離散型確率変数(discrete random variable) の一様分布(uniform distribution)の分散(variance)
$$\begin{array}{rcl}
\displaystyle M_X^{(m)}(0)&\equiv&\frac{ \mathrm{d}^m }{ \mathrm{d}^m t } M_X(t)|_{t=0}\\
&=&\displaystyle E[X^m\mathrm{e}^{tX}]|_{t=0}\\
&=&\displaystyle E[X^m]\\
\end{array}$$
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)
$$f_X(x) =
\begin{cases}
\displaystyle \frac{1}{n} & \quad x \in \left\{1,2, \dots ,n\right\}\\
\displaystyle 0 & \quad x \notin \left\{1,2, \dots ,n\right\}
\end{cases}
$$
$$\begin{array}{rcl}
\displaystyle M_X(t)&\equiv&\displaystyle E[\mathrm{e}^{tX}]\\
&=&\displaystyle \sum_{x=1}^{n}\mathrm{e}^{tx}\left(\frac{1}{n}\right)\\
&=&\displaystyle \frac{1}{n} \sum_{x=1}^{n}\mathrm{e}^{tx}\\
&=&\displaystyle \frac{1}{n} \left(\mathrm{e}^t+\mathrm{e}^{2t}+\dotsb+\mathrm{e}^{nt}\right)\\
&=&\displaystyle \frac{1}{n} \frac{\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{\mathrm{e}^{t}-1}
\,\dotso\,a+ar+ar^2+\dotsb+ar^{n-1}=\frac{a(1-r^n)}{1-r}=\frac{a(r^n-1)}{r-1}\;(r \neq 1)\\
&=&\displaystyle \frac{\mathrm{e}^{t}(\mathrm{e}^{nt}-1)}{n(\mathrm{e}^{t}-1)}\\
\end{array}$$
離散型確率変数(discrete random variable) / 分散(variance)
$$\begin{array}{rcl}
母集団(population)の確率変数&:&X,Y\\
確率質量凾数(probability\,mass\,function,\,PMF)&:&f_X(x)=P(X=x_i)\\
&&\displaystyle\sum_{i=1}^{\infty} f_X(x_i)=1\\
累積分布凾数(cumulative\,distribution\,function,\,CDF)&:&\displaystyle F_X(x)=\sum_{i:x_i \leq x} f_X(x_i)\\
同時確率質量凾数(joint\,probability\,mass\,function)&:&f_X(x, y)=P(X=x_i,\,Y=y_j)\\
&&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} f_{XY}(x_i, y_j)=1\\
周辺確率質量凾数(marginal\, probability\, mass\, function)&:&\displaystyle f_X(x)=\sum_{j=1}^{\infty} f_{XY}(x, y_i)\\
\end{array}$$
$$\begin{array}{rcl}
V[X]&\equiv&E[(X-E[X])^2]\\
&=&\displaystyle\sum_{i=1}^{\infty} (x_i-E[X])^2 f_X(x_i) \\
&=&\displaystyle\sum_{i=1}^{\infty} (x_i^2-2E[X]x_i+E[X]^2) f_X(x_i) \\
&=&\displaystyle\sum_{i=1}^{\infty} (x_i^2f_X(x_i)-2E[X]x_if_X(x_i)+E[X]^2f_X(x_i)) \\
&=&\displaystyle\sum_{i=1}^{\infty} (x_i^2) f_X(x_i)
-2E[X]\displaystyle\sum_{i=1}^{\infty} (x_i) f_X(x_i)
+E[X]^2\displaystyle\sum_{i=1}^{\infty} f_X(x_i)\\
&=&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\sum_{i=1}^{\infty} (c x_i - E[c X])^2 f_X(x_i) \\
&=&\displaystyle\sum_{i=1}^{\infty} (c x_i - c E[X])^2 f_X(x_i) \\
&=&\displaystyle\sum_{i=1}^{\infty} (c (x_i - E[X]))^2 f_X(x_i) \\
&=&\displaystyle\sum_{i=1}^{\infty} c^2 (x_i-E[X])^2 f_X(x_i) \\
&=&c^2\displaystyle\sum_{i=1}^{\infty} (x_i-E[X])^2 f_X(x_i) \\
&=&c^2 V[X] \\
V[X \pm t]&=&E[((X \pm t)-E[X \pm t])^2]\,\dotso\,tは定数\\
