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ラベル 確率変数 の投稿を表示しています。 すべての投稿を表示
ラベル 確率変数 の投稿を表示しています。 すべての投稿を表示

単回帰モデルの最小二乗推定量の分布

単回帰モデルの最小二乗推定量\(\hat{\alpha},\hat{\beta}\)の分布

単回帰モデル

$$ \begin{eqnarray} y_i&=&\alpha+\beta x_i +\epsilon_i\;(i=1,\cdots,n)\;\dots\;\epsilon_i \overset{iid}{\sim} \mathrm{N}\left(0,\sigma^2\right) \\\mathrm{E}\left[y_i\right]&=&\mathrm{E}\left[\alpha+\beta x_i +\epsilon_i\right] \\&=&\alpha+\beta x_i +\mathrm{E}\left[\epsilon_i\right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/continuous-random-variable-expected.html}{\mathrm{E}\left[X+t\right]=\mathrm{E}\left[X\right]+t} \\&=&\alpha+\beta x_i+0 \;\cdots\;\epsilon_i \overset{iid}{\sim} \mathrm{N}\left(0,\sigma^2\right) \\&=&\alpha+\beta x_i \\\mathrm{V}\left[y_i\right]&=&\mathrm{V}\left[\alpha+\beta x_i +\epsilon_i\right] \\&=&\mathrm{V}\left[\epsilon_i\right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/continuous-random-variable-variance.html}{\mathrm{V}\left[X+t\right]=\mathrm{V}\left[X\right]} \\&=&\sigma^2 \;\cdots\;\epsilon_i \overset{iid}{\sim} \mathrm{N}\left(0,\sigma^2\right) \\y_i&\sim&\mathrm{N}(\alpha+\beta x_i,\sigma^2) \end{eqnarray} $$ \(y_i\)は\(\mathrm{N}\left(\alpha+\beta x_i,\sigma^2\right)\)に従う確率変数である.

\(\hat{\beta}\)を\(\sum_{i=1}^n c_iy_i\)の形で表す

推定量が\(\sum_{i=1}^n c_ix_i\;(x_i:標本,\;c_i:定数)\)の形で表現できるとき,この推定量を線形推定量(linear estimate)という.
(よく知られる線形推定量の例として平均\(\bar{x}\)があり,\(\bar{x}=\sum_{i=1}^n \frac{1}{n} x_i\)で表現される) $$ \begin{eqnarray} \hat{\beta}&=&\frac{S_{xy}}{S_{xx}} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/03/blog-post.html}{\hat{\beta}=\frac{S_{xy}}{S_{xx}}} ,\;S_{xx}=\sum_{i=1}^n\left(x_i-\bar{x}\right)^2,\;\bar{x}=\frac{1}{n}\sum_{i=1}^nx_i \\&=&\frac{1}{S_{xx}} \sum_{i=1}^n \left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right) \\&=&\frac{1}{S_{xx}} \sum_{i=1}^n \left(x_i-\bar{x}\right)y_i \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/10/sxy.html}{\sum_{i=1}^n\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)=S_{xy}= \sum_{i=1}^n \left(x_i-\bar{x}\right)y_i} \\&=& \sum_{i=1}^n \frac{x_i-\bar{x}}{S_{xx}}y_i \\&=& \sum_{i=1}^n c_iy_i \;\cdots\;c_i=\frac{x_i-\bar{x}}{S_{xx}} \end{eqnarray} $$

\(\hat{\beta}\)の期待値を\(\sum_{i=1}^n c_iy_i\)から求めてみる

$$ \begin{eqnarray} \mathrm{E}\left[\sum_{i=1}^n c_iy_i\right] &=&\mathrm{E}\left[\sum_{i=1}^n \frac{x_i-\bar{x}}{S_{xx}}y_i\right] \\&=&\mathrm{E}\left[\frac{S_{xy}}{S_{xx}}\right] \;\cdots\;上記,\;\frac{S_{xy}}{S_{xx}}=\sum_{i=1}^n \frac{x_i-\bar{x}}{S_{xx}}y_i \\&=&\mathrm{E}\left[\hat{\beta}\right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/03/blog-post.html}{\hat{\beta}=\frac{S_{xy}}{S_{xx}}} \\&=&\beta \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/2.html}{\mathrm{E}\left[\hat{\beta}\right]=\beta} \end{eqnarray} $$

\(\hat{\beta}\)の分散を\(\sum_{i=1}^n c_iy_i\)から求めてみる

$$ \begin{eqnarray} \mathrm{V}\left[\sum_{i=1}^n c_iy_i\right] &=&\mathrm{V}\left[\sum_{i=1}^n \frac{x_i-\bar{x}}{S_{xx}}y_i\right] \\&=&\sum_{i=1}^n \mathrm{V}\left[\frac{x_i-\bar{x}}{S_{xx}}y_i\right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/continuous-random-variable-variance.html}{y_iは互いに独立\mathrm{Cov}\left[y_i, y_j\right]=0,\;互いに独立の場合\mathrm{V}\left[X+Y\right]=\mathrm{V}\left[X\right]+\mathrm{V}\left[Y\right]} \\&=&\sum_{i=1}^n \left(\frac{x_i-\bar{x}}{S_{xx}}\right)^2\mathrm{V}\left[y_i\right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/continuous-random-variable-variance.html}{\mathrm{V}\left[cX\right]=c^2\mathrm{V}\left[X\right]} \\&=&\sum_{i=1}^n \left(\frac{x_i-\bar{x}}{S_{xx}}\right)^2\sigma^2 \\&=&\frac{\sigma^2}{S_{xx}^2}\sum_{i=1}^n \left(x_i-\bar{x}\right)^2 \\&=&\frac{\sigma^2}{S_{xx}^2}S_{xx} \;\cdots\;S_{xx}=\sum_{i=1}^n \left(x_i-\bar{x}\right)^2 \\&=&\frac{\sigma^2}{S_{xx}} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/2variancecovariance.html}{\mathrm{V}\left[\frac{S_{xy}}{S_{xx}}\right]=\frac{\sigma^2}{S_{xx}}}と同じ結果 \end{eqnarray} $$

\(\hat{\beta}\)の分布

以上のように,\(\hat{\beta}\)は線形推定量であり,正規分布に従う\(y_i\)の定数倍の和で表すことができた.よって\(\hat{\beta}\)は同様に正規分布に従い,その期待値と分散はそれぞれ上記で求めたとおりである \(\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/09/zc1xc2y-2.html}{Z=c_1X+c_2Y(X\sim\mathrm{N}(\mu_1,\sigma_1^2),Y\sim\mathrm{N}(\mu_2,\sigma_2^2),Z\sim\mathrm{N}(c_1\mu_1+c_2\mu_2,c_1^2\sigma_1^2+c_2^2\sigma_2^2))}\). $$ \begin{eqnarray} \hat{\beta}&\sim& \mathrm{N}\left(\beta,\frac{\sigma^2}{S_{xx}}\right) \end{eqnarray} $$

\(\hat{\alpha}\)を\(\sum_{i=1}^n c_iy_i\)の形で表す

$$ \begin{eqnarray} \hat{\alpha}&=&\bar{y}-\hat{\beta}\bar{x} \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/03/blog-post.html}{\hat{\alpha}=\bar{y}-\hat{\beta}\bar{x}} \\&=&\sum_{i=1}^n\frac{1}{n}y_i-\frac{S_{xy}}{S_{xx}}\bar{x} \\&=&\sum_{i=1}^n\frac{1}{n}y_i-\left(\sum_{i=1}^n\frac{x_i-\bar{x}}{S_{xx}}y_i\right)\bar{x} \;\cdots\;上記,\;\frac{S_{xy}}{S_{xx}}=\sum_{i=1}^n \frac{x_i-\bar{x}}{S_{xx}}y_i \\&=&\sum_{i=1}^n\frac{1}{n}y_i-\bar{x}\sum_{i=1}^n\frac{x_i-\bar{x}}{S_{xx}}y_i \\&=&\sum_{i=1}^n\frac{1}{n}y_i-\sum_{i=1}^n\frac{\bar{x}\left(x_i-\bar{x}\right)}{S_{xx}}y_i \\&=&\sum_{i=1}^n\left(\frac{1}{n}-\frac{\bar{x}\left(x_i-\bar{x}\right)}{S_{xx}}\right)y_i \\&=& \sum_{i=1}^n c_iy_i \;\cdots\;c_i=\frac{1}{n}-\frac{\bar{x}\left(x_i-\bar{x}\right)}{S_{xx}} \end{eqnarray} $$

\(\hat{\alpha}\)の期待値を\(\sum_{i=1}^n c_iy_i\)から求めてみる

$$ \begin{eqnarray} \mathrm{E}\left[\sum_{i=1}^n c_iy_i\right] &=&\mathrm{E}\left[\sum_{i=1}^n\left(\frac{1}{n}-\frac{\bar{x}\left(x_i-\bar{x}\right)}{S_{xx}}\right)y_i\right] \\&=&\mathrm{E}\left[\sum_{i=1}^n\left(\frac{1}{n}y_i-\frac{\bar{x}\left(x_i-\bar{x}\right)}{S_{xx}}y_i\right)\right] \\&=&\mathrm{E}\left[\sum_{i=1}^n\frac{1}{n}y_i-\sum_{i=1}^n\frac{\bar{x}\left(x_i-\bar{x}\right)}{S_{xx}}y_i\right] \\&=&\mathrm{E}\left[\sum_{i=1}^n\frac{1}{n}y_i-\sum_{i=1}^n\frac{\bar{x}\left(x_i-\bar{x}\right)}{S_{xx}}y_i\right] \\&=&\mathrm{E}\left[\sum_{i=1}^n\frac{1}{n}y_i\right]-\mathrm{E}\left[\sum_{i=1}^n\frac{\bar{x}\left(x_i-\bar{x}\right)}{S_{xx}}y_i\right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/continuous-random-variable-expected.html}{\mathrm{E}\left[X+Y\right]=\mathrm{E}\left[X\right]+\mathrm{E}\left[Y\right]} \\&=&\mathrm{E}\left[\frac{1}{n}\sum_{i=1}^ny_i\right]-\mathrm{E}\left[\bar{x}\sum_{i=1}^n\frac{\left(x_i-\bar{x}\right)}{S_{xx}}y_i\right] \\&=&\frac{1}{n}\mathrm{E}\left[\sum_{i=1}^ny_i\right]-\bar{x}\mathrm{E}\left[\sum_{i=1}^n\frac{\left(x_i-\bar{x}\right)}{S_{xx}}y_i\right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/continuous-random-variable-expected.html}{\mathrm{E}\left[cX\right]=c\mathrm{E}\left[X\right]} \\&=&\frac{1}{n}\sum_{i=1}^n\mathrm{E}\left[y_i\right]-\bar{x}\mathrm{E}\left[\frac{S_{xy}}{S_{xx}}\right] \;\cdots\;上記,\;\frac{S_{xy}}{S_{xx}}=\sum_{i=1}^n \frac{x_i-\bar{x}}{S_{xx}}y_i \\&=&\frac{1}{n}\sum_{i=1}^n\left(\alpha+\beta x_i\right)-\bar{x}\mathrm{E}\left[\frac{S_{xy}}{S_{xx}}\right] \;\cdots\;\mathrm{E}\left[y_i\right]=\alpha+\beta x_i \\&=&\frac{1}{n}\left(\alpha\sum_{i=1}^n1+\beta\sum_{i=1}^n x_i\right)-\bar{x}\mathrm{E}\left[\hat{\beta}\right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/03/blog-post.html}{\hat{\beta}=\frac{S_{xy}}{S_{xx}}} \\&=&\frac{1}{n}\left(n\alpha+\beta\;n\bar{x}\right)-\bar{x}\beta \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/2.html}{\mathrm{E}\left[\hat{\beta}\right]=\beta} \\&=&\frac{1}{n}n\left(\alpha+\beta\bar{x}\right)-\bar{x}\beta \\&=&\left(\alpha+\beta\bar{x}\right)-\bar{x}\beta \\&=&\alpha \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/2.html}{\mathrm{E}\left[\hat{\alpha}\right]=\alpha} \end{eqnarray} $$

