共分散(covariance)

共分散(covariance)

$$\begin{gathered} \begin{array}{rcl}\mathrm{Cov}[X,Y]& =& \mathrm{E}\left[(X-\mathrm{E}\left[X\right])(Y-\mathrm{E}\left[Y\right])\right]\cdots \mathrm{共}\mathrm{分}\mathrm{散}(covariance)\\ \mathrm{Cov}[c_0{X}_i,c_1{X}_j]& =& \mathrm{E}\left[(c_0{X}_i-\mathrm{E}\left[c_0{X}_i\right])(c_1{X}_j-\mathrm{E}\left[c_1{X}_j\right])\right]\cdots \mathrm{Cov}[X,Y]=\mathrm{E}\left[(X-\mathrm{E}\left[X\right])(Y-\mathrm{E}\left[Y\right])\right]\\ & =& \mathrm{E}\left[(c_0{X}_i-c_0\mathrm{E}\left[\overline{X}\right])(c_1{X}_j-c_1\mathrm{E}\left[\overline{X}\right])\right]\cdots \mathrm{E}\left[cX\right]=c\mathrm{E}\left[X\right]\\ & =& \mathrm{E}\left[c_0(X_i-\mathrm{E}\left[\overline{X}\right])c_1(X_j-\mathrm{E}\left[\overline{X}\right])\right]\\ & =& c_0{c}_1\mathrm{E}\left[(X_i-\mathrm{E}\left[\overline{X}\right])(X_j-\mathrm{E}\left[\overline{X}\right])\right]\cdots \mathrm{E}\left[cX\right]=c\mathrm{E}\left[X\right]\\ & =& c_0{c}_1\mathrm{Cov}[X_i,X_j]\\ \mathrm{V}\left[\sum _{i=1}^n{X}_i\right]& =& \mathrm{V}[X_1+\cdots +X_i+\cdots +X_n]\\ & =& \mathrm{E}\left[{\{(X_1+\cdots +X_i+\cdots +X_n)-\mathrm{E}[X_1+\cdots +X_i+\cdots +X_n]\}}^2\right]\\ & =& \mathrm{E}\left[{\{X_1+\cdots +X_i+\cdots +X_n-\mathrm{E}\left[X_1\right]-\cdots -\mathrm{E}\left[X_i\right]-\cdots -\mathrm{E}\left[X_n\right]\}}^2\right]\\ & =& \mathrm{E}\left[{\{(X_1-\mathrm{E}\left[X_1\right])+\cdots +(X_i-\mathrm{E}\left[X_i\right])+\cdots +(X_n-\mathrm{E}\left[X_n\right])\}}^2\right]\\ & =& \mathrm{E}[{(X_1-\mathrm{E}\left[X_1\right])}^2+\cdots +(X_1-\mathrm{E}\left[X_1\right])(X_j-\mathrm{E}\left[X_j\right])+\cdots +(X_1-\mathrm{E}\left[X_1\right])(X_n-\mathrm{E}\left[X_n\right]) \\ +\cdots +(X_i-\mathrm{E}\left[X_i\right])(X_1-\mathrm{E}\left[X_1\right])+\cdots +(X_i-\mathrm{E}\left[X_i\right])(X_j-\mathrm{E}\left[X_j\right])+\cdots +(X_i-\mathrm{E}\left[X_i\right])(X_n-\mathrm{E}\left[X_n\right]) \\ +\cdots +(X_n-\mathrm{E}\left[X_n\right])(X_1-\mathrm{E}\left[X_1\right])+\cdots +(X_n-\mathrm{E}\left[X_n\right])(X_j-\mathrm{E}\left[X_j\right])+\cdots +{(X_n-\mathrm{E}\left[X_n\right])}^2]\\ & =& \mathrm{E}\left[{(X_1-\mathrm{E}\left[X_1\right])}^2\right]+\cdots +\mathrm{E}\left[(X_1-\mathrm{E}\left[X_1\right])(X_j-\mathrm{E}\left[X_j\right])\right]+\cdots +\mathrm{E}\left[(X_1-\mathrm{E}\left[X_1\right])(X_n-\mathrm{E}\left[X_n\right])\right]\\ & & +\cdots +\mathrm{E}\left[(X_i-\mathrm{E}\left[X_i\right])(X_1-\mathrm{E}\left[X_1\right])\right]+\cdots +\mathrm{E}\left[(X_i-\mathrm{E}\left[X_i\right])(X_j-\mathrm{E}\left[X_j\right])\right]+\cdots +\mathrm{E}\left[(X_i-\mathrm{E}\left[X_i\right])(X_n-\mathrm{E}\left[X_n\right])\right]\\ & & +\cdots +\mathrm{E}\left[(X_n-\mathrm{E}\left[X_n\right])(X_1-\mathrm{E}\left[X_1\right])\right]+\cdots +\mathrm{E}\left[(X_n-\mathrm{E}\left[X_n\right])(X_j-\mathrm{E}\left[X_j\right])\right]+\cdots +\mathrm{E}\left[{(X_n-\mathrm{E}\left[X_n\right])}^2\right]\\ & & \cdots \mathrm{E}[X+Y]=\mathrm{E}\left[X\right]+\mathrm{E}\left[Y\right]\\ & =& \mathrm{Cov}[X_1,X_1]+\cdots +\mathrm{Cov}[X_1,X_j]+\cdots +\mathrm{Cov}[X_1,X_n]\\ & & +\cdots +\mathrm{Cov}[X_i,X_1]+\cdots +\mathrm{Cov}[X_i,X_j]+\cdots +\mathrm{Cov}[X_i,X_n]\\ & & +\cdots +\mathrm{Cov}[X_n,X_1]+\cdots +\mathrm{Cov}[X_n,X_j]+\cdots +\mathrm{Cov}[X_n,X_n]\\ & =& \sum _{i=1}^n\sum _{j=1}^n\mathrm{Cov}[X_i,X_j]\\ & & \cdots \mathrm{E}\left[(X_i-\mathrm{E}\left[X_i\right])(X_j-\mathrm{E}\left[X_j\right])\right]=\mathrm{Cov}[X_i,X_j]\\ & =& \sum _{i=1}^n\mathrm{Cov}[X_i,X_i]+2\sum _{i<j}\mathrm{Cov}[X_i,X_j]\\ & & \cdots \mathrm{Cov}\mathrm{の}\mathrm{項}\mathrm{は}\mathrm{Cov}[X_i,X_j]=\mathrm{Cov}[X_j,X_i]\mathrm{で}i<j\mathrm{と}i>j\mathrm{で}2\mathrm{回}\mathrm{あ}\mathrm{る}\mathrm{の}\mathrm{で}2\mathrm{倍}\\ & =& \sum _{i=1}^n\mathrm{V}\left[X_i\right]+2\sum _{i<j}\mathrm{Cov}[X_i,X_j]\\ & & \cdots \mathrm{Cov}[X_i,X_i]=\mathrm{V}\left[X_i\right]\end{array} \end{gathered}$$

単回帰における最小二乗推定量の分散(variance)・共分散(covariance)

単回帰における最小二乗推定量$\hat{\alpha },\hat{\beta }$の分散(variance)・共分散(covariance)

