathrm{d}x\mathrm{d}y=\left(\int_{-\infty}^{\infty}f(x)\mathrm{d}x\right)\left(\int_{-\infty}^{\infty}g(y)\mathrm{d}y\right)\\ &=&E[X]E[Y] \end{array}$$

連続型確率変数(continuous random variable) / 分散(variance)

$$\begin{array}{rcl} 母集団(population)の確率変数&:&X,Y\\ 確率密度凾数(probability \, density \, function, \, PDF)&:&\displaystyle\int_{a}^{b}f_X(x)\mathrm{d}x=P(a \leq x \leq b)\\ &&\displaystyle\int_{-\infty}^{\infty}f_X(x)\mathrm{d}x=1\\ 累積分布凾数(cumulative \, distribution \, function, \, CDF)&:&\displaystyle F_X(x)=\int_{-\infty}^{x}f_X(t)\mathrm{d}t\\ &&\displaystyle f_X(x)=\frac{\mathrm{d}}{\mathrm{d}x}F_X(x)\\ 同時確率密度凾数(joint\,probability\,density\,function)&:&\displaystyle\int_{a}^{b}\int_{c}^{d}f_{XY}(x, y)\mathrm{d}x\mathrm{d}y=P(a \leq x \leq b,\,c \leq y \leq d)\\ &&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f_{XY}(x, y)\mathrm{d}x\mathrm{d}y=1\\ 周辺確率密度凾数(marginal\, probability\, density\, function)&:&\displaystyle f_X(x)=\int_{-\infty}^{\infty}f_{XY}(x,y)\mathrm{d}y \end{array}$$ $$\begin{array}{rcl} V[X]&\equiv&E[(X-E[X])^2]\\ &=&\displaystyle\int_{-\infty}^{\infty} (x-E[X])^2 f_X(x) \mathrm{d}x\\ &=&\displaystyle\int_{-\infty}^{\infty} (x^2-2E[X]x+E[X]^2) f_X(x) \mathrm{d}x\\ &=&\displaystyle\int_{-\infty}^{\infty} (x^2f_X(x)-2E[X]xf_X(x)+E[X]^2f_X(x)) \mathrm{d}x\\ &=&\displaystyle\int_{-\infty}^{\infty} (x^2) f_X(x)\mathrm{d}x -2E[X]\displaystyle\int_{-\infty}^{\infty} (x) f_X(x)\mathrm{d}x +E[X]^2\displaystyle\int_{-\infty}^{\infty} f_X(x)\mathrm{d}x\\ &=&E[X^2] -2E[X]E[X] +E[X]^2 1\\ &=&E[X^2] -2E[X]^2 +E[X]^2\\ &=&E[X^2]-E[X]^2\\ V[cX] &=&E[(cX-E[cX])^2]\,\dotso\,cは定数\\ &=&\displaystyle\int_{-\infty}^{\infty} (c x - E[c X])^2 f_X(x) \mathrm{d}x\\ &=&\displaystyle\int_{-\infty}^{\infty} (c x - c E[X])^2 f_X(x) \mathrm{d}x\\ &=&\displaystyle\int_{-\infty}^{\infty} (c (x - E[X]))^2 f_X(x) \mathrm{d}x\\ &=&\displaystyle\int_{-\infty}^{\infty} c^2 (x-E[X])^2 f_X(x) \mathrm{d}x\\ &=&c^2\displaystyle\int_{-\infty}^{\infty} (x-E[X])^2 f_X(x) \mathrm{d}x\\ &=&c^2 V[X]\\ V[X \pm t]&=&E[((X \pm t)-E[X \pm t])^2]\,\dotso\,tは定数\\ &=&\displaystyle\int_{-\infty}^{\infty} ((x \pm t)-E[X \pm t])^2 f_X(x) \mathrm{d}x\\ &=&\displaystyle\int_{-\infty}^{\infty} ((x \pm t)-(E[X] \pm t))^2 f_X(x) \mathrm{d}x\\ &=&\displaystyle\int_{-\infty}^{\infty} (x \pm t - E[X] \mp t))^2 f_X(x) \mathrm{d}x\\ &=&\displaystyle\int_{-\infty}^{\infty} (x-E[X])^2 f_X(x) \mathrm{d}x\\ &=&V[X]\\ V[X \pm Y]&=&E[((X \pm Y)-E[X \pm Y])^2]\\ &=&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} ((x \pm y)-E[X \pm Y])^2 f_{XY}(x, y) \mathrm{d}x\mathrm{d}y\\ &=&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} ((x \pm y)-(E[X] \pm E[Y]))^2 f_{XY}(x, y) \mathrm{d}x\mathrm{d}y\\ &=&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} ((x - E[X]) \pm (y - E[Y]))^2 f_{XY}(x, y) \mathrm{d}x\mathrm{d}y\\ &=&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} ((x - E[X])^2 \pm 2(x - E[X])(y - E[Y]) +(y - E[Y])^2) f_{XY}(x, y) \mathrm{d}x\mathrm{d}y\\ &=&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} ((x - E[X])^2 f_{XY}(x, y) \pm 2(x - E[X])(y - E[Y]) f_{XY}(x, y) +(y - E[Y])^2 f_{XY}(x, y)) \mathrm{d}x\mathrm{d}y\\ &=&\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} (x - E[X])^2 f_{XY}(x, y) \mathrm{d}x\mathrm{d}y \pm 2\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(x - E[X])(y - E[Y]) f_{XY}(x, y) \mathrm{d}x\mathrm{d}y + \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} (y - E[Y])^2 f_{XY}(x, y) \mathrm{d}x\mathrm{d}y\\ &=&\displaystyle\int_{-\infty}^{\infty} (x - E[X])^2 \int_{-\infty}^{\infty} f_{XY}(x, y) \mathrm{d}y\, \mathrm{d}x \pm 2\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(x - E[X])(y - E[Y]) f_{XY}(x, y) \mathrm{d}x\mathrm{d}y + \displaystyle\int_{-\infty}^{\infty} (y - E[Y])^2 \int_{-\infty}^{\infty} f_{XY}(x, y) \mathrm{d}x\,\mathrm{d}y\\ &=&\displaystyle\int_{-\infty}^{\infty} (x - E[X])^2 f_{X}(x) \mathrm{d}x \pm 2\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(x - E[X])(y - E[Y]) f_{XY}(x, y) \mathrm{d}x\mathrm{d}y + \displaystyle\int_{-\infty}^{\infty} (y - E[Y])^2 f_{Y}(y) \mathrm{d}y \,\dotso\,周辺確率密度凾数を適用\\ &=&V[X] \pm 2Cov[X,Y] +V[Y]\,\dotso\,Cov[X,Y]=\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(x - E[X])(y - E[Y]) f_{XY}(x, y) \mathrm{d}x\mathrm{d}y\\ \end{array}$$