\(\hat{\alpha}\)の分散を\(\sum_{i=1}^n c_iy_i\)から求めてみる

$$ \begin{eqnarray} \mathrm{V}\left[\sum_{i=1}^n c_iy_i\right] &=&\mathrm{V}\left[\sum_{i=1}^n\left(\frac{1}{n}-\frac{\bar{x}\left(x_i-\bar{x}\right)}{S_{xx}}\right)y_i\right] \\&=&\sum_{i=1}^n\mathrm{V}\left[\left(\frac{1}{n}-\frac{\bar{x}\left(x_i-\bar{x}\right)}{S_{xx}}\right)y_i\right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/2variancecovariance.html}{\mathrm{V}\left[\frac{S_{xy}}{S_{xx}}\right]=\frac{\sigma^2}{S_{xx}}}と同じ結果 \\&=&\sum_{i=1}^n\left(\frac{1}{n}-\frac{\bar{x}\left(x_i-\bar{x}\right)}{S_{xx}}\right)^2\mathrm{V}\left[y_i\right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/continuous-random-variable-variance.html}{\mathrm{V}\left[cX\right]=c^2\mathrm{V}\left[X\right]} \\&=&\sum_{i=1}^n\left(\frac{1}{n}-\frac{\bar{x}\left(x_i-\bar{x}\right)}{S_{xx}}\right)^2\sigma^2 \\&=&\sigma^2\sum_{i=1}^n\left(\frac{1}{n}-\frac{\bar{x}\left(x_i-\bar{x}\right)}{S_{xx}}\right)^2 \\&=&\sigma^2\sum_{i=1}^n\left( \frac{1}{n^2} -2\frac{1}{n}\frac{\bar{x}\left(x_i-\bar{x}\right)}{S_{xx}} +\left(\frac{\bar{x}\left(x_i-\bar{x}\right)}{S_{xx}}\right)^2 \right) \\&=&\sigma^2\left( \sum_{i=1}^n\frac{1}{n^2} -\sum_{i=1}^n2\frac{1}{n}\frac{\bar{x}\left(x_i-\bar{x}\right)}{S_{xx}} +\sum_{i=1}^n\left(\frac{\bar{x}\left(x_i-\bar{x}\right)}{S_{xx}}\right)^2 \right) \\&=&\sigma^2\left( \frac{1}{n^2}\sum_{i=1}^n1 -\frac{2\bar{x}}{nS_{xx}}\sum_{i=1}^n\left(x_i-\bar{x}\right) +\frac{\bar{x}^2}{S_{xx}^2}\sum_{i=1}^n\left(x_i-\bar{x}\right)^2 \right) \\&=&\sigma^2\left( \frac{1}{n^2}n -\frac{2\bar{x}}{nS_{xx}}\cdot 0 +\frac{\bar{x}^2}{S_{xx}^2}S_{xx} \right) \\&=&\sigma^2\left( \frac{1}{n^2}\cdot n -\frac{2\bar{x}}{nS_{xx}}\cdot 0 +\frac{\bar{x}^2}{S_{xx}^2}\cdot S_{xx} \right) \\&=&\sigma^2\left( \frac{1}{n} +\frac{\bar{x}^2}{S_{xx}} \right) \;\cdots\;\href{https://shikitenkai.blogspot.com/2020/08/2variancecovariance.html}{\mathrm{V}\left[\bar{y}-\hat{\beta}\bar{x}\right]=\left(\frac{1}{n}+\frac{\bar{x}^2}{S_{xx}}\right)\sigma^2と同じ結果} \end{eqnarray} $$

\(\hat{\alpha}\)の分布

以上のように,\(\hat{\alpha}\)は線形推定量であり,正規分布に従う\(y_i\)の定数倍の和で表すことができた.よって\(\hat{\alpha}\)は同様に正規分布に従い,その期待値と分散はそれぞれ上記で求めたとおりである \(\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/09/zc1xc2y-2.html}{Z=c_1X+c_2Y(X\sim\mathrm{N}(\mu_1,\sigma_1^2),Y\sim\mathrm{N}(\mu_2,\sigma_2^2),Z\sim\mathrm{N}(c_1\mu_1+c_2\mu_2,c_1^2\sigma_1^2+c_2^2\sigma_2^2))}\). $$ \begin{eqnarray} \hat{\alpha}&\sim& \mathrm{N}\left(\alpha,\sigma^2\left( \frac{1}{n} +\frac{\bar{x}^2}{S_{xx}} \right)\right) \end{eqnarray} $$

確率変数の変数変換 Z=c1X+c2Y / 正規分布の定数倍同士の和

確率変数の変数変換 \(Z=c_1X+c_2Y\) / 正規分布の定数倍同士の和

確率変数の変数変換 \(Z=c_1X+c_2Y\)

$$ \begin{eqnarray} p_{c_1X+c_2Y}(z) &=&\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \delta(z-(c_1x+c_2y))f_X(x)g_Y(y)\mathrm{d}x\mathrm{d}y \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/09/delta-function.html}{\int_{-\infty}^{\infty}\delta(x)f(x)\mathrm{d}x=f(0)} \\&&\;\cdots\;X=x,Y=yの同時確率f(x)g(x)のうちz-\left(c_1x+c_2y\right)=0を満たすものだけを足し合わせる. \\&=&\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \delta(z-(c_1x+c_2y))f_X(x)g_Y(y)\mathrm{d}x\mathrm{d}y \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/09/delta-function.html}{\delta(u(x))=\sum_{\alpha\in u^{-1}(0)}\frac{1}{\left|u^{\prime}(\alpha)\right|}\delta\left(x-\alpha\right)} \\&&\;\cdots\;u(x)=z-(c_1x+c_2y) \\&&\;\cdots\;u(x=\alpha)=0,\;\alpha=\frac{z-c_2y}{c_1} \\&&\;\cdots\;u^\prime=\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}\left(z-(c_1x+c_2y)\right)=-c_1 \\&&\;\cdots\;\delta\left(z-(c_1x+c_2y)\right) =\frac{1}{|u^\prime(\alpha)|}\delta\left(x-\alpha\right) =\frac{1}{\left|-c_1\right|}\delta\left(x-\frac{z-c_2y}{c_1}\right) =\frac{1}{\left|c_1\right|}\delta\left(x-\frac{z-c_2y}{c_1}\right) \\&=&\int_{-\infty}^{\infty}\frac{1}{\left|c_1\right|}\delta\left(x-\frac{z-c_2y}{c_1}\right)f_X\left(x\right)g_Y(y)\mathrm{d}y \\&=&\int_{-\infty}^{\infty}\frac{1}{\left|c_1\right|}\delta\left(\frac{z-c_2y}{c_1}-\frac{z-c_2y}{c_1}\right)f_X\left(\frac{z-c_2y}{c_1}\right)g_Y(y)\mathrm{d}y \;\cdots\;z=c_1x+c_2y,\;x=\frac{z-c_2y}{c_1} \\&=&\frac{1}{\left|c_1\right|}\int_{-\infty}^{\infty}\delta(0)f_X\left(\frac{z-c_2y}{c_1}\right)g_Y(y)\mathrm{d}y \;\cdots\;\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x \\&=&\frac{1}{\left|c_1\right|}\int_{-\infty}^{\infty}f_X\left(\frac{z-c_2y}{c_1}\right)g_Y(y)\mathrm{d}y \end{eqnarray} $$