単回帰における観測値$y_i$の分散・共分散について

$$\begin{array}{rcl}{y}_i& =& \alpha +\beta x_i+{ϵ}_i(i=1,\cdots ,n)\\ \{{ϵ}_i|i=1,\cdots ,n\}& :& {ϵ}_i\stackrel{iid}{\sim }N(0,{\sigma }^2)\\ & & \cdots \mathrm{独}\mathrm{立}\mathrm{同}\mathrm{一}\mathrm{分}\mathrm{布}(independentandidenticallydistributed;IID,i.i.d.,iid)\\ & & \cdots \mathrm{E}\left[{ϵ}_i\right]=0,\mathrm{V}\left[{ϵ}_i\right]={\sigma }^2,\mathrm{互}\mathrm{い}\mathrm{に}\mathrm{独}\mathrm{立}(\mathrm{Cov}[{ϵ}_i,{ϵ}_j]=\{\begin{array}{}\mathrm{V}\left[{ϵ}_i\right]={\sigma }^2& (i=j)\\ 0& (i\ne j)\end{array})\\ \mathrm{V}\left[y_i\right]& =& \mathrm{V}[\alpha +\beta x_i+{ϵ}_i]\\ & =& \mathrm{V}\left[{ϵ}_i\right]\cdots \mathrm{V}[X\pm t]=\mathrm{V}\left[X\right](t:\mathrm{分}\mathrm{散}\mathrm{を}\mathrm{と}\mathrm{る}\mathrm{こ}\mathrm{と}\mathrm{に}\mathrm{つ}\mathrm{い}\mathrm{て}\mathrm{定}\mathrm{数})\\ & =& {\sigma }^2\\ \mathrm{Cov}[y_i,y_j]& =& \mathrm{E}\left[(y_i-\mathrm{E}\left[y_i\right])(y_j-\mathrm{E}\left[y_j\right])\right]\cdots \mathrm{Cov}[X,Y]=\mathrm{E}\left[(X-\mathrm{E}\left[X\right])(Y-\mathrm{E}\left[Y\right])\right]\\ & =& \mathrm{E}\left[(\alpha +\beta x_i+{ϵ}_i-\mathrm{E}[\alpha +\beta x_i+{ϵ}_i])(\alpha +\beta x_j+{ϵ}_j-\mathrm{E}[\alpha +\beta x_j+{ϵ}_j])\right]\cdots y_i=\alpha +\beta x_i+{ϵ}_i\\ & =& \mathrm{E}\left[(\alpha +\beta x_i+{ϵ}_i-\alpha -\beta x_i-\mathrm{E}\left[{ϵ}_i\right])(\alpha +\beta x_j+{ϵ}_j-\alpha -\beta x_j-\mathrm{E}\left[{ϵ}_j\right])\right]\cdots \mathrm{E}[X\pm t]=\mathrm{E}\left[X\right]\pm t\\ & =& \mathrm{E}\left[({ϵ}_i-\mathrm{E}\left[{ϵ}_i\right])({ϵ}_j-\mathrm{E}\left[{ϵ}_j\right])\right]\\ & =& \mathrm{Cov}[{ϵ}_i,{ϵ}_j]\end{array}$$ 上記を踏まえて$(x_i-\overline{x})$を加えた分散・共分散について $$\begin{array}{rcl}\mathrm{Cov}[(x_i-\overline{x})(y_i-\overline{y}),(x_j-\overline{x})(y_j-\overline{y})]& =& (x_i-\overline{x})(x_j-\overline{x})\mathrm{Cov}[(y_i-\overline{y}),(y_j-\overline{y})]\cdots \mathrm{Cov}[c_0{X}_i,c_1{X}_j]=c_0{c}_1\mathrm{Cov}[X_i,X_j]\\ & =& (x_i-\overline{x})(x_j-\overline{x})\mathrm{E}\left[\{(y_i-\overline{y})-\mathrm{E}[y_i-\overline{y}]\}\{(y_j-\overline{y})-\mathrm{E}[y_j-\overline{y}]\}\right]\cdots \mathrm{Cov}[X,Y]=\mathrm{E}\left[(X-\mathrm{E}\left[X\right])(Y-\mathrm{E}\left[Y\right])\right]\\ & =& (x_i-\overline{x})(x_j-\overline{x})\mathrm{E}\left[\{y_i-\overline{y}-\mathrm{E}\left[y_i\right]+\overline{y}\}\{y_j-\overline{y}-\mathrm{E}\left[y_j\right]+\overline{y}\}\right]\cdots \mathrm{E}[X\pm t]=\mathrm{E}\left[X\right]\pm t\\ & =& (x_i-\overline{x})(x_j-\overline{x})\mathrm{E}\left[(y_i-\mathrm{E}\left[\overline{y}\right])(y_j-\mathrm{E}\left[\overline{y}\right])\right]\\ & =& (x_i-\overline{x})(x_j-\overline{x})\mathrm{Cov}[y_i,y_j]\cdots \mathrm{Cov}[X,Y]=\mathrm{E}\left[(X-\mathrm{E}\left[X\right])(Y-\mathrm{E}\left[Y\right])\right]\\ & =& \{\begin{array}{}(x_i-\overline{x}{)}^2{\sigma }^2& (i=j)\\ (x_i-\overline{x})(x_j-\overline{x})0& (i\ne j)\end{array}\cdots \mathrm{Cov}[y_i,y_j]=\mathrm{Cov}[{ϵ}_i,{ϵ}_j]=\{\begin{array}{}\mathrm{V}\left[{ϵ}_i\right]={\sigma }^2& (i=j)\\ 0& (i\ne j)\end{array}\\ \mathrm{V}[(x_i-\overline{x})y_i]& =& (x_i-\overline{x}{)}^2\mathrm{V}\left[y_i\right]\cdots \mathrm{V}\left[cX\right]=c^2\mathrm{V}\left[X\right]\\ & =& (x_i-\overline{x}{)}^2{\sigma }^2\cdots \mathrm{上}\mathrm{記}i=j\mathrm{の}\mathrm{ケ}\mathrm{ー}\mathrm{ス}\end{array}$$

$S_{xy}$の分散

$$\begin{array}{rcl}\mathrm{V}\left[S_{xy}\right]& =& \mathrm{V}\left[\sum _{i=1}^n(x_i-\overline{x})(y_i-\overline{y})\right]\cdots S_{xy}=\sum _{i=1}^n(x_i-\overline{x})(y_i-\overline{y})\\ & =& \sum _{i=1}^n\mathrm{V}\left[(x_i-\overline{x})(y_i-\overline{y})\right]+2\sum _{i<j}\mathrm{Cov}[(x_i-\overline{x})(y_i-\overline{y}),(x_j-\overline{x})(y_j-\overline{y})]\\ & & \cdots \mathrm{V}\left[\sum _{i=1}^n{X}_i\right]=\sum _{i=1}^n\mathrm{V}\left[X_i\right]+2\sum _{i<j}\mathrm{Cov}[X_i,X_j]\\ & =& \sum _{i=1}^n{(x_i-\overline{x})}^2{\sigma }^2+2\sum _{i<j}0\\ & & \cdots \mathrm{V}\left[(x_i-\overline{x})(y_i-\overline{y})\right]={(x_i-\overline{x})}^2{\sigma }^2\\ & & \cdots \mathrm{Cov}[(x_i-\overline{x})(y_i-\overline{y}),(x_j-\overline{x})(y_j-\overline{y})]=0(i\ne j)\\ & =& {\sigma }^2\sum _{i=1}^n{(x_i-\overline{x})}^2\cdots \sum _{i=0}^{n}c{X}_i=c\sum _{i=0}^n{X}_i\\ & =& {\sigma }^2{S}_{xx}\cdots S_{xx}=\sum _{i=1}^n{(x_i-\overline{x})}^2\end{array}$$

最小2乗推定量$\hat{\beta }$の分散

$$\begin{array}{rcl}\mathrm{V}\left[\hat{\beta }\right]& =& \mathrm{V}\left[\frac{S_{xy}}{S_{xx}}\right]\cdots \hat{\beta }=\frac{S_{xy}}{S_{xx}},S_{xx}=\sum _{i=1}^n{(x_i-\overline{x})}^2,S_{xy}=\sum _{i=1}^n(x_i-\overline{x})(y_i-\overline{y})\\ & =& \frac{1}{S_{xx}^2}\mathrm{V}\left[S_{xy}\right]\cdots \mathrm{V}\left[cX\right]=c^2\mathrm{V}\left[X\right]\\ & =& \frac{1}{S_{xx}^2}{\sigma }^2{S}_{xx}\cdots \mathrm{V}\left[S_{xy}\right]={\sigma }^2{S}_{xx}\\ & =& \frac{1}{S_{xx}}{\sigma }^2\end{array}$$

最小2乗推定量$\hat{\alpha }$の分散

$$\begin{array}{rcl}\mathrm{V}\left[\hat{\alpha }\right]& =& \mathrm{V}[\overline{y}-\hat{\beta }\overline{x}]\cdots \hat{\alpha }=\overline{y}-\hat{\beta }\overline{x}\\ & =& \mathrm{V}[\overline{y}-\frac{S_{xy}}{S_{xx}}\overline{x}]\cdots \hat{\beta }=\frac{S_{xy}}{S_{xx}},S_{xx}=\sum _{i=1}^n{(x_i-\overline{x})}^2,S_{xy}=\sum _{i=1}^n(x_i-\overline{x})(y_i-\overline{y})\\ & =& \mathrm{V}\left[\overline{y}\right]+\mathrm{V}\left[\frac{S_{xy}}{S_{xx}}\overline{x}\right]-2\mathrm{Cov}[\overline{y},\frac{S_{xy}}{S_{xx}}\overline{x}]\\ & & \cdots \mathrm{V}[X\pm Y]=\mathrm{V}\left[X\right]\pm 2\mathrm{Cov}[X,Y]+\mathrm{V}\left[Y\right]\\ & =& \mathrm{V}\left[\overline{y}\right]+\mathrm{V}\left[\frac{S_{xy}}{S_{xx}}\overline{x}\right]-2\cdot 0\cdots \mathrm{Cov}[\overline{y},\frac{S_{xy}}{S_{xx}}\overline{x}]=0(\mathrm{後}\mathrm{述})\\ & =& \mathrm{V}\left[\overline{y}\right]+\frac{{\overline{x}}^2}{S_{xx}^2}\mathrm{V}\left[S_{xy}\right]\\ & =& \mathrm{V}\left[\frac{1}{n}\sum _{i=1}^n{y}_i\right]+\frac{{\overline{x}}^2}{S_{xx}^2}{\sigma }^2{S}_{xx}\cdots \mathrm{V}\left[S_{xy}\right]={\sigma }^2{S}_{xx}\\ & =& \frac{1}{n^2}\mathrm{V}\left[\sum _{i=1}^n{y}_i\right]+\frac{{\overline{x}}^2}{S_{xx}}{\sigma }^2\cdots \mathrm{V}\left[cX\right]=c^2\mathrm{V}\left[X\right]\\ & =& \frac{1}{n^2}\{\sum _{i=1}^n\mathrm{V}\left[y_i\right]+2\sum _{i<j}\mathrm{Cov}[y_i,y_j]\}+\frac{{\overline{x}}^2}{S_{xx}}{\sigma }^2\\ & & \cdots \mathrm{V}\left[\sum _{i=1}^n{X}_i\right]=\sum _{i=1}^n\mathrm{V}\left[X_i\right]+2\sum _{i<j}\mathrm{Cov}[X_i,X_j]\\ & =& \frac{1}{n^2}\{\sum _{i=1}^n{\sigma }^2+2\sum _{i<j}0\}+\frac{{\overline{x}}^2}{S_{xx}}{\sigma }^2\cdots \mathrm{V}\left[y_i\right]={\sigma }^2,\mathrm{Cov}[y_i,y_j]=0\\ & =& \frac{1}{n^2}n{\sigma }^2+\frac{{\overline{x}}^2}{S_{xx}}{\sigma }^2\cdots \sum _{i=0}^{n}c=nc\\ & =& (\frac{1}{n}+\frac{{\overline{x}}^2}{S_{xx}}){\sigma }^2\end{array}$$