&=&\displaystyle\sum_{i=1}^{\infty} ((x_i \pm t)-E[X \pm t])^2 f_X(x_i) \\
&=&\displaystyle\sum_{i=1}^{\infty} ((x_i \pm t)-(E[X] \pm t))^2 f_X(x_i) \\
&=&\displaystyle\sum_{i=1}^{\infty} (x_i \pm t - E[X] \mp t))^2 f_X(x_i) \\
&=&\displaystyle\sum_{i=1}^{\infty} (x_i-E[X])^2 f_X(x_i) \\
&=&V[X] \\
V[X \pm Y]&=&E[((X \pm Y)-E[X \pm Y])^2]\\
&=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} ((x_i \pm y_j)-E[X \pm Y])^2 f_{XY}(x_i, y_j) \\
&=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} ((x_i \pm y_j)-(E[X] \pm E[Y]))^2 f_{XY}(x_i, y_j) \\
&=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} ((x_i - E[X]) \pm (y_j - E[Y]))^2 f_{XY}(x_i, y_j) \\
&=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} ((x_i - E[X])^2 \pm 2(x_i - E[X])(y_j - E[Y]) +(y_j - E[Y])^2) f_{XY}(x_i, y_j) \\
&=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} ((x_i - E[X])^2 f_{XY}(x_i, y_j) \pm 2(x_i - E[X])(y_j - E[Y]) f_{XY}(x_i, y_j) +(y_j - E[Y])^2 f_{XY}(x_i, y_j)) \\
&=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} (x_i - E[X])^2 f_{XY}(x_i, y_j)
\pm 2\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}(x_i - E[X])(y_j - E[Y]) f_{XY}(x_i, y_j)
+ \displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} (y_j - E[Y])^2 f_{XY}(x_i, y_j) \\
&=&\displaystyle\sum_{i=1}^{\infty}(x_i - E[X])^2\sum_{j=1}^{\infty}f_{XY}(x_i, y_j)
\pm 2\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}(x_i - E[X])(y_j - E[Y]) f_{XY}(x_i, y_j)
+ \displaystyle\sum_{j=1}^{\infty} (y_j - E[Y])^2 \sum_{i=1}^{\infty} f_{XY}(x_i, y_j) \\
&=&\displaystyle\sum_{i=1}^{\infty} (x_i - E[X])^2 f_{X}(x_i)
\pm 2\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}(x_i - E[X])(y_j - E[Y]) f_{XY}(x_i, y_j)
+ \displaystyle\sum_{j=1}^{\infty} (y_j - E[Y])^2 f_{Y}(y_j)\,\dotso\,周辺確率質量凾数を適用 \\
&=&V[X] \pm 2Cov[X,Y] +V[Y]\,\dotso\,Cov[X,Y]=\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}(x_i - E[X])(y_j - E[Y]) f_{XY}(x_i, y_j)\\
\end{array}$$
X,Yが独立の場合
$$\begin{array}{rcl} V[X \pm Y]&=&\displaystyle V[X]+V[Y]\,\dotso\,X,Yが独立なのでCov[X,Y]=0\\ V[XY]&=&\displaystyle E[(XY)^2]-E[XY]^2\\ &=&\displaystyle E[X^2Y^2]-E[XY]^2\,\dotso\,(ab)^2=a^2b^2\\ &=&\displaystyle E[X^2]E[Y^2]-(E[X]E[Y])^2\,\dotso\,X,Yが独立なのでE[XY]=E[X]E[Y]\\ &=&\displaystyle E[X^2]E[Y^2]-E[X]^2E[Y]^2\,\dotso\,(ab)^2=a^2b^2\\ &=&\displaystyle (V[X]+E[X]^2)(V[Y]+E[Y]^2)-E[X]^2E[Y]^2\,\dotso\,V[X]=E[X^2]-E[X]^2よりE[X^2]=V[X]+E[X]^2\\ &=&\displaystyle V[X]V[Y]+V[X]E[Y]^2+V[Y]E[X]^2+E[X]^2E[Y]^2-E[X]^2E[Y]^2\\ &=&\displaystyle V[X]V[Y]+V[X]E[Y]^2+V[Y]E[X]^2\,\dotso\,X,Yが独立でもV[X]V[Y]とはならない\\ \end{array}$$離散型確率変数(discrete random variable) / 期待値(expected value)