正規分布の定数倍同士の和の例

$$ \begin{eqnarray} f_X(x)&=&\frac{1}{\sqrt{2\pi\sigma_1^2}}\mathrm{e}^{\frac{-(x-\mu_1)^2}{2\sigma_1^2}} \;\cdots\;\mathrm{N}(\mu_1,\sigma_1^2)(\href{https://shikitenkai.blogspot.com/2019/06/binomial-distributionnormal-distribution.html}{正規分布}) \\g_Y(y)&=&\frac{1}{\sqrt{2\pi\sigma_2^2}}\mathrm{e}^{\frac{-(y-\mu_2)^2}{2\sigma_2^2}} \;\cdots\;\mathrm{N}(\mu_2,\sigma_2^2)(\href{https://shikitenkai.blogspot.com/2019/06/binomial-distributionnormal-distribution.html}{正規分布}) \\p_{c_1X+c_2Y}(z) &=&\frac{1}{\left|c_1\right|}\int_{-\infty}^{\infty}f_X\left(\frac{z-c_2y}{c_1}\right)g_Y(y)\mathrm{d}y \\&=&\frac{1}{\left|c_1\right|}\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi\sigma_1^2}}\mathrm{e}^{\frac{-\left\{\left(\frac{z-c_2y}{c_1}\right)-\mu_1\right\}^2}{2\sigma_1^2}}\frac{1}{\sqrt{2\pi\sigma_2^2}}\mathrm{e}^{\frac{-(y-\mu_2)^2}{2\sigma_2^2}} \mathrm{d}y \\&=&\frac{1}{\left|c_1\right|}\frac{1}{\sqrt{2\pi\sigma_1^2}}\frac{1}{\sqrt{2\pi\sigma_2^2}}\int_{-\infty}^{\infty}\mathrm{e}^{\frac{-\left\{\left(\frac{z-c_2y}{c_1}\right)-\mu_1\right\}^2}{2\sigma_1^2}+\frac{-(y-\mu_2)^2}{2\sigma_2^2}} \mathrm{d}y \;\cdots\;\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x \\&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}}\int_{-\infty}^{\infty}\mathrm{e}^{\frac{-\left\{\left(\frac{z-c_2y}{c_1}\right)-\mu_1\right\}^2}{2\sigma_1^2}+\frac{-(y-\mu_2)^2}{2\sigma_2^2}} \mathrm{d}y \\&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \int_{-\infty}^{\infty} \mathrm{e}^{f_1(c_1,c_2,y,z,\mu_1,\mu_2,\sigma_1,\sigma_2)} \mathrm{d}y \;\cdots\;f_1はyが引数にある. \end{eqnarray} $$
$$ \begin{eqnarray} f_1(c_1,c_2,y,z,\mu_1,\mu_2,\sigma_1,\sigma_2) &=&\frac{-\left\{\left(\frac{z-c_2y}{c_1}\right)-\mu_1\right\}^2}{2\sigma_1^2}+\frac{-(y-\mu_2)^2}{2\sigma_2^2} \\&=&-\frac{1}{2} \left\{ \frac{\left(\frac{1}{c_1}z-\frac{c_2}{c_1}y-\mu_1\right)^2}{\sigma_1^2} +\frac{(y-\mu_2)^2}{\sigma_2^2} \right\} \\&=&-\frac{1}{2} \left\{ \frac{\sigma_2^2\left(\frac{1}{c_1}z-\frac{c_2}{c_1}y-\mu_1\right)^2 +\sigma_1^2(y-\mu_2)^2}{\sigma_1^2\sigma_2^2} \right\} \\&=&-\frac{1}{2\sigma_1^2\sigma_2^2} \left\{ \sigma_2^2\left(\frac{1}{c_1}z-\frac{c_2}{c_1}y-\mu_1\right)^2 +\sigma_1^2(y-\mu_2)^2 \right\} \\&=&-\frac{1}{2\sigma_1^2\sigma_2^2} \left[ \sigma_2^2\left\{\left(\frac{1}{c_1}z\right)^2-2\left(\frac{1}{c_1}z\right)\left(\frac{c_2}{c_1}y\right)-2\left(\frac{1}{c_1}z\right)\mu_1+\left(\frac{c_2}{c_1}y\right)^2+2\left(\frac{c_2}{c_1}y\right)\mu_1+\mu_1^2\right\} +\sigma_1^2(y^2-2y\mu_2+\mu_2^2) \right] \\&=&-\frac{1}{2\sigma_1^2\sigma_2^2} \left\{ \frac{1}{c_1^2}\sigma_2^2z^2 -2\frac{c_2}{c_1^2}\sigma_2^2zy -2\frac{1}{c_1}\sigma_2^2z\mu_1 +\frac{c_2^2}{c_1^2}\sigma_2^2y^2 +2\frac{c_2}{c_1}\sigma_2^2y\mu_1 +\sigma_2^2\mu_1^2 +\sigma_1^2y^2 -2\sigma_1^2y\mu_2 +\sigma_1^2\mu_2^2 \right\} \\&=&-\frac{1}{2\sigma_1^2\sigma_2^2} \frac{1}{c_1^2} \left\{ \sigma_2^2z^2 -2c_2\sigma_2^2zy -2c_1\sigma_2^2z\mu_1 +c_2^2\sigma_2^2y^2 +2c_1c_2\sigma_2^2y\mu_1 +c_1^2\sigma_2^2\mu_1^2 +c_1^2\sigma_1^2y^2 -2c_1^2\sigma_1^2y\mu_2 +c_1^2\sigma_1^2\mu_2^2 \right\} \\&=&-\frac{1}{2c_1^2\sigma_1^2\sigma_2^2} \left\{ \left( c_1^2\sigma_1^2 +c_2^2\sigma_2^2 \right)y^2 +\left( -2c_2\sigma_2^2z +2c_1c_2\sigma_2^2\mu_1 -2c_1^2\sigma_1^2\mu_2 \right)y +\sigma_2^2z^2 -2c_1\sigma_2^2z\mu_1 +c_1^2\sigma_2^2\mu_1^2 +c_1^2\sigma_1^2\mu_2^2 \right\} \\&=&-\frac{1}{2c_1^2\sigma_1^2\sigma_2^2} \left\{ \left( c_1^2\sigma_1^2 +c_2^2\sigma_2^2 \right)y^2 -2\left( c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 \right)y +\sigma_2^2z^2 -2c_1\sigma_2^2z\mu_1 +c_1^2\sigma_2^2\mu_1^2 +c_1^2\sigma_1^2\mu_2^2 \right\} \\&=&-\frac{1}{2c_1^2\sigma_1^2\sigma_2^2} \left\{ \left( c_1^2\sigma_1^2 +c_2^2\sigma_2^2 \right) \left( y^2 -2\frac{ c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2}y \right) +\sigma_2^2z^2 -2c_1\sigma_2^2z\mu_1 +c_1^2\sigma_2^2\mu_1^2 +c_1^2\sigma_1^2\mu_2^2 \right\} \\&=& \\&=&-\frac{1}{2c_1^2\sigma_1^2\sigma_2^2} \left\{ \left( c_1^2\sigma_1^2 +c_2^2\sigma_2^2 \right) \left( y^2 -2\frac{ c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2}y \right) +A-A +\sigma_2^2z^2 -2c_1\sigma_2^2z\mu_1 +c_1^2\sigma_2^2\mu_1^2 +c_1^2\sigma_1^2\mu_2^2 \right\} \\&&\;\cdots\;平方完成させるための項Aを考える \\&=&-\frac{1}{2c_1^2\sigma_1^2\sigma_2^2} \left\{ \left( c_1^2\sigma_1^2 +c_2^2\sigma_2^2 \right) \left( y^2 -2\frac{ c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2}y +\frac{1}{c_1^2\sigma_1^2 +c_2^2\sigma_2^2}A \right) -A +\sigma_2^2z^2 -2c_1\sigma_2^2z\mu_1 +c_1^2\sigma_2^2\mu_1^2 +c_1^2\sigma_1^2\mu_2^2 \right\} \\&=&-\frac{1}{2c_1^2\sigma_1^2\sigma_2^2} \left[ \left( c_1^2\sigma_1^2 +c_2^2\sigma_2^2 \right) \left\{ y^2 -2\frac{ c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2}y +\frac{1}{c_1^2\sigma_1^2 +c_2^2\sigma_2^2} \frac{ \left( c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 \right)^2 }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2} \right\} -\frac{ \left( c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 \right)^2 }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2} +\sigma_2^2z^2 -2c_1\sigma_2^2z\mu_1 +c_1^2\sigma_2^2\mu_1^2 +c_1^2\sigma_1^2\mu_2^2 \right] \\&&\;\cdots\;A=\frac{ \left( c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 \right)^2 } { c_1^2\sigma_1^2 +c_2^2\sigma_2^2 } \\&=&-\frac{1}{2c_1^2\sigma_1^2\sigma_2^2} \left\{ \left( c_1^2\sigma_1^2 +c_2^2\sigma_2^2 \right) \left( y^2 -\frac{ c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2} \right)^2 -\frac{ \left( c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 \right)^2 }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2} +\sigma_2^2z^2 -2c_1\sigma_2^2z\mu_1 +c_1^2\sigma_2^2\mu_1^2 +c_1^2\sigma_1^2\mu_2^2 \right\} \\&=& -\frac{\left( c_1^2\sigma_1^2 +c_2^2\sigma_2^2 \right)}{2c_1^2\sigma_1^2\sigma_2^2} \left( y^2 -\frac{ c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2} \right)^2 -\frac{1}{2c_1^2\sigma_1^2\sigma_2^2} \left\{ -\frac{ \left( c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 \right)^2 }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2} +\sigma_2^2z^2 -2c_1\sigma_2^2z\mu_1 +c_1^2\sigma_2^2\mu_1^2 +c_1^2\sigma_1^2\mu_2^2 \right\} \\&=& -f_2(c_1,c_2,\sigma_1^2,\sigma_2^2) \left\{ y-f_3(c_1,c_2,z,\mu_1,\mu_2,\sigma_1^2,\sigma_2^2) \right\}^2 -f_4(c_1,c_2,z,\mu_1,\mu_2,\sigma_1^2,\sigma_2^2) \;\cdots\;f_2,f_3,f_4はyを引数としない. \\&=& -f_2\left(y-f_3\right)^2-f_4 \;\cdots\;yが平方完成の中の一つだけになっている. \\ \\f_2(c_1,c_2,\sigma_1^2,\sigma_2^2)&=& \frac{\left( c_1^2\sigma_1^2 +c_2^2\sigma_2^2 \right)}{2c_1^2\sigma_1^2\sigma_2^2} \\ \\f_3(c_1,c_2,z,\mu_1,\mu_2,\sigma_1^2,\sigma_2^2)&=&\frac{ c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2} \\ \\f_4(c_1,c_2,z,\mu_1,\mu_2,\sigma_1^2,\sigma_2^2) &=&\frac{1}{2c_1^2\sigma_1^2\sigma_2^2} \left\{ -\frac{ \left( c_2\sigma_2^2z -c_1c_2\sigma_2^2\mu_1 +c_1^2\sigma_1^2\mu_2 \right)^2 }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2} +\sigma_2^2z^2 -2c_1\sigma_2^2z\mu_1 +c_1^2\sigma_2^2\mu_1^2 +c_1^2\sigma_1^2\mu_2^2 \right\} \\&=&\frac{1}{2c_1^2\sigma_1^2\sigma_2^2} \left\{ \frac{ -\left( c_2\sigma_2^2 \left(z -c_1\mu_1 \right) +c_1^2\sigma_1^2\mu_2 \right)^2 +\left(c_1^2\sigma_1^2+c_2^2\sigma_2^2\right) \left(\sigma_2^2\left( z-c_1\mu_1 \right)^2 +c_1^2\sigma_1^2\mu_2^2\right) }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2} \right\} \\&=&\frac{1}{2c_1^2\sigma_1^2\sigma_2^2} \frac{ -\left( c_2\sigma_2^2 B +c_1^2\sigma_1^2\mu_2 \right)^2 +\left(c_1^2\sigma_1^2+c_2^2\sigma_2^2\right) \left(\sigma_2^2B^2 +c_1^2\sigma_1^2\mu_2^2\right) }{c_1^2\sigma_1^2 +c_2^2\sigma_2^2} \;\cdots\;B=z-c_1\mu_1 \\&=&\frac{1}{2c_1^2\sigma_1^2\sigma_2^2(c_1^2\sigma_1^2 +c_2^2\sigma_2^2)} \left\{ -\left( c_2\sigma_2^2 B +c_1^2\sigma_1^2\mu_2 \right)^2 +\left(c_1^2\sigma_1^2+c_2^2\sigma_2^2\right) \left(\sigma_2^2B^2 +c_1^2\sigma_1^2\mu_2^2\right) \right\} \\&=&\frac{1}{2c_1^2\sigma_1^2\sigma_2^2(c_1^2\sigma_1^2 +c_2^2\sigma_2^2)} \left( \color{red}{-c_2^2\sigma_2^4B^2} \color{black}{-2c_1^2c_2\sigma_1^2\sigma_2^2\mu_2B} \color{green}{-c_1^4\sigma_1^4\mu_2^2} \color{black}{+c_1^2\sigma_1^2\sigma_2^2B^2} \color{red}{+c_2^2\sigma_2^4B^2} \color{green}{+c_1^4\sigma_1^4\mu_2^2} \color{black}{+c_1^2c_2^2\sigma_1^2\sigma_2^2\mu_2^2} \right) \\&=&\frac{1}{2c_1^2\sigma_1^2\sigma_2^2(c_1^2\sigma_1^2 +c_2^2\sigma_2^2)} c_1^2\sigma_1^2 \sigma_2^2\left\{ -2 c_2 \mu_2 B + B^2 + c_2^2 \mu_2^2 \right\} \\&=&\frac{1}{2(c_1^2\sigma_1^2+c_2^2\sigma_2^2)}\left(B-c_2\mu_2\right)^2 \\&=&\frac{\left(z-c_1\mu_1-c_2\mu_2\right)^2}{2(c_1^2\sigma_1^2+c_2^2\sigma_2^2)} \;\cdots\;B=z-c_1\mu_1 \\&=&\frac{\left(z-\left(c_1\mu_1+c_2\mu_2\right)\right)^2}{2(c_1^2\sigma_1^2+c_2^2\sigma_2^2)} \end{eqnarray} $$
$$ \begin{eqnarray} p_{c_1X+c_2Y}(z)&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \int_{-\infty}^{\infty} \mathrm{e}^{f_1(c_1,c_2,y,z,\mu_1,\mu_2,\sigma_1,\sigma_2)} \mathrm{d}y \\&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \int_{-\infty}^{\infty} \mathrm{e}^{-f_2\left(y-f_3\right)^2-f_4} \mathrm{d}y \\&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \int_{-\infty}^{\infty} \mathrm{e}^{-f_2\left(y-f_3\right)^2} \mathrm{e}^{-f_4} \mathrm{d}y \\&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \mathrm{e}^{-f_4} \int_{-\infty}^{\infty} \mathrm{e}^{-f_2\left(y-f_3\right)^2} \mathrm{d}y \\&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \mathrm{e}^{-f_4} \int_{-\infty}^{\infty} \mathrm{e}^{-u^2} \frac{1}{\sqrt{f_2}}\mathrm{d}u \\&&\;\cdots\;u=\sqrt{f_2}\left(y-f_3\right),\;\frac{\mathrm{d}u}{\mathrm{d}y}=\sqrt{f_2},\;\mathrm{d}y=\frac{1}{\sqrt{f_2}}\mathrm{d}u \\&&\;\cdots\;y:-\infty \rightarrow \infty,\;u:-\infty \rightarrow \infty \\&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \mathrm{e}^{-f_4} \frac{1}{\sqrt{f_2}} \int_{-\infty}^{\infty} \mathrm{e}^{-u^2} \mathrm{d}u \;\cdots\;\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x \\&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \mathrm{e}^{-f_4} \frac{1}{\sqrt{f_2}} \sqrt{\pi} \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/gaussian-integral.html}{\int_{-\infty}^{\infty}\mathrm{e}^{-u^2}\mathrm{d}u=\sqrt{\pi}} \\&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \mathrm{e}^{-f_4} \frac{\sqrt{\pi}}{\sqrt{f_2}} \\&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \sqrt{\frac{\pi}{f_2}} \mathrm{e}^{-f_4} \\&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \sqrt{\frac{\pi}{\frac{\left(c_1^2\sigma_1^2+c_2^2\sigma_2^2\right)}{2c_1^2\sigma_1^2\sigma_2^2}}} \mathrm{e}^{-\frac{\left(z-\left(c_1\mu_1+c_2\mu_2\right)\right)^2}{2(c_1^2\sigma_1^2+c_2^2\sigma_2^2)}} \\&=&\frac{1}{\left|c_1\right|}\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \sqrt{\frac{2\pi c_1^2\sigma_1^2\sigma_2^2}{\left(c_1^2\sigma_1^2+c_2^2\sigma_2^2\right)}} \mathrm{e}^{-\frac{\left(z-\left(c_1\mu_1+c_2\mu_2\right)\right)^2}{2(c_1^2\sigma_1^2+c_2^2\sigma_2^2)}} \\&=&\frac{1}{\left|c_1\right|}\frac{\sqrt{c_1^2}}{\sqrt{2\pi\left(c_1^2\sigma_1^2+c_2^2\sigma_2^2\right)}} \mathrm{e}^{-\frac{\left(z-\left(c_1\mu_1+c_2\mu_2\right)\right)^2}{2(c_1^2\sigma_1^2+c_2^2\sigma_2^2)}} \\&=&\frac{1}{\left|c_1\right|}\frac{\left|c_1\right|}{\sqrt{2\pi\left(c_1^2\sigma_1^2+c_2^2\sigma_2^2\right)}} \mathrm{e}^{-\frac{\left(z-\left(c_1\mu_1+c_2\mu_2\right)\right)^2}{2(c_1^2\sigma_1^2+c_2^2\sigma_2^2)}} \;\cdots\;\sqrt{A^2}= \left\{ \begin{array} \\A&(A\geq0) \\-A&(A\lt0) \end{array} \right. =\left|A\right| \\&=&\frac{1}{\sqrt{2\pi\left(c_1^2\sigma_1^2+c_2^2\sigma_2^2\right)}} \mathrm{e}^{-\frac{\left(z-\left(c_1\mu_1+c_2\mu_2\right)\right)^2}{2(c_1^2\sigma_1^2+c_2^2\sigma_2^2)}} \\&\sim&\mathrm{N}\left(c_1\mu_1+c_2\mu_2,\;c_1^2\sigma_1^2+c_2^2\sigma_2^2\right) \end{eqnarray} $$