最小2乗推定量$\hat{\alpha }$と$\hat{\beta }$の共分散

$$\begin{array}{rcl}\mathrm{Cov}[\hat{\alpha },\hat{\beta }]& =& \mathrm{E}\left[(\hat{\alpha }-\mathrm{E}\left[\hat{\alpha }\right])(\hat{\beta }-\mathrm{E}\left[\hat{\beta }\right])\right]\cdots \mathrm{Cov}[X,Y]=\mathrm{E}\left[(X-\mathrm{E}\left[X\right])(Y-\mathrm{E}\left[Y\right])\right]\\ & =& \mathrm{E}\left[((\overline{y}-\hat{\beta }\overline{x})-\mathrm{E}[\overline{y}-\hat{\beta }\overline{x}])(\hat{\beta }-\mathrm{E}\left[\hat{\beta }\right])\right]\cdots \alpha =\overline{y}-\hat{\beta }\overline{x}\\ & =& \mathrm{E}\left[(\overline{y}-\frac{S_{xy}}{S_{xx}}\overline{x}-\mathrm{E}[\overline{y}-\frac{S_{xy}}{S_{xx}}\overline{x}])(\frac{S_{xy}}{S_{xx}}-\mathrm{E}\left[\frac{S_{xy}}{S_{xx}}\right])\right]\cdots \hat{\beta }=\frac{S_{xy}}{S_{xx}},S_{xx}=\sum _{i=1}^n{(x_i-\overline{x})}^2,S_{xy}=\sum _{i=1}^n(x_i-\overline{x})(y_i-\overline{y})\\ & =& \mathrm{E}\left[(\overline{y}-\frac{S_{xy}}{S_{xx}}\overline{x}-\mathrm{E}\left[\overline{y}\right]+\mathrm{E}\left[\frac{S_{xy}}{S_{xx}}\overline{x}\right])(\frac{S_{xy}}{S_{xx}}-\mathrm{E}\left[\frac{S_{xy}}{S_{xx}}\right])\right]\cdots \mathrm{E}[X\pm Y]=\mathrm{E}\left[X\right]\pm \mathrm{E}\left[Y\right]\\ & =& \mathrm{E}\left[(\overline{y}-\frac{S_{xy}}{S_{xx}}\overline{x}-\mathrm{E}\left[\overline{y}\right]+\frac{\overline{x}}{S_{xx}}\mathrm{E}\left[S_{xy}\right])(\frac{S_{xy}}{S_{xx}}-\frac{1}{S_{xx}}\mathrm{E}\left[S_{xy}\right])\right]\cdots \mathrm{E}\left[cX\right]=c\mathrm{E}\left[X\right]\\ & =& \mathrm{E}\left[\{\overline{y}-\mathrm{E}\left[\overline{y}\right]-\frac{\overline{x}}{S_{xx}}(S_{xy}-\mathrm{E}\left[S_{xy}\right])\}\left\{\frac{1}{S_{xx}}(S_{xy}-\mathrm{E}\left[S_{xy}\right])\right\}\right]\\ & =& \mathrm{E}[-\frac{\overline{x}}{S_{xx}}(S_{xy}-\mathrm{E}\left[S_{xy}\right])\frac{1}{S_{xx}}(S_{xy}-\mathrm{E}\left[S_{xy}\right])]\cdots \overline{y}-\mathrm{E}\left[\overline{y}\right]=\overline{y}-\overline{y}=0\\ & =& \mathrm{E}[-\frac{\overline{x}}{S_{xx}^2}{(S_{xy}-\mathrm{E}\left[S_{xy}\right])}^2]\\ & =& -\frac{\overline{x}}{S_{xx}^2}\mathrm{E}\left[{(S_{xy}-\mathrm{E}\left[S_{xy}\right])}^2\right]\cdots \mathrm{E}\left[cX\right]=c\mathrm{E}\left[X\right]\\ & =& -\frac{\overline{x}}{S_{xx}^2}{\sigma }^2{S}_{xx}\cdots \mathrm{E}\left[{(S_{xy}-\mathrm{E}\left[S_{xy}\right])}^2\right]=\mathrm{V}\left[S_{xy}\right]={\sigma }^2{S}_{xx}\\ & =& -\frac{\overline{x}}{S_{xx}}{\sigma }^2\end{array}$$

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$\mathrm{Cov}[\overline{y},\frac{S_{xy}}{S_{xx}}\overline{x}]=0$について

$$\begin{array}{rcl}\mathrm{Cov}[\overline{y},\frac{S_{xy}}{S_{xx}}\overline{x}]& =& \mathrm{E}\left[(\overline{y}-\mathrm{E}\left[\overline{y}\right])(\frac{S_{xy}}{S_{xx}}\overline{x}-\mathrm{E}\left[\frac{S_{xy}}{S_{xx}}\overline{x}\right])\right]\cdots \mathrm{Cov}[X,Y]=\mathrm{E}\left[(X-\mathrm{E}\left[X\right])(Y-\mathrm{E}\left[Y\right])\right]\\ & =& \mathrm{E}\left[(\overline{y}-\overline{y})(\frac{S_{xy}}{S_{xx}}\overline{x}-\frac{\overline{x}}{S_{xx}}\mathrm{E}\left[S_{xy}\right])\right]\cdots \mathrm{E}\left[\overline{y}\right]=\overline{y},\mathrm{E}\left[cX\right]=c\mathrm{E}\left[X\right]\\ & =& \mathrm{E}[0\cdot \frac{\overline{x}}{S_{xx}}(S_{xy}-\mathrm{E}\left[S_{xy}\right])]\\ & =& \mathrm{E}\left[0\right]\\ & =& 0\end{array}$$

単回帰における最小二乗推定量が不偏推定量であることの証明

線形推定量

以前に単回帰における最小2乗推定量(least squares estimator; LSE)を求めた際に利用したのは以下の式である．． $$\begin{array}{r}\{\begin{array}{}\hat{\alpha }+\overline{x}\hat{\beta }-\overline{y}& =& 0\\ {\displaystyle n\overline{x}\hat{\alpha }+\left(\sum _{i=1}^n{x}_i^2\right)\hat{\beta }-\left(\sum _{i=1}^n{x}_i{y}_i\right)}& =& 0\end{array}\cdots \mathrm{線}\mathrm{形}\mathrm{単}\mathrm{回}\mathrm{帰}\mathrm{の}\mathrm{回}\mathrm{帰}\mathrm{直}\mathrm{線}(\hat{\alpha },\hat{\beta }\mathrm{を}\mathrm{求}\mathrm{め}\mathrm{る})\end{array}$$ 上記は以下の形にでき，これは正規方程式(normal equation)と呼ばれる． $$\begin{array}{r}\{\begin{array}{}\hat{\alpha }+\overline{x}\hat{\beta }& =& \overline{y}& =& {\displaystyle \left(\sum _{i=1}^n\frac{1}{n}{y}_i\right)}\\ {\displaystyle n\overline{x}\hat{\alpha }+\left(\sum _{i=1}^n{x}_i^2\right)\hat{\beta }}& =& {\displaystyle \left(\sum _{i=1}^n{x}_i{y}_i\right)}\end{array}\end{array}$$ 観測値$y_i$の一次式$\sum _{i=1}^n{c}_i{y}_i(c_i:\mathrm{定}\mathrm{数})$で表される推定量を線形推定量(linear estimator)と呼ぶ．
よって$\hat{\alpha },\hat{\beta }$は$y_i$の線形推定量である．