$$\begin{array}{rcl}
母集団(population)の確率変数&:&X,Y\\
確率質量凾数(probability\,mass\,function,\,PMF)&:&f_X(x)=P(X=x_i)\\
&&\displaystyle\sum_{i=1}^{\infty} f_X(x_i)=1\\
累積分布凾数(cumulative\,distribution\,function,\,CDF)&:&\displaystyle F_X(x)=\sum_{i:x_i \leq x} f_X(x_i)\\
同時確率質量凾数(joint\,probability\,mass\,function)&:&f_X(x, y)=P(X=x_i,\,Y=y_j)\\
&&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} f_{XY}(x_i, y_j)=1\\
周辺確率質量凾数(marginal\, probability\, mass\, function)&:&\displaystyle f_X(x)=\sum_{j=1}^{\infty} f_{XY}(x, y_i)\\
\end{array}$$
$$\begin{array}{rcl}
E[g(X)]&\equiv&\displaystyle\sum_{i=1}^{\infty} g(x_i) f_X(x_i) \\
E[X]&=&\displaystyle\sum_{i=1}^{\infty} (x_i) f_X(x_i) \\
E[cX] &=&\displaystyle\sum_{i=1}^{\infty} (c x_i) f_X(x_i) \,\dotso\,cは定数\\
&=&c\displaystyle\sum_{i=1}^{\infty} x_i f_X(x_i) \\
&=&c E[X] \\
E[X \pm t]&=&\displaystyle\sum_{i=1}^{\infty} (x_i \pm t) f_X(x_i) \,\dotso\,tは定数\\
&=&\displaystyle\sum_{i=1}^{\infty} (x_i f_X(x_i) \pm t f_X(x_i)) \\
&=&\displaystyle\sum_{i=1}^{\infty} x_i f_X(x_i)
\pm \displaystyle\sum_{i=1}^{\infty} t f_X(x_i) \\
&=&\displaystyle\sum_{i=1}^{\infty} x_i f_X(x_i)
\pm t \displaystyle\sum_{i=1}^{\infty} f_X(x_i) \\
&=&E[X] \pm t\\
E[X \pm Y]&=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} (x_i \pm y_j) f_{XY}(x_i, y_j) \\
&=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} (x_i f_{XY}(x_i, y_j) \pm y_j f_{XY}(x_i, y_j)) \\
&=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} x_i f_{XY}(x_i, y_j)
\pm \displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty} y_j f_{XY}(x_i, y_j)\\
&=&\displaystyle\sum_{i=1}^{\infty} x_i \sum_{j=1}^{\infty} f_{XY}(x_i, y_j)
\pm \displaystyle\sum_{j=1}^{\infty} y_j \sum_{i=1}^{\infty}f_{XY}(x_i, y_j)\\
&=&\displaystyle\sum_{i=1}^{\infty} x_i f_{X}(x_i)
\pm \displaystyle\sum_{j=1}^{\infty} y_j f_{Y}(y_j)\,\dotso\,周辺確率質量凾数を適用\\
&=& E[X] \pm E[Y] \\
\end{array}$$
X,Yが独立の場合
$$\begin{array}{rcl} E[XY]&=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}(x_i y_j)f_{XY}(x_i, y_j)\\ &=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}x_iy_jf_{X}(x_i)f_{Y}(y_i) \,\dotso\,X,Yが独立\,f_{XY}(x, y)=f_{X}(x)f_{Y}(y)\\ &=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}x_if_{X}(x_i)y_jf_{Y}(y_i)\\ &=&\displaystyle\left(\sum_{i=1}^{\infty}x_if_{X}(x_i)\right)\left(\sum_{j=1}^{\infty} y_jf_{Y}(y_i)\right) \,\dotso\,\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_ib_j=\sum_{i=1}^{\infty}a_i\sum_{j=1}^{\infty}b_j\\ &=&E[X]E[Y] \end{array}$$
登録:
投稿 (Atom)