確率変数の変数変換 Z=X+Y その2 / 正規分布の再生性

確率変数の変数変換 Z=X+Y その2 / 正規分布の再生性

Z=X+Y 正規分布同士の例

$$ \begin{eqnarray} f_X(x)&=&\frac{1}{\sqrt{2\pi\sigma_1^2}}\mathrm{e}^{\frac{-(x-\mu_1)^2}{2\sigma_1^2}} \;\cdots\;\mathrm{N}(\mu_1,\sigma_1^2)(\href{https://shikitenkai.blogspot.com/2019/06/binomial-distributionnormal-distribution.html}{正規分布}) \\g_Y(x)&=&\frac{1}{\sqrt{2\pi\sigma_2^2}}\mathrm{e}^{\frac{-(y-\mu_2)^2}{2\sigma_2^2}} \;\cdots\;\mathrm{N}(\mu_2,\sigma_2^2)(\href{https://shikitenkai.blogspot.com/2019/06/binomial-distributionnormal-distribution.html}{正規分布}) \\p_{X+Y}(z)&=&\href{https://shikitenkai.blogspot.com/2020/09/zxy.html}{\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \delta(z-(x+y))f_X(x)g_Y(y)\mathrm{d}x\mathrm{d}y} \\&=&\int_{-\infty}^{\infty}\delta(z-((z-y)+y))f_X(z-y)g_Y(y)\mathrm{d}y \;\cdots\;z=x+y,\;x=z-y \\&=&\int_{-\infty}^{\infty}\delta(0)f_X(z-y)g_Y(y)\mathrm{d}y \\&=&\int_{-\infty}^{\infty}f_X(z-y)g_Y(y)\mathrm{d}y \\&=&\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi\sigma_1^2}}\mathrm{e}^{\frac{-(z-y-\mu_1)^2}{2\sigma_1^2}}\frac{1}{\sqrt{2\pi\sigma_2^2}}\mathrm{e}^{\frac{-(y-\mu_2)^2}{2\sigma_2^2}} \mathrm{d}y \\&=&\frac{1}{\sqrt{2\pi\sigma_1^2}}\frac{1}{\sqrt{2\pi\sigma_2^2}}\int_{-\infty}^{\infty}\mathrm{e}^{\frac{-(z-y-\mu_1)^2}{2\sigma_1^2}+\frac{-(y-\mu_2)^2}{2\sigma_2^2}} \mathrm{d}y \\&=&\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}}\int_{-\infty}^{\infty}\mathrm{e}^{\frac{-(z-y-\mu_1)^2}{2\sigma_1^2}+\frac{-(y-\mu_2)^2}{2\sigma_2^2}} \mathrm{d}y \\&=&\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \int_{-\infty}^{\infty} \mathrm{e}^{f_1(y,z,\mu_1,\mu_2,\sigma_1,\sigma_2)} \mathrm{d}y \;\cdots\;f_1はyが引数にある. \end{eqnarray} $$
$$ \begin{eqnarray} f_1(y,z,\mu_1,\mu_2,\sigma_1,\sigma_2) &=&\frac{-(z-y-\mu_1)^2}{2\sigma_1^2}+\frac{-(y-\mu_2)^2}{2\sigma_2^2} \\&=&-\frac{1}{2} \left( \frac{(z-y-\mu_1)^2}{\sigma_1^2} +\frac{(y-\mu_2)^2}{\sigma_2^2} \right) \\&=&-\frac{1}{2} \left( \frac{\sigma_2^2(z-y-\mu_1)^2 +\sigma_1^2(y-\mu_2)^2}{\sigma_1^2\sigma_2^2} \right) \\&=&-\frac{1}{2\sigma_1^2\sigma_2^2} \left( \sigma_2^2(z-y-\mu_1)^2 +\sigma_1^2(y-\mu_2)^2 \right) \\&=&-\frac{1}{2\sigma_1^2\sigma_2^2} \left\{ \sigma_2^2(z^2-2zy-2z\mu_1+y^2+2y\mu_1+\mu_1^2) +\sigma_1^2(y^2-2y\mu_2+\mu_2^2) \right\} \\&=&-\frac{1}{2\sigma_1^2\sigma_2^2} \left\{ \sigma_2^2z^2 -2\sigma_2^2zy -2\sigma_2^2z\mu_1 +\sigma_2^2y^2 +2\sigma_2^2y\mu_1 +\sigma_2^2\mu_1^2 +\sigma_1^2y^2 -2\sigma_1^2y\mu_2 +\sigma_1^2\mu_2^2 \right\} \\&=& -\frac{1}{2\sigma_1^2\sigma_2^2} \left\{ \left( \sigma_1^2 +\sigma_2^2 \right)y^2 +\left( -2\sigma_2^2z +2\sigma_2^2\mu_1 -2\sigma_1^2\mu_2 \right)y +\sigma_2^2z^2 -2\sigma_2^2z\mu_1 +\sigma_2^2\mu_1^2 +\sigma_1^2\mu_2^2 \right\} \\&=& -\frac{1}{2\sigma_1^2\sigma_2^2} \left\{ \left( \sigma_1^2 +\sigma_2^2 \right)y^2 -2\left( \sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2 \right)y +\sigma_2^2z^2 -2\sigma_2^2z\mu_1 +\sigma_2^2\mu_1^2 +\sigma_1^2\mu_2^2 \right\} \\&=& -\frac{1}{2\sigma_1^2\sigma_2^2} \left\{ \left( \sigma_1^2 +\sigma_2^2 \right) \left(y^2 -2\frac{ \sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2 }{\sigma_1^2 +\sigma_2^2} y \right) +A -A +\sigma_2^2z^2 -2\sigma_2^2z\mu_1 +\sigma_2^2\mu_1^2 +\sigma_1^2\mu_2^2 \right\} \\&&\;\cdots\;平方完成させるための項Aを考える \\&=& -\frac{1}{2\sigma_1^2\sigma_2^2} \left\{ \left( \sigma_1^2 +\sigma_2^2 \right) \left(y^2 -2\frac{ \sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2 }{\sigma_1^2 +\sigma_2^2} y +\frac{1}{\sigma_1^2+\sigma_2^2}A \right) -A +\sigma_2^2z^2 -2\sigma_2^2z\mu_1 +\sigma_2^2\mu_1^2 +\sigma_1^2\mu_2^2 \right\} \\&=& -\frac{1}{2\sigma_1^2\sigma_2^2} \left[ \left( \sigma_1^2 +\sigma_2^2 \right) \left\{y^2 -2\frac{ \sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2 }{\sigma_1^2 +\sigma_2^2} y +\frac{1}{\sigma_1^2+\sigma_2^2} \frac{ (\sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2)^2 }{\sigma_1^2 +\sigma_2^2} \right\}-\frac{ (\sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2)^2 }{\sigma_1^2 +\sigma_2^2} +\sigma_2^2z^2 -2\sigma_2^2z\mu_1 +\sigma_2^2\mu_1^2 +\sigma_1^2\mu_2^2 \right] \\&&\;\cdots\;A=\frac{ (\sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2)^2 }{\sigma_1^2 +\sigma_2^2} \\&=& -\frac{1}{2\sigma_1^2\sigma_2^2} \left\{ \left( \sigma_1^2 +\sigma_2^2 \right) \left(y-\frac{ \sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2 }{\sigma_1^2 +\sigma_2^2} \right)^2 -\frac{ (\sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2)^2 }{\sigma_1^2 +\sigma_2^2} +\sigma_2^2z^2 -2\sigma_2^2z\mu_1 +\sigma_2^2\mu_1^2 +\sigma_1^2\mu_2^2 \right\} \\&=&-\frac{\sigma_1^2+\sigma_2^2}{2\sigma_1^2\sigma_2^2} \left(y-\frac{ \sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2 }{\sigma_1^2 +\sigma_2^2} \right)^2 -\frac{1}{2\sigma_1^2\sigma_2^2} \left\{ -\frac{ (\sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2)^2 }{\sigma_1^2+\sigma_2^2} +\sigma_2^2z^2 -2\sigma_2^2z\mu_1 +\sigma_2^2\mu_1^2 +\sigma_1^2\mu_2^2 \right\} \\&=& -f_2(\sigma_1^2,\sigma_2^2) \left\{ y-f_3(z,\mu_1,\mu_2,\sigma_1^2,\sigma_2^2) \right\}^2 -f_4(z,\mu_1,\mu_2,\sigma_1^2,\sigma_2^2) \;\cdots\;f_2,f_3,f_4はyを引数としない. \\&=& -f_2\left(y-f_3\right)^2-f_4 \;\cdots\;yが平方完成の中の一つだけになっている. \\ \\f_2(\sigma_1^2,\sigma_2^2)&=& \frac{\sigma_1^2+\sigma_2^2}{2\sigma_1^2\sigma_2^2} \\ \\f_3(z,\mu_1,\mu_2,\sigma_1^2,\sigma_2^2)&=&\frac{ \sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2 }{\sigma_1^2 +\sigma_2^2 } \\ \\f_4(z,\mu_1,\mu_2,\sigma_1^2,\sigma_2^2)&=&\frac{1}{2\sigma_1^2\sigma_2^2}\left\{ -\frac{ (\sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2)^2 }{\sigma_1^2 +\sigma_2^2} +\sigma_2^2z^2 -2\sigma_2^2z\mu_1 +\sigma_2^2\mu_1^2 +\sigma_1^2\mu_2^2 \right\} \\&=&\frac{1}{2\sigma_1^2\sigma_2^2}\left\{ -\frac{ (\sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2)^2 +(\sigma_1^2+\sigma_2^2)(\sigma_2^2z^2 -2\sigma_2^2z\mu_1 +\sigma_2^2\mu_1^2 +\sigma_1^2\mu_2^2) }{\sigma_1^2+\sigma_2^2} \right\} \\&=&\frac{1}{2\sigma_1^2\sigma_2^2} \frac{1}{\sigma_1^2+\sigma_2^2} \left\{ -(\sigma_2^2z -\sigma_2^2\mu_1 +\sigma_1^2\mu_2)^2 +(\sigma_1^2+\sigma_2^2)(\sigma_2^2z^2 -2\sigma_2^2z\mu_1 +\sigma_2^2\mu_1^2 +\sigma_1^2\mu_2^2) \right\} \\&=&\frac{1}{2\sigma_1^2\sigma_2^2} \frac{1}{\sigma_1^2+\sigma_2^2} \left\{ -\sigma_2^4z^2 -\sigma_2^4\mu_1^2 -\sigma_1^4\mu_2^2 +2\sigma_2^4z\mu_1 -2\sigma_1^2\sigma_2^2z\mu_2 +2\sigma_1^2\sigma_2^2\mu_1\mu_2 +(\sigma_1^2+\sigma_2^2)\sigma_2^2z^2 -2(\sigma_1^2+\sigma_2^2)\sigma_2^2z\mu_1 +(\sigma_1^2+\sigma_2^2)\sigma_2^2\mu_1^2 +(\sigma_1^2+\sigma_2^2)\sigma_1^2\mu_2^2 \right\} \\&=&\frac{1}{2\sigma_1^2\sigma_2^2} \frac{1}{\sigma_1^2+\sigma_2^2} \left( \color{red }{-\sigma_2^4 z^2} \color{green}{-\sigma_2^4 \mu_1^2} \color{blue }{-\sigma_1^4 \mu_2^2} \color{cyan }{+2\sigma_2^4 z\mu_1} \color{black}{-2\sigma_1^2 \sigma_2^2z\mu_2} \color{black}{+2\sigma_1^2 \sigma_2^2\mu_1\mu_2} \color{black}{+ \sigma_1^2 \sigma_2^2 z^2} \color{red }{+ \sigma_2^4 z^2} \color{black}{-2\sigma_1^2 \sigma_2^2 z\mu_1} \color{cyan}{-2\sigma_2^4 z\mu_1} \color{black}{+ \sigma_1^2 \sigma_2^2 \mu_1^2} \color{green}{+ \sigma_2^4 \mu_1^2} \color{blue }{+ \sigma_1^4 \mu_2^2} \color{black}{+ \sigma_1^2 \sigma_2^2 \mu_2^2} \right) \\&=&\frac{1}{2\sigma_1^2\sigma_2^2} \frac{1}{\sigma_1^2+\sigma_2^2} \left( \color{black}{-2\sigma_1^2 \sigma_2^2z\mu_2} \color{black}{+2\sigma_1^2 \sigma_2^2\mu_1\mu_2} \color{black}{+ \sigma_1^2 \sigma_2^2 z^2} \color{black}{-2\sigma_1^2 \sigma_2^2 z\mu_1} \color{black}{+ \sigma_1^2 \sigma_2^2 \mu_1^2} \color{black}{+ \sigma_1^2 \sigma_2^2 \mu_2^2} \right) \\&=&\frac{1}{2\sigma_1^2\sigma_2^2} \frac{1}{\sigma_1^2+\sigma_2^2} \left(\sigma_1^2 \sigma_2^2\right) \left( \color{black}{-2z\mu_2} \color{black}{+2\mu_1\mu_2} \color{black}{+ z^2} \color{black}{-2z\mu_1} \color{black}{+ \mu_1^2} \color{black}{+ \mu_2^2} \right) \\&=&\frac{1}{2\sigma_1^2\sigma_2^2} \frac{\sigma_1^2\sigma_2^2}{\sigma_1^2+\sigma_2^2} \left(z-\mu_1-\mu_2\right)^2 \\&=&\frac{\left(z-\mu_1-\mu_2\right)^2}{2\left(\sigma_1^2+\sigma_2^2\right)} \\&=&\frac{\left\{z-\left(\mu_1+\mu_2\right)\right\}^2}{2\left(\sigma_1^2+\sigma_2^2\right)} \end{eqnarray} $$
$$ \begin{eqnarray} p_{X+Y}(z)&=&\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \int_{-\infty}^{\infty} \mathrm{e}^{f_1(y,z,\mu_1,\mu_2,\sigma_1,\sigma_2)} \mathrm{d}y \\&=&\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \int_{-\infty}^{\infty}\mathrm{e}^{ -f_2 \left( y-f_3 \right)^2 -f_4} \mathrm{d}y \\&=&\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \int_{-\infty}^{\infty}\mathrm{e}^{ -f_2 \left( y-f_3 \right)^2 } \mathrm{e}^{-f_4} \mathrm{d}y \;\cdots\;A^{B+C}=A^BA^C \\&=&\frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \mathrm{e}^{-f_4} \int_{-\infty}^{\infty}\mathrm{e}^{ -f_2 \left( y-f_3 \right)^2 } \mathrm{d}y \;\cdots\;\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x \\&=& \frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \mathrm{e}^{-f_4} \int_{-\infty}^{\infty}\mathrm{e}^{-f_2\left(y-f_3\right)^2} \mathrm{d}y \\&=& \frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \mathrm{e}^{-f_4} \int_{-\infty}^{\infty}\mathrm{e}^{-u^2} \left(\frac{1}{\sqrt{f_2}}\right)\mathrm{d}u \\&&\;\cdots\;u=\sqrt{f_2}\left(y-f_3\right),\;\frac{\mathrm{d}u}{\mathrm{d}y}=\sqrt{f_2},\;\mathrm{d}y=\frac{1}{\sqrt{f_2}}\mathrm{d}u, \\&&\;\cdots\;y:-\infty \rightarrow \infty,\;u:-\infty \rightarrow \infty \\&=& \frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \mathrm{e}^{-f_4} \frac{1}{\sqrt{f_2}} \int_{-\infty}^{\infty}\mathrm{e}^{-u^2}\mathrm{d}u \\&=& \frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \frac{1}{\sqrt{f_2}} \mathrm{e}^{-f_4} \sqrt{\pi} \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/gaussian-integral.html}{\int_{-\infty}^{\infty}\mathrm{e}^{-u^2}\mathrm{d}u=\sqrt{\pi}} \\&=& \frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \frac{1}{\sqrt{\frac{\sigma_1^2+\sigma_2^2}{2\sigma_1^2\sigma_2^2}}} \mathrm{e}^{-\frac{\left\{z-\left(\mu_1+\mu_2\right)\right\}^2}{2\left(\sigma_1^2+\sigma_2^2\right)}} \sqrt{\pi} \\&=& \frac{1}{2\pi\sqrt{\sigma_1^2\sigma_2^2}} \sqrt{\frac{2\pi\sigma_1^2\sigma_2^2}{\sigma_1^2+\sigma_2^2}} \mathrm{e}^{-\frac{\left\{z-\left(\mu_1+\mu_2\right)\right\}^2}{2\left(\sigma_1^2+\sigma_2^2\right)}} \\&=& \frac{1}{\sqrt{2\pi\left(\sigma_1^2+\sigma_2^2\right)}} \mathrm{e}^{-\frac{\left\{z-\left(\mu_1+\mu_2\right)\right\}^2}{2\left(\sigma_1^2+\sigma_2^2\right)}} \;\cdots\;\mathrm{N}(\mu_1+\mu_2,\sigma_1^2+\sigma_2^2) (再生性) \end{eqnarray} $$ 再生性は,同じ確率分布族に含まれる確率分布\(F_1\),\(F_2\)に対して \(X_1\sim F_1\),\(X_2\sim F_2\)とする互いに独立な確率変数に対して, \(X_1+X_2\)がやはり同一の確率分布族に含まれる性質.