観測値$y_i$の期待値

$$\begin{array}{rcl}{y}_i& =& \alpha +\beta x_i+{ϵ}_i(i=1,\cdots ,n)\\ \{{ϵ}_i|i=1,\cdots ,n\}& :& {ϵ}_i\stackrel{iid}{\sim }N(0,{\sigma }^2)\\ & & \cdots \mathrm{独}\mathrm{立}\mathrm{同}\mathrm{一}\mathrm{分}\mathrm{布}(independentandidenticallydistributed;IID,i.i.d.,iid)\\ & & \cdots \mathrm{E}\left[{ϵ}_i\right]=0,\mathrm{V}\left[{ϵ}_i\right]={\sigma }^2,\mathrm{互}\mathrm{い}\mathrm{に}\mathrm{独}\mathrm{立}\\ \mathrm{E}\left[y_i\right]& =& \mathrm{E}[\alpha +\beta x_i+{ϵ}_i]\\ & =& \mathrm{E}\left[\alpha \right]+\mathrm{E}\left[\beta x_i\right]+\mathrm{E}\left[{ϵ}_i\right]\\ & =& \alpha +\beta x_i+0\cdots \mathrm{E}\left[C\right]=C(C:\mathrm{期}\mathrm{待}\mathrm{値}\mathrm{を}\mathrm{と}\mathrm{る}\mathrm{こ}\mathrm{と}\mathrm{に}\mathrm{つ}\mathrm{い}\mathrm{て}\mathrm{定}\mathrm{数}),\mathrm{E}\left[{ϵ}_i\right]=0\\ & =& \alpha +\beta x_i\end{array}$$

最小二乗推定量$\hat{\beta }$が不偏推定量であることの証明

$$\begin{array}{rcl}\mathrm{E}\left[\hat{\beta }\right]& =& \mathrm{E}\left[\frac{S_{xy}}{S_{xx}}\right]\cdots \hat{\beta }=\frac{S_{xy}}{S_{xx}}\\ & =& \mathrm{E}\left[\frac{\sum _{i=1}^n(x_i-\overline{x})(y_i-\overline{y})}{S_{xx}}\right]\cdots S_{xy}=\sum _{i=1}^n(x_i-\overline{x})(y_i-\overline{y})\\ & =& \frac{1}{S_{xx}}\mathrm{E}\left[\sum _{i=1}^n(x_i-\overline{x})(y_i-\overline{y})\right]\cdots \mathrm{E}\left[cX\right]=c\mathrm{E}\left[X\right]\\ & =& \frac{1}{S_{xx}}\sum _{i=1}^n\mathrm{E}\left[(x_i-\overline{x})(y_i-\overline{y})\right]\\ & & \cdots \mathrm{E}\left[\sum _{i=1}^n{A}_i\right]=\mathrm{E}[A_i+\cdots +A_i+\cdots +A_n]=\mathrm{E}\left[A_i\right]+\cdots +\mathrm{E}\left[A_i\right]+\cdots +\mathrm{E}\left[A_n\right]=\sum _{i=1}^n\mathrm{E}\left[A_i\right]\\ & & \cdots \mathrm{E}[X\pm Y]=\mathrm{E}\left[X\right]\pm \mathrm{E}\left[Y\right]\\ & =& \frac{1}{S_{xx}}\sum _{i=1}^n(x_i-\overline{x})\mathrm{E}[y_i-\overline{y}]\cdots \mathrm{E}\left[cX\right]=c\mathrm{E}\left[X\right]\\ & =& \frac{1}{S_{xx}}\sum _{i=1}^n(x_i-\overline{x})(\mathrm{E}\left[y_i\right]-\mathrm{E}\left[\overline{y}\right])\cdots \mathrm{E}[X\pm Y]=\mathrm{E}\left[X\right]\pm \mathrm{E}\left[Y\right]\\ & =& \frac{1}{S_{xx}}\sum _{i=1}^n(x_i-\overline{x})\{(\alpha +\beta x_i)-(\alpha +\beta \overline{x})\}\cdots \mathrm{E}\left[y_i\right]=\alpha +\beta x_i,\mathrm{E}\left[\overline{y}\right]=\alpha +\beta \overline{x}\\ & =& \frac{1}{S_{xx}}\sum _{i=1}^n(x_i-\overline{x})(\alpha +\beta x_i-\alpha -\beta \overline{x})\\ & =& \frac{1}{S_{xx}}\sum _{i=1}^n(x_i-\overline{x})\beta (x_i-\overline{x})\\ & =& \frac{1}{S_{xx}}\sum _{i=1}^n{(x_i-\overline{x})}^2\beta \\ & =& \frac{1}{S_{xx}}\beta \sum _{i=1}^n{(x_i-\overline{x})}^2\\ & & \cdots \sum _{i=1}^{n}c{X}_i=c{X}_1+\cdots +c{X}_i+\cdots +c{X}_n=c(X_1+\cdots +X_i+\cdots +X_n)=c\sum _{i=1}^n{X}_i\\ & =& \frac{1}{S_{xx}}\beta S_{xx}\cdots S_{xx}=\sum _{i=1}^n{(x_i-\overline{x})}^2\\ & =& \beta \end{array}$$

最小二乗推定量$\hat{\alpha }$が不偏推定量であることの証明

$$\begin{array}{r}\\ \mathrm{E}\left[\hat{\alpha }\right]& =& \mathrm{E}[\overline{y}-\hat{\beta }\overline{x}]\cdots \hat{\alpha }=\overline{y}-\hat{\beta }\overline{x}\\ & =& \mathrm{E}\left[\overline{y}\right]-\mathrm{E}\left[\hat{\beta }\overline{x}\right]\cdots \mathrm{E}[X\pm Y]=\mathrm{E}\left[X\right]\pm \mathrm{E}\left[Y\right]\\ & =& \mathrm{E}\left[\overline{y}\right]-\overline{x}\mathrm{E}\left[\hat{\beta }\right]\cdots \mathrm{E}\left[cX\right]=c\mathrm{E}\left[X\right]\\ & =& \alpha +\beta \overline{x}-\beta \overline{x}\cdots \mathrm{E}\left[\overline{y}\right]=\alpha +\beta \overline{x},\mathrm{E}\left[\hat{\beta }\right]=\beta \\ & =& \alpha \end{array}$$

標本平均の母平均まわりの4次モーメント (標本平均の4次の中心(化)モーメント)

標本平均$\bar{X}$の母平均$\mu$まわりの4次モーメント(=標本平均$\bar{X}$の4次の中心(化)モーメント)