X,Yが独立の場合

$$\begin{array}{rcl} V[X \pm Y]&=&\displaystyle V[X]+V[Y]\,\dotso\,X,Yが独立なのでCov[X,Y]=0\\ V[XY]&=&\displaystyle E[(XY)^2]-E[XY]^2\\ &=&\displaystyle E[X^2Y^2]-E[XY]^2\,\dotso\,(ab)^2=a^2b^2\\ &=&\displaystyle E[X^2]E[Y^2]-(E[X]E[Y])^2\,\dotso\,X,Yが独立なのでE[XY]=E[X]E[Y]\\ &=&\displaystyle E[X^2]E[Y^2]-E[X]^2E[Y]^2\,\dotso\,(ab)^2=a^2b^2\\ &=&\displaystyle (V[X]+E[X]^2)(V[Y]+E[Y]^2)-E[X]^2E[Y]^2\,\dotso\,V[X]=E[X^2]-E[X]^2よりE[X^2]=V[X]+E[X]^2\\ &=&\displaystyle V[X]V[Y]+V[X]E[Y]^2+V[Y]E[X]^2+E[X]^2E[Y]^2-E[X]^2E[Y]^2\\ &=&\displaystyle V[X]V[Y]+V[X]E[Y]^2+V[Y]E[X]^2\,\dotso\,X,Yが独立でもV[X]V[Y]とはならない\\ \end{array}$$