確率変数の標準化

確率変数の標準化

期待値(平均)が\(\mu\), 分散が\(\sigma^2\)の確率変数\(X\)

$$ \begin{eqnarray} \mathrm{E}\left[X\right]&=&\mu \\\mathrm{V}\left[X\right]&=&\sigma^2 \end{eqnarray} $$

確率変数の変換\(Z=\frac{X-\mu}{\sigma}\)

$$ \begin{eqnarray} \\Z&=&\frac{X-\mu}{\sigma} \end{eqnarray} $$

変換後の確率変数\(Z\)の期待値(平均)と分散

$$ \begin{eqnarray} \\\mathrm{E}\left[Z\right]&=&\mathrm{E}\left[\frac{X-\mu}{\sigma}\right] \\&=&\frac{1}{\sigma}\mathrm{E}\left[X-\mu\right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-expected-value.html}{\mathrm{E}\left[cX\right]=c\mathrm{E}\left[X\right]} \\&=&\frac{1}{\sigma}\left(\mathrm{E}\left[X\right]-\mu\right) \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-expected-value.html}{\mathrm{E}\left[X\pm t\right]=\mathrm{E}\left[X\right] \pm t} \\&=&\frac{1}{\sigma}\left(\mu-\mu\right) \;\cdots\;\mathrm{E}\left[X\right]=\mu \\&=&\frac{1}{\sigma}\left(0\right) \\&=&0 \\\mathrm{V}\left[Z\right]&=&\mathrm{V}\left[\frac{X-\mu}{\sigma}\right] \\&=&\frac{1}{\sigma^2}\mathrm{V}\left[X-\mu\right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-variance.html}{\mathrm{V}\left[cX\right]=c^2\mathrm{V}\left[X\right]} \\&=&\frac{1}{\sigma^2}\mathrm{V}\left[X\right] \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-variance.html}{\mathrm{V}\left[X\pm t\right]=\mathrm{V}\left[X\right]} \\&=&\frac{1}{\sigma^2}\sigma^2 \;\cdots\;\mathrm{V}\left[X\right]=\sigma^2 \\&=&1 \end{eqnarray} $$ \(X\)の分布によらず,\(X\)の期待値(平均)と分散が\(\mu\)と\(\sigma^2\)であることから\(Z\)の期待値(平均)と分散が\(0\), \(1\)と標準化される.

確率変数の変数変換 Z=X/Y

確率変数の変数変換 Z=X/Y

Z=X/Y

$$ \begin{eqnarray} p_{X/Y}(z) &=&\int \int \delta\left(z-\frac{x}{y}\right)f(x)g(y)\mathrm{d}x\mathrm{d}y \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/09/delta-function.html}{\int_{-\infty}^{\infty}\delta(x)f(x)\mathrm{d}x=f(0)} \\&&\;\cdots\;X=x,Y=yの同時確率f(x)g(x)のうちz-\left(\frac{x}{y}\right)=0を満たすものだけを足し合わせる. \\&=&\int \int \left|y\right|\delta\left(x-yz\right)\;f\left(x\right)g(y)\mathrm{d}x\mathrm{d}y \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/09/delta-function.html}{\delta(u(x))=\sum_{\alpha\in u^{-1}(0)}\frac{1}{\left|u^{\prime}(\alpha)\right|}\delta\left(x-\alpha\right)} \\&&\;\cdots\;u(x)=z-\frac{x}{y} \\&&\;\cdots\;u(x=\alpha)=0,\;\alpha=yz \\&&\;\cdots\;u^\prime=\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(z-\frac{x}{y})=-\frac{1}{y} \\&&\;\cdots\;\delta\left(z-\frac{x}{y}\right) =\frac{1}{|u^\prime(\alpha)|}\delta\left(x-\alpha\right) =\frac{1}{\left|-\frac{1}{y}\right|}\delta\left(x-yz\right) =\frac{1}{\left|\frac{1}{y}\right|}\delta\left(x-yz\right) =\left|y\right|\delta\left(x-yz\right) \\&=&\int |y|\delta\left(yz-yz\right)f\left(yz\right)g(y)\mathrm{d}y \;\cdots\;z=\frac{x}{y},x=yz \\&=&\int |y|\delta\left(0\right)f\left(yz\right)g(y)\mathrm{d}y \\&=&\int |y|f(yz)g(y)\mathrm{d}y \;\cdots\;\delta凾数の引数が全域で0なので,全域で|y|f(yz)g(y)を足し合わせる積分になる. \end{eqnarray} $$

標準正規分布同士の例

$$ \begin{eqnarray} f_X(x)&=&\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} \;\cdots\;N(0,1)\;(標準正規分布) \\g_Y(y)&=&\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}} \;\cdots\;N(0,1)\;(標準正規分布) \\p_{X/Y}(z)&=&\int \int \delta(z-\frac{x}{y})f_X(x)g_Y(y)\mathrm{d}x\mathrm{d}y \\&=&\int \int \delta(z-\frac{x}{y})\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}\mathrm{d}x\mathrm{d}y \\&=&\int_{-\infty}^{\infty} \left|y\right| \frac{1}{\sqrt{2\pi}}e^{-\frac{\left(yz\right)^2}{2}}\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}} \mathrm{d}y \;\cdots\;z=\frac{x}{y},x=yz \\&=&\frac{1}{2\pi}\int_{-\infty}^{\infty} \left|y\right| e^{-\frac{\left(yz\right)^2}{2}}e^{-\frac{y^2}{2}} \mathrm{d}y \;\cdots\;\int c f(x) \mathrm{d}x = c \int f(x) \mathrm{d}x \\&=&\frac{1}{2\pi}\int_{-\infty}^{\infty} \left|y\right| e^{-\frac{\left(yz\right)^2}{2}-\frac{y^2}{2}} \mathrm{d}y \;\cdots\;A^BA^C=A^{B+C} \\&=&\frac{1}{2\pi}\int_{-\infty}^{\infty} \left|y\right| e^{-\frac{y^2\left(z^2+1\right)}{2}} \mathrm{d}y \\&=&\frac{1}{2\pi}\left\{ \int_{-\infty}^{0} \left|y\right| e^{-\frac{y^2\left(z^2+1\right)}{2}} \mathrm{d}y + \int_{0}^{\infty} \left|y\right| e^{-\frac{y^2\left(z^2+1\right)}{2}} \mathrm{d}y \right\} \\&&\;\cdots\;\int_{A}^{B} f(x)\mathrm{d}x=\int_{A}^{C} f(x)\mathrm{d}x+\int_{C}^{B} f(x)\mathrm{d}x\;(ただしA\leq C \leq B) \\&=&\frac{1}{2\pi}\left\{ \int_{-\infty}^{0} -y e^{-\frac{y^2\left(z^2+1\right)}{2}} \mathrm{d}y + \int_{0}^{\infty} y e^{-\frac{y^2\left(z^2+1\right)}{2}} \mathrm{d}y \right\} \;\cdots\;\left|A\right|=\left\{ \begin{array} \\A&(A\geq0) \\-A&(A\lt0) \end{array} \right. \\&=&\frac{1}{2\pi}2\int_{0}^{\infty} y e^{-\frac{y^2\left(z^2+1\right)}{2}} \mathrm{d}y \\&&\;\cdots\;y=xが奇凾数でy=e^{-\frac{x^2\left(z^2+1\right)}{2}}が偶凾数なのでその積は奇凾数. \\&&\;\cdots\;負の区間(-\infty,0)と正の区間(0,\infty)で符号が逆なので全体としては偶凾数と同じに扱える. \\&&\;\cdots\;-f_{奇凾数(-\infty,0)}(x) + f_{奇凾数(0,\infty)}(x) = f_{偶凾数}(x)\mathrm{d}x \\&&\;\cdots\;偶凾数の(-a,a)での積分は正の区間(0,a)の2倍として計算できる. \\&&\;\cdots\;\int_{-a}^a f_{奇凾数}(x) \mathrm{d}x=2\int_0^a f_{偶凾数}(x)\mathrm{d}x \\&=&\frac{1}{\pi}\int_{0}^{\infty} y e^{-\frac{y^2\left(z^2+1\right)}{2}} \mathrm{d}y \\&=&\frac{1}{\pi}\int_{0}^{\infty} \color{red}{y} e^{-t}\cdot\frac{1}{\left(z^2+1\right)\color{red}{y}}\mathrm{d}t \\&&\;\cdots\;t=\frac{y^2\left(z^2+1\right)}{2},\frac{\mathrm{d}t}{\mathrm{d}y}=\frac{\left(z^2+1\right)}{2}2y=\left(z^2+1\right)y,\mathrm{d}y=\frac{1}{\left(z^2+1\right)y}\mathrm{d}t \\&=&\frac{1}{\pi}\frac{1}{\left(z^2+1\right)}\int_{0}^{\infty} e^{-t} \mathrm{d}t \;\cdots\;\int c f(x) \mathrm{d}x = c \int f(x) \mathrm{d}x \\&=&\frac{1}{\pi}\frac{1}{\left(z^2+1\right)}\cdot1 \;\cdots\;\int_{0}^{\infty} e^{-t} \mathrm{d}t=\left[-e^{-t}\right]_{0}^{\infty}=\left[\left(-e^{-\infty}\right)-\left(-e^0\right)\right]=\left[0+1\right]=1 \\&=&\frac{1}{\pi}\frac{1}{\left(z^2+1\right)}\;\cdots\;標準コーシー分布に等しい. \\f(x;x_0, \gamma)&=&\frac{1}{\pi}\frac{\gamma}{(x-x_0)^2+\gamma^2}\;\cdots\;コーシー分布(Cauchy\;distribution)の確率密度凾数 \\f(x;0, 1)&=&\frac{1}{\pi}\frac{1}{(x-0)^2+1^2}\;\cdots\;標準コーシー分布の確率密度凾数 \\&=&\frac{1}{\pi}\frac{1}{x^2+1} \end{eqnarray} $$

指数分布同士の例

$$ \begin{eqnarray} f_X(x)&=&\lambda e^{-\lambda x} (x \geq 0, \lambda \gt 0) \\g_Y(y)&=&\lambda e^{-\lambda y} (x \geq 0, \lambda \gt 0) \\p_{X/Y}(z)&=&\int \int \delta(x-\frac{z}{y})f_X(x)g_Y(y)\mathrm{d}x\mathrm{d}y \\&=&\int \int \delta(x-\frac{z}{y})\lambda e^{-\lambda x} \lambda e^{-\lambda y} \mathrm{d}x\mathrm{d}y \\&=&\int_0^{\infty} \left|y\right|\lambda e^{-\lambda yz} \lambda e^{-\lambda y}\mathrm{d}y \;\cdots\;z=\frac{x}{y},x=yz \\&=&\lambda^2 \int_0^{\infty} \left|y\right|e^{-\lambda yz} e^{-\lambda y}\mathrm{d}y \;\cdots\;\int c f(x) \mathrm{d}x = c \int f(x) \mathrm{d}x \\&=&\lambda^2 \int_0^{\infty} \left|y\right|e^{-\lambda yz-\lambda y}\mathrm{d}y \;\cdots\;A^BA^C=A^{B+C} \\&=&\lambda^2 \int_0^{\infty} \left|y\right|e^{-\lambda y\left(z+1\right)}\mathrm{d}y \\&=&\lambda^2 \int_0^{\infty} ye^{-\lambda y\left(z+1\right)}\mathrm{d}y \;\cdots\;y \gt 0\;(積分範囲より) \\&=&\lambda^2 \int_0^{\infty} \frac{t}{\lambda\left(z+1\right)}e^{-t}\frac{1}{\lambda\left(z+1\right)}\mathrm{d}t \\&&\;\cdots\;t=\lambda\left(z+1\right)y, y=\frac{t}{\lambda\left(z+1\right)} \\&&\;\cdots\;\frac{\mathrm{d}t}{\mathrm{d}y}=\lambda\left(z+1\right),\mathrm{d}y=\frac{1}{\lambda\left(z+1\right)}\mathrm{d}t \\&=&\lambda^2 \left\{\frac{1}{\lambda\left(z+1\right)}\right\}^2\int_0^{\infty} te^{-t}\mathrm{d}t \;\cdots\;\int c f(x) \mathrm{d}x = c \int f(x) \mathrm{d}x \\&=&\frac{1}{\left(z+1\right)^2}\int_0^{\infty} te^{-t}\mathrm{d}t \\&=&\frac{1}{\left(z+1\right)^2}\int_0^{\infty} t\left(-e^{-t}\right)^\prime\mathrm{d}t \;\cdots\;\left(-e^{-t}\right)^\prime=e^{-t} \\&=&\frac{1}{\left(z+1\right)^2}\left\{\left[-te^{-t}\right]_0^{\infty}-\int_0^{\infty} -e^{-t}\mathrm{d}t\right\} \;\cdots\;\int_a^b f^\prime(x)g(x) \mathrm{d}x=\left[f(x)g(x)\right]_a^b -\int_a^b f(x)g^\prime(x)\mathrm{d}x \\&=&\frac{1}{\left(z+1\right)^2}\left\{\left[-(\infty)e^{-\infty}-\left(-0e^{-0}\right)\right]+\int_0^{\infty} e^{-t}\mathrm{d}t\right\} \\&=&\frac{1}{\left(z+1\right)^2}\left(0+1\right) \;\cdots\;\int_{0}^{\infty} e^{-t} \mathrm{d}t=\left[-e^{-t}\right]_{0}^{\infty}=\left[\left(-e^{-\infty}\right)-\left(-e^0\right)\right]=\left[0+1\right]=1 \\&=&\frac{1}{\left(z+1\right)^2} \end{eqnarray} $$