$$\begin{array}{rcl}\mathrm{E}[(\bar{X}-\mu {)}^4]& =& \mathrm{E}\left[{\{\left(\frac{1}{n}\sum _{k=1}^n{X}_k\right)-\mu \}}^4\right]\cdots \bar{X}=\frac{1}{n}\sum _{i=1}^n{X}_i\\ & =& \mathrm{E}\left[{\{\left(\frac{1}{n}\sum _{k=1}^n{X}_k\right)-\left(\frac{1}{n}\sum _{k=1}^n\mu \right)\}}^4\right]\cdots C=\frac{n}{n}C=\frac{1}{n}C\sum _{i=1}^{n}1=\frac{1}{n}\sum _{i=1}^{n}C(C:i\mathrm{に}\mathrm{よ}\mathrm{ら}\mathrm{な}\mathrm{い}\mathrm{数}，\sum \mathrm{に}\mathrm{と}\mathrm{っ}\mathrm{て}\mathrm{定}\mathrm{数})\\ & =& \mathrm{E}\left[{\left[\frac{1}{n}\{\left(\sum _{k=1}^n{X}_k\right)-\left(\sum _{k=1}^n\mu \right)\}\right]}^4\right]\\ & =& \mathrm{E}\left[\frac{1}{n^4}{\{\left(\sum _{k=1}^n{X}_k\right)-\left(\sum _{k=1}^n\mu \right)\}}^4\right]\cdots (AB{)}^C=A^C{B}^C\\ & =& \mathrm{E}\left[\frac{1}{n^4}{\left\{\sum _{k=1}^n(X_k-\mu )\right\}}^4\right]\cdots \sum _{i=1}^n{X}_i-\sum _{i=1}^n{Y}_i=\sum _{i=1}^n(X_i-Y_i)\end{array}$$ 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{array}{rcl}{\left(\sum _{k=1}^n{A}_k\right)}^4& =& \left(\sum _{k=1}^n{A}_k\right)\left(\sum _{l=1}^n{A}_l\right)\left(\sum _{m=1}^n{A}_m\right)\left(\sum _{s=1}^n{A}_s\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)(A_1+A_2+\cdots +A_s+\cdots +A_n)\\ & =& {}_4{\mathrm{P}}_0\times \left(\sum _{k=1}^n{A}_k^4\right)\cdots 4\mathrm{つ}\mathrm{と}\mathrm{も}\mathrm{同}\mathrm{じ}\mathrm{添}\mathrm{え}\mathrm{字}(\mathrm{ど}\mathrm{れ}\mathrm{か}0\mathrm{個}\mathrm{の}\mathrm{添}\mathrm{え}\mathrm{字}\mathrm{が}\mathrm{異}\mathrm{な}\mathrm{る}\mathrm{ケ}\mathrm{ー}\mathrm{ス})\\ & & {+}_4{\mathrm{P}}_1\times \left(\sum _{k\ne l}{A}_k^3{A}_l\right)\cdots \mathrm{い}\mathrm{ず}\mathrm{れ}\mathrm{か}3\mathrm{つ}\mathrm{が}\mathrm{同}\mathrm{じ}\mathrm{添}\mathrm{え}\mathrm{字}(\mathrm{ど}\mathrm{れ}\mathrm{か}1\mathrm{個}\mathrm{の}\mathrm{添}\mathrm{え}\mathrm{字}\mathrm{が}\mathrm{異}\mathrm{な}\mathrm{る}\mathrm{の}\mathrm{ケ}\mathrm{ー}\mathrm{ス})\\ & & {+}_4{\mathrm{C}}_2\times \left(\sum _{k<l}{A}_k^2{A}_l^2\right)\cdots \mathrm{い}\mathrm{ず}\mathrm{れ}\mathrm{か}2\mathrm{つ}\mathrm{の}\mathrm{添}\mathrm{え}\mathrm{字}\mathrm{が}\mathrm{同}\mathrm{じ}\mathrm{で}\mathrm{残}\mathrm{り}\mathrm{の}2\mathrm{つ}\mathrm{の}\mathrm{添}\mathrm{え}\mathrm{字}\mathrm{同}\mathrm{士}\mathrm{も}\mathrm{同}\mathrm{じ}\mathrm{ケ}\mathrm{ー}\mathrm{ス}(\mathrm{ど}\mathrm{れ}\mathrm{か}2\mathrm{個}\mathrm{の}\mathrm{添}\mathrm{え}\mathrm{字}\mathrm{が}\mathrm{異}\mathrm{な}\mathrm{る}\mathrm{の}\mathrm{ケ}\mathrm{ー}\mathrm{ス}(\mathrm{異}\mathrm{な}\mathrm{っ}\mathrm{た}\mathrm{添}\mathrm{え}\mathrm{字}\mathrm{同}\mathrm{士}\mathrm{も}\mathrm{同}\mathrm{じ}))\\ & & {+}_4{\mathrm{P}}_2\times \left(\sum _{k\ne l,m\mathrm{か}\mathrm{つ}l<m}{A}_k^2{A}_l{A}_m\right)\cdots \mathrm{い}\mathrm{ず}\mathrm{れ}\mathrm{か}2\mathrm{つ}\mathrm{の}\mathrm{添}\mathrm{え}\mathrm{字}\mathrm{が}\mathrm{同}\mathrm{じ}\mathrm{で}\mathrm{残}\mathrm{り}\mathrm{の}2\mathrm{つ}\mathrm{の}\mathrm{添}\mathrm{え}\mathrm{字}\mathrm{同}\mathrm{士}\mathrm{は}\mathrm{異}\mathrm{な}\mathrm{る}\mathrm{ケ}\mathrm{ー}\mathrm{ス}(\mathrm{ど}\mathrm{れ}\mathrm{か}2\mathrm{個}\mathrm{の}\mathrm{添}\mathrm{え}\mathrm{字}\mathrm{が}\mathrm{異}\mathrm{な}\mathrm{る}\mathrm{の}\mathrm{ケ}\mathrm{ー}\mathrm{ス}(\mathrm{異}\mathrm{な}\mathrm{っ}\mathrm{た}\mathrm{添}\mathrm{え}\mathrm{字}\mathrm{同}\mathrm{士}\mathrm{は}\mathrm{異}\mathrm{な}\mathrm{る}))\\ & & {+}_4{\mathrm{P}}_3\times \left(\sum _{k<l<m<s}{A}_k{A}_l{A}_m{A}_s\right)\cdots \mathrm{す}\mathrm{べ}\mathrm{て}\mathrm{の}\mathrm{添}\mathrm{え}\mathrm{字}\mathrm{が}\mathrm{異}\mathrm{な}\mathrm{る}\mathrm{ケ}\mathrm{ー}\mathrm{ス}(\mathrm{ど}\mathrm{れ}\mathrm{か}3\mathrm{個}\mathrm{の}\mathrm{添}\mathrm{え}\mathrm{字}\mathrm{が}\mathrm{異}\mathrm{な}\mathrm{る}\mathrm{の}\mathrm{ケ}\mathrm{ー}\mathrm{ス})\\ & & \cdots \mathrm{各}\mathrm{ケ}\mathrm{ー}\mathrm{ス}\mathrm{で}\mathrm{の}(\mathrm{重}\mathrm{複}\mathrm{す}\mathrm{る}\mathrm{数}\times \mathrm{組}\mathrm{合}\mathrm{せ}\mathrm{で}\mathrm{総}\mathrm{和})\mathrm{の}\mathrm{和}\\ & =& \frac{4!}{(4-0)!}\left(\sum _{k=1}^n{A}_k^4\right)+\frac{4!}{(4-1)!}\left(\sum _{k\ne l}{A}_k^3{A}_l\right)+\frac{4!}{(4-2)!2!}\left(\sum _{k<l}{A}_k^2{A}_l^2\right)+\frac{4!}{(4-2)!}\left(\sum _{k\ne l,m\mathrm{か}\mathrm{つ}l<m}{A}_k^2{A}_l{A}_m\right)+\frac{4!}{(4-3)!}\left(\sum _{k<l<m<s}{A}_k{A}_l{A}_m{A}_s\right)\\ & =& \frac{4\times 3\times 2\times 1}{4\times 3\times 2\times 1}\left(\sum _{k=1}^n{A}_k^4\right)+\frac{4\times 3\times 2\times 1}{3\times 2\times 1}\left(\sum _{k\ne l}{A}_k^3{A}_l\right)+\frac{4\times 3\times 2\times 1}{2\times 1\cdot 2\times 1}\left(\sum _{k<l}{A}_k^2{A}_l^2\right)+\frac{4\times 3\times 2\times 1}{2\times 1}\left(\sum _{k\ne l,m\mathrm{か}\mathrm{つ}l<m}{A}_k^2{A}_l{A}_m\right)+\frac{4\times 3\times 2\times 1}{1}\left(\sum _{k<l<m<s}{A}_k{A}_l{A}_m{A}_s\right)\\ & =& 1\cdot \left(\sum _{k=1}^n{A}_k^4\right)+4\cdot \left(\sum _{k\ne l}{A}_k^3{A}_l\right)+6\cdot \left(\sum _{k<l}{A}_k^2{A}_l^2\right)+12\cdot \left(\sum _{k\ne l,m\mathrm{か}\mathrm{つ}l<m}{A}_k^2{A}_l{A}_m\right)+24\cdot \left(\sum _{k<l<m<s}{A}_k{A}_l{A}_m{A}_s\right)\end{array}$$ よって， $$\begin{array}{rcl}\mathrm{E}[(\bar{X}-\mu {)}^4]& =& \mathrm{E}\left[\frac{1}{n^4}{\left\{\sum _{k=1}^n(X_k-\mu )\right\}}^4\right]\\ & =& \mathrm{E}\left[\frac{1}{n^4}\{\sum _{k=1}^n{(X_k-\mu )}^4+4\sum _{k\ne l}{(X_k-\mu )}^3(X_l-\mu )+6\sum _{k<l}{(X_k-\mu )}^2{(X_l-\mu )}^2+12\sum _{k\ne l,m\mathrm{か}\mathrm{つ}l<m}{(X_k-\mu )}^2(X_l-\mu )(X_m-\mu )+24\sum _{k<l<m<s}(X_k-\mu )(X_l-\mu )(X_m-\mu )(X_s-\mu )\}\right]\\ & =& \frac{1}{n^4}\mathrm{E}[\sum _{k=1}^n{(X_k-\mu )}^4+4\sum _{k\ne l}{(X_k-\mu )}^3(X_l-\mu )+6\sum _{k<l}{(X_k-\mu )}^2{(X_l-\mu )}^2+12\sum _{k\ne l,m\mathrm{か}\mathrm{つ}l<m}{(X_k-\mu )}^2(X_l-\mu )(X_m-\mu )+24\sum _{k<l<m<s}(X_k-\mu )(X_l-\mu )(X_m-\mu )(X_s-\mu )]\cdots \mathrm{E}[cX]=c\mathrm{E}[X]\\ & =& \frac{1}{n^4}[\mathrm{E}\left[\sum _{k=1}^n{(X_k-\mu )}^4\right]+\mathrm{E}\left[4\sum _{k\ne l}{(X_k-\mu )}^3(X_l-\mu )\right]+\mathrm{E}\left[6\sum _{k<l}{(X_k-\mu )}^2{(X_l-\mu )}^2\right]+\mathrm{E}\left[12\sum _{k\ne l,m\mathrm{か}\mathrm{つ}l<m}{(X_k-\mu )}^2(X_l-\mu )(X_m-\mu )\right]+\mathrm{E}\left[24\sum _{k<l<m<s}(X_k-\mu )(X_l-\mu )(X_m-\mu )(X_s-\mu )\right]]\cdots \mathrm{E}[X+Y]=\mathrm{E}[X]+\mathrm{E}[Y]\\ & =& \frac{1}{n^4}[\mathrm{E}\left[\sum _{k=1}^n{(X_k-\mu )}^4\right]+4\mathrm{E}\left[\sum _{k\ne l}{(X_k-\mu )}^3(X_l-\mu )\right]+6\mathrm{E}\left[\sum _{k<l}{(X_k-\mu )}^2{(X_l-\mu )}^2\right]+12\mathrm{E}\left[\sum _{k\ne l,m\mathrm{か}\mathrm{つ}l<m}{(X_k-\mu )}^2(X_l-\mu )(X_m-\mu )\right]+24\mathrm{E}\left[\sum _{k<l<m<s}(X_k-\mu )(X_l-\mu )(X_m-\mu )(X_s-\mu )\right]]\cdots \mathrm{E}[cX]=c\mathrm{E}[X]\\ & =& \frac{1}{n^4}[\sum _{k=1}^n\mathrm{E}\left[{(X_k-\mu )}^4\right]+4\sum _{k\ne l}\mathrm{E}\left[{(X_k-\mu )}^3(X_l-\mu )\right]+6\sum _{k<l}\mathrm{E}\left[{(X_k-\mu )}^2{(X_l-\mu )}^2\right]+12\sum _{k\ne l,m\mathrm{か}\mathrm{つ}l<m}\mathrm{E}\left[{(X_k-\mu )}^2(X_l-\mu )(X_m-\mu )\right]+24\sum _{k<l<m<s}\mathrm{E}\left[(X_k-\mu )(X_l-\mu )(X_m-\mu )(X_s-\mu )\right]]\\ & & \cdots \mathrm{E}\left[\sum _{i=1}^n{A}_i\right]=\mathrm{E}[A_1+A_2+\cdots +A_i+\cdots +A_n]=\mathrm{E}\left[A_1\right]+\mathrm{E}\left[A_2\right]+\cdots +\mathrm{E}\left[A_i\right]+\cdots +\mathrm{E}\left[A_n\right]=\sum _{i=1}^n\mathrm{E}\left[A_i\right]\\ & =& \frac{1}{n^4}[\sum _{k=1}^n\mathrm{E}\left[{(X_k-\mu )}^4\right]+4\sum _{k\ne l}\mathrm{E}\left[{(X_k-\mu )}^3\right]\mathrm{E}\left[(X_l-\mu )\right]+6\sum _{k<l}\mathrm{E}\left[{(X_k-\mu )}^2\right]\mathrm{E}\left[{(X_l-\mu )}^2\right]+12\sum _{k\ne l,m\mathrm{か}\mathrm{つ}l<m}\mathrm{E}\left[{(X_k-\mu )}^2\right]\mathrm{E}\left[(X_l-\mu )\right]\mathrm{E}\left[(X_m-\mu )\right]+24\sum _{k<l<m<s}\mathrm{E}\left[(X_k-\mu )\right]\mathrm{E}\left[(X_l-\mu )\right]\mathrm{E}\left[(X_m-\mu )\right]\mathrm{E}\left[(X_s-\mu )\right]]\cdots X,Y\mathrm{が}\mathrm{独}\mathrm{立}\mathrm{の}\mathrm{場}\mathrm{合}\mathrm{E}[XY]=\mathrm{E}[X]\mathrm{E}[Y]\end{array}$$ 1次の中心(化)モーメントについて考える． $$\begin{array}{rcl}\mathrm{E}[X_i-\mu ]& =& \mathrm{E}\left[X_i\right]-\mathrm{E}\left[\mu \right]\cdots \mathrm{E}[X-Y]=\mathrm{E}[X]-\mathrm{E}[Y]\\ & =& \mu -\mu \cdots \mathrm{E}[X_i]=\mathrm{E}[X]=\mu ,\mathrm{E}[C]=C(C\mathrm{定}\mathrm{数})\\ & =& 0\end{array}$$ これを用いて $$\begin{array}{r}\mathrm{E}[(\bar{X}-\mu {)}^4]\\ & =& \frac{1}{n^4}[\sum _{k=1}^n\mathrm{E}\left[{(X_k-\mu )}^4\right]+4\sum _{k\ne l}\mathrm{E}\left[{(X_k-\mu )}^3\right]\mathrm{E}\left[(X_l-\mu )\right]+6\sum _{k<l}\mathrm{E}\left[{(X_k-\mu )}^2\right]\mathrm{E}\left[{(X_l-\mu )}^2\right]+12\sum _{k\ne l,m\mathrm{か}\mathrm{つ}l<m}\mathrm{E}\left[{(X_k-\mu )}^2\right]\mathrm{E}\left[(X_l-\mu )\right]\mathrm{E}\left[(X_m-\mu )\right]+24\sum _{k<l<m<s}\mathrm{E}\left[(X_k-\mu )\right]\mathrm{E}\left[(X_l-\mu )\right]\mathrm{E}\left[(X_m-\mu )\right]\mathrm{E}\left[(X_s-\mu )\right]]\\ & =& \frac{1}{n^4}[\sum _{k=1}^n\mathrm{E}\left[{(X_k-\mu )}^4\right]+4\sum _{k\ne l}(\mathrm{E}\left[{(X_k-\mu )}^3\right]\cdot 0)+6\sum _{k<l}\mathrm{E}\left[{(X_k-\mu )}^2\right]\mathrm{E}\left[{(X_l-\mu )}^2\right]+12\sum _{k\ne l,m\mathrm{か}\mathrm{つ}l<m}(\mathrm{E}\left[{(X_k-\mu )}^2\right]\cdot 0\cdot 0)+24\sum _{k<l<m<s}(0\cdot 0\cdot 0\cdot 0)]\cdots \mathrm{E}[X_i-\mu ]=0\\ & =& \frac{1}{n^4}[\sum _{k=1}^n\mathrm{E}\left[{(X_k-\mu )}^4\right]+0+6\sum _{k<l}\mathrm{E}\left[{(X_k-\mu )}^2\right]\mathrm{E}\left[{(X_l-\mu )}^2\right]+0+0]\\ & =& \frac{1}{n^4}\{\sum _{k=1}^n\mathrm{E}\left[{(X_k-\mu )}^4\right]+6\sum _{k<l}\mathrm{E}\left[{(X_k-\mu )}^2\right]\mathrm{E}\left[{(X_l-\mu )}^2\right]\}\\ & =& \frac{1}{n^4}\{\sum _{k=1}^n{\mu }_4+6\sum _{k<l}({\sigma }^2\cdot {\sigma }^2)\}\cdots \mathrm{E}\left[{(X_i-\mu )}^4\right]={\mu }_4:4\mathrm{次}\mathrm{の}\mathrm{中}\mathrm{心}(\mathrm{化})\mathrm{モ}\mathrm{ー}\mathrm{メ}\mathrm{ン}\mathrm{ト},\mathrm{E}\left[{(X_i-\mu )}^2\right]={\sigma }^2:2\mathrm{次}\mathrm{の}\mathrm{中}\mathrm{心}(\mathrm{化})\mathrm{モ}\mathrm{ー}\mathrm{メ}\mathrm{ン}\mathrm{ト}\\ & =& \frac{1}{n^4}\{\sum _{k=1}^n{\mu }_4+6\sum _{k<l}{\sigma }^4\}\\ & =& \frac{1}{n^4}\{\sum _{k=1}^n{\mu }_4+6{\sigma }^4\sum _{k<l}1\}\\ & =& \frac{1}{n^4}\{n{\mu }_4+6\frac{n(n-1)}{2}{\sigma }^4\}\cdots \sum _{k<l}1=\frac{n(n-1)}{2}(\mathrm{各}k=1〜n-1\mathrm{に}\mathrm{対}\mathrm{し}\mathrm{て}l=k+1\mathrm{か}\mathrm{ら}l=n\mathrm{ま}\mathrm{で}\mathrm{の}\mathrm{和})\\ & =& \frac{1}{n^3}({\mu }_4+3(n-1){\sigma }^4)\\ & =& {\mu }_4\left(\bar{X}\right)\cdots {\mu }_4\left(\bar{X}\right):\mathrm{標}\mathrm{本}\mathrm{平}\mathrm{均}\bar{X}\mathrm{の}\mathrm{母}\mathrm{平}\mathrm{均}\mu \mathrm{ま}\mathrm{わ}\mathrm{り}\mathrm{の}4\mathrm{次}\mathrm{モ}\mathrm{ー}\mathrm{メ}\mathrm{ン}\mathrm{ト}(4\mathrm{次}\mathrm{の}\mathrm{中}\mathrm{心}(\mathrm{化})\mathrm{モ}\mathrm{ー}\mathrm{メ}\mathrm{ン}\mathrm{ト})\end{array}$$