確率変数の変数変換 Z=XY

確率変数の変数変換 Z=XY

Z=XY

$$ \begin{eqnarray} p_{XY}(z) &=&\int \int \delta(z-xy)f(x)g(y)\mathrm{d}x\mathrm{d}y \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/09/delta-function.html}{\int_{-\infty}^{\infty}\delta(x)f(x)\mathrm{d}x=f(0)} \\&&\;\cdots\;X=x,Y=yの同時確率f(x)g(x)のうちz-(xy)=0を満たすものだけを足し合わせる. \\&=&\int \int \frac{1}{|y|}\delta\left(x-\frac{z}{y}\right)\;f\left(x\right)g(y)\mathrm{d}x\mathrm{d}y \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/09/delta-function.html}{\delta(u(x))=\sum_{\alpha\in u^{-1}(0)}\frac{1}{\left|u^{\prime}(\alpha)\right|}\delta\left(x-\alpha\right)} \\&&\;\cdots\;u(x)=z-xy \\&&\;\cdots\;u(x=\alpha)=0,\;\alpha=\frac{z}{y} \\&&\;\cdots\;u^\prime=\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}(z-xy)=-y \\&&\;\cdots\;\delta\left(z-xy\right) =\frac{1}{|u^\prime(\alpha)|}\delta\left(x-\alpha\right) =\frac{1}{|-y|}\delta\left(x-\frac{z}{y}\right) =\frac{1}{|y|}\delta\left(x-\frac{z}{y}\right) \\&=&\int \frac{1}{|y|}\delta\left(\frac{z}{y}-\frac{z}{y}\right)\;f\left(\frac{z}{y}\right)g(y)\mathrm{d}y \;\cdots\;z=xy,x=\frac{z}{y} \\&=&\int \frac{1}{|y|}\delta\left(0\right)\;f\left(\frac{z}{y}\right)g(y)\mathrm{d}y \\&=&\int \frac{1}{|y|}f\left(\frac{z}{y}\right)g(y)\mathrm{d}y \;\cdots\;\delta凾数の引数が全域で0なので,全域で\frac{1}{|y|}f\left(\frac{z}{y}\right)g(y)を足し合わせる積分になる. \end{eqnarray} $$

標準正規分布同士の例

$$ \begin{eqnarray} f_X(x)&=&\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} \;\cdots\;N(0,1)\;(標準正規分布) \\g_Y(y)&=&\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}} \;\cdots\;N(0,1)\;(標準正規分布) \\p_{XY}(z)&=&\int \int \delta(z-xy)f_X(x)g_Y(y)\mathrm{d}x\mathrm{d}y \\&=&\int \int \delta(z-xy)\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}\mathrm{d}x\mathrm{d}y \\&=&\int_{-\infty}^{\infty} \frac{1}{|y|} \frac{1}{\sqrt{2\pi}}e^{-\frac{\left(\frac{z}{y}\right)^2}{2}}\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}} \mathrm{d}y \;\cdots\;z=xy,x=\frac{z}{y} \\&=&\frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{1}{|y|} e^{-\frac{\left(\frac{z}{y}\right)^2}{2}}e^{-\frac{y^2}{2}} \mathrm{d}y \;\cdots\;\int c f(x) \mathrm{d}x = c \int f(x) \mathrm{d}x \\&=&\frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{1}{|y|} e^{-\frac{1}{2}\left(\frac{z^2}{y^2}+y^2\right)} \mathrm{d}y \\&=&\frac{1}{2\pi}\left\{ \int_{-\infty}^{0} \frac{1}{|y|} e^{-\frac{1}{2}\left(\frac{z^2}{y^2}+y^2\right)} \mathrm{d}y + \int_{0}^{\infty} \frac{1}{|y|} e^{-\frac{1}{2}\left(\frac{z^2}{y^2}+y^2\right)} \mathrm{d}y \right\} \\&&\;\cdots\;\int_{A}^{B} f(x)\mathrm{d}x=\int_{A}^{C} f(x)\mathrm{d}x+\int_{C}^{B} f(x)\mathrm{d}x\;(ただしA\leq C \leq B) \\&=&\frac{1}{2\pi}\left\{ \int_{-\infty}^{0} -\frac{1}{y} e^{-\frac{1}{2}\left(\frac{z^2}{y^2}+y^2\right)} \mathrm{d}y + \int_{0}^{\infty} \frac{1}{y} e^{-\frac{1}{2}\left(\frac{z^2}{y^2}+y^2\right)} \mathrm{d}y \right\} \\&&\;\cdots\;\frac{1}{|y|}=-\frac{1}{y}\;(y\lt0),\frac{1}{|y|}=\frac{1}{y}\;(0\lt y) \\&=&\frac{1}{2\pi}2\int_{0}^{\infty} \frac{1}{y} e^{-\frac{1}{2}\left(\frac{z^2}{y^2}+y^2\right)} \mathrm{d}y \\&&\;\cdots\;y=\frac{1}{x}が奇凾数でy=e^{-\frac{1}{2}\left(\frac{z^2}{x^2}+x^2\right)}が偶凾数なのでその積は奇凾数. \\&&\;\cdots\;負の区間(-\infty,0)と正の区間(0,\infty)で符号が逆なので全体としては偶凾数と同じに扱える. \\&&\;\cdots\;-f_{奇凾数(-\infty,0)}(x) + f_{奇凾数(0,\infty)}(x) = f_{条件付きで偶凾数}(x)\mathrm{d}x \\&&\;\cdots\;偶凾数の(-a,a)での積分は正の区間(0,a)の2倍として計算できる. \\&&\;\cdots\;\int_{-a}^a f_{奇凾数}(x) \mathrm{d}x=2\int_0^a f_{偶凾数}(x)\mathrm{d}x \\&=&\frac{1}{\pi} \int_{0}^{\infty} \color{red}{\frac{1}{y}} e^{-\frac{1}{2}\left(\frac{z^2}{y^2}+y^2\right)} \color{red}{\mathrm{d}y} \\&=&\frac{1}{\pi} \int_{0}^{\infty} e^{-\frac{1}{2}\left\{\frac{z^2}{y^2}+y^2\right\}} \frac{\mathrm{d}t}{2t} \\&&\;\cdots\;t=\frac{y^2}{2},y^2=2t,\;y:0\rightarrow\infty,t:0\rightarrow\infty \\&&\;\cdots\;\frac{\mathrm{d}t}{\mathrm{d}y}=y=\frac{y^2}{y}=\frac{2t}{y},\frac{\mathrm{d}y}{y}=\frac{\mathrm{d}t}{2t} \\&=&\frac{1}{\pi}\frac{1}{2}\int_{0}^{\infty} \frac{1}{t} e^{-\frac{1}{2}\left\{\frac{z^2}{2t}+2t\right\}} \mathrm{d}t \\&=&\frac{1}{2\pi}\int_{0}^{\infty} \frac{1}{t} e^{-t-\frac{z^2}{4t}} \mathrm{d}t \\&&\;\cdots\;K_n(z)=\frac{1}{2}\left(\frac{z}{2}\right)^n\int_0^{\infty} t^{-n-1} e^{-t-\frac{z^2}{4t}} \mathrm{d}t\;(第二種変形ベッセル凾数) \\&&\;\cdots\;K_0(z)=\frac{1}{2}\int_0^{\infty} t^{-1} e^{-t-\frac{z^2}{4t}} \mathrm{d}t \\&=&\frac{1}{\pi}K_0(z) \end{eqnarray} $$

指数分布同士の例

$$ \begin{eqnarray} f_X(x)&=&\lambda e^{-\lambda x} (x \geq 0, \lambda \gt 0) \\g_Y(y)&=&\lambda e^{-\lambda y} (x \geq 0, \lambda \gt 0) \\p_{XY}(z)&=&\int \int \delta(x-xy)f_X(x)g_Y(y)\mathrm{d}x\mathrm{d}y \\&=&\int \int \delta(x-xy)\lambda e^{-\lambda x} \lambda e^{-\lambda y} \mathrm{d}x\mathrm{d}y \\&=&\int_0^{\infty} \frac{1}{|y|}\lambda e^{-\lambda \left(\frac{z}{y}\right)} \lambda e^{-\lambda y}\mathrm{d}y \;\cdots\;z=xy,x=\frac{z}{y} \\&=&\lambda^2\int_0^{\infty} \frac{1}{|y|} e^{-\lambda \left(\frac{z}{y}\right)} e^{-\lambda y} \mathrm{d}y \;\cdots\;\int c f(x) \mathrm{d}x = c \int f(x) \mathrm{d}x \\&=&\lambda^2\int_0^{\infty} \frac{1}{|y|} e^{-\lambda \left(\frac{z}{y}\right)-\lambda y} \mathrm{d}y \;\cdots\;A^BA^C=A^{B+C} \\&=&\lambda^2\int_0^{\infty} \frac{1}{|y|} e^{-\lambda \left(\frac{z}{y}+y\right)} \mathrm{d}y \\&=&\lambda^2\int_0^{\infty} \frac{1}{|\frac{t}{\lambda}|} e^{-\lambda \left(\frac{z}{\frac{t}{\lambda}}+\frac{t}{\lambda}\right)} \frac{1}{\lambda}\mathrm{d}t \\&&\;\cdots\;t=\lambda y, y=\frac{t}{\lambda}, \frac{\mathrm{d}y}{\mathrm{d}t}=\frac{1}{\lambda},\mathrm{d}y=\frac{1}{\lambda}\mathrm{d}t,\;y:0\rightarrow\infty, t:0\rightarrow\infty \\&=&\lambda^2\int_0^{\infty} \frac{|\lambda|}{|t|} e^{-\lambda^2\frac{z}{t}-t} \frac{1}{\lambda}\mathrm{d}t \\&=&\lambda^2\int_0^{\infty} \frac{|\lambda|}{\lambda}\frac{1}{|t|} e^{-\lambda^2\frac{z}{t}-t} \mathrm{d}t \\&=&\lambda^2\int_0^{\infty} \frac{1}{|t|} e^{-\lambda^2\frac{z}{t}-t} \mathrm{d}t \;\cdots\;\lambda \gt 0 \\&=&\lambda^2\int_0^{\infty} \frac{1}{t} e^{-\lambda^2\frac{z}{t}-t} \mathrm{d}t \;\cdots\;t \gt 0\;(積分範囲より) \\&=&2\lambda^2K_0(2\lambda\sqrt{z}) \\&&\;\cdots\;K_n(z)=\frac{1}{2}\left(\frac{z}{2}\right)^n\int_0^{\infty} t^{-n-1} e^{-t-\frac{z^2}{4t}} \mathrm{d}t\;(第二種変形ベッセル凾数) \\&&\;\cdots\;K_0(z)=\frac{1}{2}\int_0^{\infty} t^{-1} e^{-t-\frac{z^2}{4t}} \mathrm{d}t \end{eqnarray} $$

確率変数の変数変換 Z=X+Y

確率変数の変数変換 Z=X+Y

Z=X+Y

$$ \begin{eqnarray} p_{X+Y}(z) &=&\int\int \delta(z-(x+y))f(x)g(y)\mathrm{d}x\mathrm{d}y \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/09/delta-function.html}{\int_{-\infty}^{\infty}\delta(x)f(x)\mathrm{d}x=f(0)} \\&&\;\cdots\;X=x,Y=yの同時確率f(x)g(x)のうちz-(x+y)=0を満たすものだけを足し合わせる. \\&=&\int\int \delta(z-((z-y)+y))f(z-y)g(y)\mathrm{d}y \;\cdots\;z=x+y,x=z-y \\&=&\int \delta(0)f(z-y)g(y)\mathrm{d}y \\&=&\int f(z-y)g(y)\mathrm{d}y \;\cdots\;\delta凾数の引数が全域で0なので,全域でf(z-y)g(y)を足し合わせる積分になる. \end{eqnarray} $$

標準正規分布同士の例

$$ \begin{eqnarray} f_X(x)&=&\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} \;\cdots\;N(0,1)\;(標準\href{https://shikitenkai.blogspot.com/2019/06/binomial-distributionnormal-distribution.html}{正規分布}) \\g_Y(y)&=&\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}} \;\cdots\;N(0,1)\;(標準\href{https://shikitenkai.blogspot.com/2019/06/binomial-distributionnormal-distribution.html}{正規分布}) \\p_{X+Y}(z)&=&\int\int \delta(z-(x+y))f_X(x)g_Y(y)\mathrm{d}x\mathrm{d}y \\&=&\int\int \delta(z-(x+y))\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}\mathrm{d}x\mathrm{d}y \\&=&\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}}e^{-\frac{(z-y)^2}{2}}\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}} \mathrm{d}y \;\cdots\;z=x+y,x=z-y \\&=&\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-\frac{(z-y)^2}{2}}e^{-\frac{y^2}{2}} \mathrm{d}y \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/continuous-random-variable-expected.html}{\int c f(x) \mathrm{d}x = c \int f(x) \mathrm{d}x} \\&=&\frac{1}{2\pi}\int e^{-\frac{(z-y)^2}{2}-\frac{y^2}{2}} \mathrm{d}y \;\cdots\;A^BA^C=A^{B+C} \\&=&\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-\frac{(z-y)^2+y^2}{2}} \mathrm{d}y \\&=&\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-\frac{z^2-2yz+y^2+y^2}{2}}\mathrm{d}y \\&=&\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-\frac{z^2-2yz+2y^2}{2}}\mathrm{d}y \\&=&\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-\frac{z^2}{2}+yz-y^2}\mathrm{d}y \\&=&\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-\frac{z^2}{2}\color{red}{+\frac{z^2}{4}-\frac{z^2}{4}}\color{black}{+yz-y^2}\mathrm{d}y} \\&=&\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-\frac{z^2}{4}-(\frac{z^2}{4}-yz+y^2)}\mathrm{d}y \\&=&\frac{1}{2\pi}\int_{-\infty}^{\infty} e^{-\frac{z^2}{4}}e^{-(\frac{z^2}{4}-yz+y^2)}\mathrm{d}y \\&=&\frac{1}{2\pi}e^{-\frac{z^2}{4}}\int_{-\infty}^{\infty} e^{-(\frac{z}{2}-y)^2}\mathrm{d}y \\&=&\frac{1}{2\pi}e^{-\frac{z^2}{4}}\int_{\infty}^{-\infty} e^{-t^2}(-1)\mathrm{d}t \\&&\;\cdots\;t=\frac{z}{2}-y,\frac{\mathrm{d}t}{\mathrm{d}y}=-1,\mathrm{d}y=-\mathrm{d}t,\;y:-\infty\rightarrow\infty, t:\infty\rightarrow -\infty \\&=&\frac{1}{2\pi}e^{-\frac{z^2}{4}}(-1)\int_{\infty}^{-\infty} e^{-t^2}\mathrm{d}t \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/continuous-random-variable-expected.html}{\int c f(x) \mathrm{d}x = c \int f(x) \mathrm{d}x} \\&=&\frac{1}{2\pi}e^{-\frac{z^2}{4}}(-1)\left(-\int_{-\infty}^{\infty} e^{-t^2}\mathrm{d}t\right) \;\cdots\;\int_{A}^{B} f(x)\mathrm{d}x=-\int_{B}^{A} f(x)\mathrm{d}x \\&=&\frac{1}{2\pi}e^{-\frac{z^2}{4}}\int_{-\infty}^{\infty} e^{-t^2}\mathrm{d}t \\&=&\frac{1}{2\pi}e^{-\frac{z^2}{4}}\sqrt{\pi} \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/gaussian-integral.html}{\int_{-\infty}^{\infty} e^{-t^2}\mathrm{d}y=\sqrt{\pi}} \\&=&\frac{1}{2\sqrt{\pi}}e^{-\frac{z^2}{4}} \\&=&\frac{1}{\sqrt{2\pi\cdot2}}e^{-\frac{z^2}{2\cdot2}} \;\cdots\;N(0,2) \end{eqnarray} $$

指数分布同士の例

$$ \begin{eqnarray} f_X(x)&=&\lambda e^{-\lambda x} (x \geq 0, \lambda \gt 0) \;\cdots\;Exp(\lambda)\;(\href{https://shikitenkai.blogspot.com/2020/05/blog-post_0.html}{指数分布}) \\g_Y(y)&=&\lambda e^{-\lambda y} (x \geq 0, \lambda \gt 0) \;\cdots\;Exp(\lambda)\;(\href{https://shikitenkai.blogspot.com/2020/05/blog-post_0.html}{指数分布}) \\p_{X+Y}(z)&=&\int \int \delta(z-(x+y))f_X(x)g_Y(y)\mathrm{d}x\mathrm{d}y \\&=&\int \int \delta(z-(x+y))\lambda e^{-\lambda x} \lambda e^{-\lambda y} \mathrm{d}x\mathrm{d}y \\&=&\int_0^z \lambda e^{-\lambda (z-y)} \lambda e^{-\lambda y}\mathrm{d}y \;\cdots\;z=x+y,x=z-y \\&=&\lambda^2\int_0^z e^{-\lambda (z-y)} e^{-\lambda y}\mathrm{d}y \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/continuous-random-variable-expected.html}{\int c f(x) \mathrm{d}x = c \int f(x) \mathrm{d}x} \\&=&\lambda^2\int_0^z e^{-\lambda (z-y)-\lambda y}\mathrm{d}y \;\cdots\;A^BA^C=A^{B+C} \\&=&\lambda^2\int_0^z e^{-\lambda (z-y+y)}\mathrm{d}y \\&=&\lambda^2\int_0^z e^{-\lambda z}\mathrm{d}y \\&=&\lambda^2 e^{-\lambda z}\int_0^z\mathrm{d}y \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/continuous-random-variable-expected.html}{\int c f(x) \mathrm{d}x = c \int f(x) \mathrm{d}x} \\&=&\lambda^2 e^{-\lambda z}\left[y\right]_0^z \\&=&\lambda^2 e^{-\lambda z}\left[z-0\right] \\&=&\lambda^2 z e^{-\lambda z} \\&=&\Gamma(2, \lambda) \end{eqnarray} $$ $$ \begin{eqnarray} \Gamma(\alpha, \lambda)&=&\frac{1}{\Gamma(\alpha)}\lambda^{\alpha} x^{\alpha-1} e^{-\lambda x}\;(\href{https://shikitenkai.blogspot.com/2020/05/n.html}{ガンマ分布;\;gamma\;disribution}) \\\Gamma(1, \lambda)&=&\frac{1}{\Gamma(1)}\lambda^1 x^{1-1} e^{-\lambda x}=\frac{1}{1}\lambda e^{-\lambda x}=\lambda e^{-\lambda x} \\&&\;\cdots\;\Gamma(1)=(1-1)!=0!=1 \\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/05/blog-post_0.html}{指数分布} \\\Gamma(2, \lambda)&=&\frac{1}{\Gamma(2)}\lambda^2 x^{2-1} e^{-\lambda x}=\frac{1}{1}\lambda^2xe^{-\lambda x}=\lambda^2xe^{-\lambda x} \\&&\;\cdots\;\Gamma(2)=(2-1)!=1!=1 \\&&\;\cdots\;指数分布に従う確率変数同士の和の分布,\;\alpha=2のガンマ分布 \end{eqnarray} $$

母集団の尖度

母集団(平均 \(\mu\), 分散 \(\sigma^2\))の尖度\(\beta_2\)

\begin{array}{rcl} \beta_2 &=& \frac{\mathrm{E}\left[\left(X-\mu\right)^4\right]}{ \mathrm{V}\left[ X \right]^{\frac{4}{2}} }-3\;\cdots\;尖度(せんど,\;kurtosis)\\ &=& \frac{\mu_4}{\sigma^4}-3\\ && \,\dotso\, \mu_4 = \mathrm{E}\left[\left(X-\mu\right)^3\right] \,\,\, \left(確率変数Xの母平均\muまわりの4次モーメント=確率変数Xの4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4次のモーメント\(\mu_4\)を尖度\(\beta_2\)で表す

\begin{array}{rcl} \mu_4 &=& (\beta_2-3)\sigma^4\\ \end{array}

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}

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)

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

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

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

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

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

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

$$\begin{array}{rcl} E[XY]&=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}(x_i y_j)f_{XY}(x_i, y_j)\\ &=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}x_iy_jf_{X}(x_i)f_{Y}(y_i) \,\dotso\,X,Yが独立\,f_{XY}(x, y)=f_{X}(x)f_{Y}(y)\\ &=&\displaystyle\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}x_if_{X}(x_i)y_jf_{Y}(y_i)\\ &=&\displaystyle\left(\sum_{i=1}^{\infty}x_if_{X}(x_i)\right)\left(\sum_{j=1}^{\infty} y_jf_{Y}(y_i)\right) \,\dotso\,\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_ib_j=\sum_{i=1}^{\infty}a_i\sum_{j=1}^{\infty}b_j\\ &=&E[X]E[Y] \end{array}$$