標本平均$\bar{X}$の母平均$\mu$まわりの4次モーメント(標本平均$\bar{X}$の4次の中心(化)モーメント)を尖度${\beta }_2$で表す

$$\begin{array}{rcl}\mathrm{E}[(\bar{X}-\mu {)}^4]& =& {\mu }_4\left(\bar{X}\right)\\ & =& \frac{1}{n^3}({\mu }_4+3(n-1){\sigma }^4)\\ & =& \frac{1}{n^3}(\frac{{\beta }_2+3}{{\sigma }^4}+3(n-1){\sigma }^4)\cdots {\mu }_4=\frac{{\beta }_2+3}{{\sigma }^4}:4\mathrm{次}\mathrm{の}\mathrm{中}\mathrm{心}(\mathrm{化})\mathrm{モ}\mathrm{ー}\mathrm{メ}\mathrm{ン}\mathrm{ト}\\ & =& \frac{1}{n^3}\left(\frac{{\beta }_2+3+3(n-1)}{{\sigma }^4}\right)\\ & =& \frac{1}{n^3}\left(\frac{{\beta }_2+3+3n-3}{{\sigma }^4}\right)\\ & =& \frac{1}{n^3}\left(\frac{{\beta }_2+3n}{{\sigma }^4}\right)\end{array}$$

標本平均$\bar{X}$の尖度${\beta }_2\left(\bar{X}\right)$

$$\begin{array}{r}{\beta }_2\left(\bar{X}\right)=\frac{{\beta }_2}{n}\end{array}$$

標本平均の分布の尖度

標本平均$\bar{X}$の尖度${\beta }_2\left(\bar{X}\right)$

$$\begin{array}{rcl}{\beta }_2\left(\bar{X}\right)& =& \frac{\mathrm{E}[(\bar{X}-\mu {)}^4]}{\mathrm{V}{\left[\bar{X}\right]}^{\frac{4}{2}}}-3\cdots {\beta }_2={\beta }_2\left(X\right)=\frac{\mathrm{E}[(X-\mu {)}^4]}{\mathrm{V}{\left[X\right]}^{\frac{4}{2}}}-3=\frac{{\mu }_4}{{\sigma }^4}-3:\mathrm{尖}\mathrm{度}\\ & =& \frac{\frac{1}{n^3}({\mu }_4+3(n-1){\sigma }^4)}{{\left(\frac{{\sigma }^2}{n}\right)}^2}-3\\ & & \cdots \mathrm{E}[(\bar{X}-\mu {)}^4]=\frac{1}{n^3}({\mu }_4+3(n-1){\sigma }^4):\mathrm{標}\mathrm{本}\mathrm{平}\mathrm{均}\mathrm{の}\mathrm{母}\mathrm{平}\mathrm{均}\mathrm{ま}\mathrm{わ}\mathrm{り}\mathrm{の}4\mathrm{次}\mathrm{モ}\mathrm{ー}\mathrm{メ}\mathrm{ン}\mathrm{ト}(\mathrm{標}\mathrm{本}\mathrm{平}\mathrm{均}\mathrm{の}4\mathrm{次}\mathrm{の}\mathrm{中}\mathrm{心}(\mathrm{化})\mathrm{モ}\mathrm{ー}\mathrm{メ}\mathrm{ン}\mathrm{ト})\\ & & \cdots \mathrm{V}\left[\bar{X}\right]=\frac{{\sigma }^2}{n}:\mathrm{標}\mathrm{本}\mathrm{平}\mathrm{均}\mathrm{の}\mathrm{分}\mathrm{散}\\ & =& \frac{n^2}{n^3}\frac{{\mu }_4+3(n-1){\sigma }^4}{{\sigma }^4}-3\\ & =& \frac{1}{n}(\frac{{\mu }_4}{{\sigma }^4}+3(n-1))-3\\ & =& \frac{1}{n}(\frac{{\mu }_4}{{\sigma }^4}+3n-3)-3\\ & =& \frac{\frac{{\mu }_4}{{\sigma }^4}-3}{n}+3-3\\ & =& \frac{\frac{{\mu }_4}{{\sigma }^4}-3}{n}\\ & =& \frac{{\beta }_2}{n}\cdots {\beta }_2=\frac{{\mu }_4}{{\sigma }^4}-3\end{array}$$ よって標本数$n$を増やすことで${\beta }_2\left(\bar{X}\right)$は$0$に近づいていく．

母集団の尖度

母集団(平均 $\mu$, 分散 ${\sigma }^2$)の尖度${\beta }_2$

$$\begin{array}{rcl}{\beta }_2& =& \frac{\mathrm{E}\left[{(X-\mu )}^4\right]}{\mathrm{V}{\left[X\right]}^{\frac{4}{2}}}-3\cdots \mathrm{尖}\mathrm{度}(\mathrm{せ}\mathrm{ん}\mathrm{ど},kurtosis)\\ & =& \frac{{\mu }_4}{{\sigma }^4}-3\\ & & \dots {\mu }_4=\mathrm{E}\left[{(X-\mu )}^3\right](\mathrm{確}\mathrm{率}\mathrm{変}\mathrm{数}X\mathrm{の}\mathrm{母}\mathrm{平}\mathrm{均}\mu \mathrm{ま}\mathrm{わ}\mathrm{り}\mathrm{の}4\mathrm{次}\mathrm{モ}\mathrm{ー}\mathrm{メ}\mathrm{ン}\mathrm{ト}=\mathrm{確}\mathrm{率}\mathrm{変}\mathrm{数}X\mathrm{の}4\mathrm{次}\mathrm{の}\mathrm{中}\mathrm{心}(\mathrm{化})\mathrm{モ}\mathrm{ー}\mathrm{メ}\mathrm{ン}\mathrm{ト})\end{array}$$

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

$$\begin{array}{rcl}{\mu }_4& =& ({\beta }_2-3){\sigma }^4\end{array}$$

カイ二乗分布の期待値と分散

カイ二乗分布の期待値と分散

カイ二乗分布の期待値(一次モーメント)

$$\begin{array}{rcl}\mathrm{E}\left[x\right]& =& {\int }_0^{\mathrm{\infty }}x{\chi }^2(x)\mathrm{d}x\\ & =& {\int }_0^{\mathrm{\infty }}x\frac{1}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{e}^{-\frac{x}{2}}{x}^{\frac{n}{2}-1}\mathrm{d}x\\ & =& \frac{1}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{\int }_0^{\mathrm{\infty }}x{e}^{-\frac{x}{2}}{x}^{\frac{n}{2}-1}\mathrm{d}x\\ & =& \frac{1}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{\int }_0^{\mathrm{\infty }}{e}^{-\frac{x}{2}}{x}^{\frac{n}{2}}\mathrm{d}x\\ & =& \frac{1}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{\int }_0^{\mathrm{\infty }}{e}^{-\frac{x}{2}}{x}^{\frac{n}{2}}{2}^{\frac{n}{2}}{\left(\frac{1}{2}\right)}^{\frac{n}{2}}\mathrm{d}x\\ & =& \frac{1}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{2}^{\frac{n}{2}}{\int }_0^{\mathrm{\infty }}{e}^{-\frac{x}{2}}{x}^{\frac{n}{2}}{\left(\frac{1}{2}\right)}^{\frac{n}{2}}\mathrm{d}x\\ & =& \frac{1}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}{\int }_0^{\mathrm{\infty }}{e}^{-\frac{x}{2}}{\left(\frac{x}{2}\right)}^{\frac{n}{2}}\mathrm{d}x\\ & =& \frac{1}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}{\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{\frac{n}{2}}2\mathrm{d}t\cdots t=\frac{x}{2},\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{1}{2},\mathrm{d}x=2\mathrm{d}t\\ & =& \frac{1}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}2{\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{\frac{n}{2}}\mathrm{d}t\cdots \int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x\\ & =& \frac{1}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}2{\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{\frac{n}{2}+1-1}\mathrm{d}t\\ & =& \frac{1}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}2\mathrm{\Gamma }(\frac{n}{2}+1)\cdots \mathrm{\Gamma }\left(s\right)={\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{s-1}\mathrm{d}t\\ & =& \frac{1}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}2\frac{n}{2}\mathrm{\Gamma }\left(\frac{n}{2}\right)\cdots \mathrm{\Gamma }(s+1)={\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^s\mathrm{d}t=s\mathrm{\Gamma }(s)\\ & =& n\end{array}$$

カイ二乗分布の二次モーメント

$$\begin{array}{rcl}\mathrm{E}\left[x^2\right]& =& {\int }_0^{\mathrm{\infty }}{x}^2{\chi }^2(x)\mathrm{d}x\\ & =& {\int }_0^{\mathrm{\infty }}{x}^2\frac{1}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{e}^{-\frac{x}{2}}{x}^{\frac{n}{2}-1}\mathrm{d}x\\ & =& \frac{1}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{\int }_0^{\mathrm{\infty }}{x}^2{e}^{-\frac{x}{2}}{x}^{\frac{n}{2}-1}\mathrm{d}x\\ & =& \frac{1}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{\int }_0^{\mathrm{\infty }}{e}^{-\frac{x}{2}}{x}^{\frac{n}{2}+1}\mathrm{d}x\\ & =& \frac{1}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{\int }_0^{\mathrm{\infty }}{e}^{-\frac{x}{2}}{x}^{\frac{n}{2}+1}{2}^{\frac{n}{2}+1}{\left(\frac{1}{2}\right)}^{\frac{n}{2}+1}\mathrm{d}x\\ & =& \frac{1}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{2}^{\frac{n}{2}+1}{\int }_0^{\mathrm{\infty }}{e}^{-\frac{x}{2}}{x}^{\frac{n}{2}+1}{\left(\frac{1}{2}\right)}^{\frac{n}{2}+1}\mathrm{d}x\\ & =& \frac{1}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{2}^{\frac{n}{2}}2{\int }_0^{\mathrm{\infty }}{e}^{-\frac{x}{2}}{x}^{\frac{n}{2}+1}{\left(\frac{1}{2}\right)}^{\frac{n}{2}+1}\mathrm{d}x\\ & =& \frac{1}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{2}^{\frac{n}{2}}2{\int }_0^{\mathrm{\infty }}{e}^{-\frac{x}{2}}{x}^{\frac{n}{2}+1}{\left(\frac{1}{2}\right)}^{\frac{n}{2}+1}\mathrm{d}x\\ & =& \frac{2}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}{\int }_0^{\mathrm{\infty }}{e}^{-\frac{x}{2}}{\left(\frac{x}{2}\right)}^{\frac{n}{2}+1}\mathrm{d}x\\ & =& \frac{2}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}{\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{\frac{n}{2}+1}2\mathrm{d}t\cdots t=\frac{x}{2},\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{1}{2},\mathrm{d}x=2\mathrm{d}t\\ & =& \frac{2}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}2{\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{\frac{n}{2}+1}\mathrm{d}t\cdots \int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x\\ & =& \frac{2}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}2{\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{\frac{n}{2}+1+1-1}\mathrm{d}t\\ & =& \frac{2}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}2{\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{\frac{n}{2}+2-1}\mathrm{d}t\\ & =& \frac{2}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}2\mathrm{\Gamma }(\frac{n}{2}+2)\cdots \mathrm{\Gamma }\left(s\right)={\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{s-1}\mathrm{d}t\\ & =& \frac{2}{\mathrm{\Gamma }\left(\frac{n}{2}\right)}2(\frac{n}{2}+1)\frac{n}{2}\mathrm{\Gamma }\left(\frac{n}{2}\right)\cdots \mathrm{\Gamma }(s+2)=(s+1)\mathrm{\Gamma }(s+1)=(s+1)s\mathrm{\Gamma }(s)\\ & =& 2n(\frac{n}{2}+1)\\ & =& n(n+2)\end{array}$$

カイ二乗分布の分散(二次の中心モーメント)

$$\begin{array}{rcl}\mathrm{V}\left[x^2\right]& =& \mathrm{E}[(x-\mathrm{E}\left[x\right]{)}^2]\\ & =& \mathrm{E}\left[x^2\right]-\mathrm{E}{\left[x\right]}^2\\ & =& n(n+2)-n^2\\ & =& n^2+2n-n^2\\ & =& 2n\end{array}$$

Γ(s+1)=sΓ(s)

$\mathrm{\Gamma }(s+1)=s\mathrm{\Gamma }(s)$

$$\begin{array}{rcl}\mathrm{\Gamma }(s+1)& =& {\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{s+1-1}\mathrm{d}t\cdots \mathrm{\Gamma }(s)={\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{s-1}\mathrm{d}t\\ & =& {\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^s\mathrm{d}t\\ & =& {\int }_0^{\mathrm{\infty }}{\{-e^{-t}\}}^{\prime }{t}^s\mathrm{d}t\cdots \frac{\mathrm{d}}{\mathrm{d}t}(-e^{-t})={\{-e^{-t}\}}^{\prime }=e^{-t}\\ & =& {[-e^{-t}{t}^s]}_0^{\mathrm{\infty }}-{\int }_0^{\mathrm{\infty }}\{-e^{-t}\}\left\{s{t}^{s-1}\right\}\mathrm{d}t\cdots \int f^{\prime }g\mathrm{d}x=fg-\int f{g}^{\prime }\mathrm{d}x\\ & =& [(\underset{t\to \mathrm{\infty }}{lim}-\frac{t^s}{e^t})-(-\frac{0^s}{e^0})]+s{\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{s-1}\mathrm{d}t\cdots \int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x\\ & =& [0-0]+s{\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{s-1}\mathrm{d}t\cdots \underset{t\to \mathrm{\infty }}{lim}\frac{t^s}{e^t}=0\\ & =& s\mathrm{\Gamma }(s)\cdots \mathrm{\Gamma }(s)={\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{s-1}\mathrm{d}t\end{array}$$

$\mathrm{\Gamma }(s+2)=(s+1)s\mathrm{\Gamma }(s)$

$$\begin{array}{rcl}\mathrm{\Gamma }(s+2)& =& (s+1)\mathrm{\Gamma }(s+1)\\ & =& (s+1)s\mathrm{\Gamma }(s)\end{array}$$

(x^n)/(e^x)の極限

(x^n)/(e^x)の極限

以下を証明する $$\begin{array}{rcl}\underset{x\to \mathrm{\infty }}{lim}\frac{x^n}{e^x}& =& 0\end{array}$$ まず$e^x$のマクローリン展開から$e^x$より小さい値を考える． $$\begin{array}{r}\\ e^x& =& \sum _{k=0}^{\mathrm{\infty }}\frac{x^k}{k!}\\ & =& \sum _{k=0}^{\mathrm{\infty }}{f}_k(x)\cdots f_k(x)=\frac{x^k}{k!}\\ & >& \frac{x^{n+1}}{(n+1)!}\cdots \mathrm{い}\mathrm{く}\mathrm{つ}\mathrm{も}\mathrm{の}{f}_k\mathrm{を}\mathrm{足}\mathrm{し}\mathrm{合}\mathrm{わ}\mathrm{せ}\mathrm{た}{e}^x\mathrm{の}\mathrm{方}\mathrm{が}\mathrm{特}\mathrm{定}\mathrm{の}{f}_k(\mathrm{例}\mathrm{え}\mathrm{ば}k=n+1)\mathrm{だ}\mathrm{け}\mathrm{よ}\mathrm{り}\mathrm{大}\mathrm{き}\mathrm{い}\end{array}$$ 上記値の逆数をとる． $$\begin{array}{r}\\ \frac{1}{e^x}& <& \frac{(n+1)!}{x^{n+1}}\cdots \mathrm{逆}\mathrm{数}\mathrm{な}\mathrm{の}\mathrm{で}\mathrm{大}\mathrm{小}\mathrm{関}\mathrm{係}\mathrm{が}\mathrm{反}\mathrm{対}\mathrm{に}\mathrm{な}\mathrm{る}\end{array}$$ 両辺にx^nを掛け，証明したい式の形にする． $$\begin{array}{r}\\ x^n\frac{1}{e^x}& <& x^n\frac{(n+1)!}{x^{n+1}}\cdots \mathrm{両}\mathrm{辺}\mathrm{に}{x}^n\mathrm{を}\mathrm{掛}\mathrm{け}\mathrm{る}\\ & <& \frac{(n+1)!}{x}\cdots \frac{x^n}{x^{n+1}}=\frac{1}{x}\end{array}$$ 両辺の極限をとることで$\frac{x^n}{e^x}$の極限の値が0より小さいことがわかる． $$\begin{array}{r}\\ \underset{x\to \mathrm{\infty }}{lim}\frac{x^n}{e^x}& <& \underset{x\to \mathrm{\infty }}{lim}\frac{(n+1)!}{x}& =& 0\end{array}$$ また，$0\le x$において$0\le x^n$及び$0<e^x$となるので$\frac{x^n}{e^x}$が0以上であることがわかる．
以上から次の関係（不等式）を得る． $$\begin{array}{rclrclr}0& \le & \underset{x\to \mathrm{\infty }}{lim}\frac{x^n}{e^x}& <& \underset{x\to \mathrm{\infty }}{lim}\frac{(n+1)!}{x}& =& 0\end{array}$$ これより“はさみうちの原理”から極限の値が0であることが証明された． $$\begin{array}{r}\\ \underset{x\to \mathrm{\infty }}{lim}\frac{x^n}{e^x}& =& 0\end{array}$$

定積分を用いた凾数同士の被積分凾数

定積分を用いた凾数同士の被積分凾数

$$\begin{array}{rcl}F(x)& =& {\int }_a^{x}f(t)\mathrm{d}t\cdots \mathrm{定}\mathrm{積}\mathrm{分}\mathrm{を}\mathrm{用}\mathrm{い}\mathrm{た}\mathrm{凾}\mathrm{数}\\ G(x)& =& {\int }_a^{x}g(t)\mathrm{d}t\\ \mathrm{た}\mathrm{だ}\mathrm{し},& & a\le x\le b\end{array}$$ とする．今，任意の$x$において($x$の取りえる範囲すべての$x$において) $$\begin{array}{rcl}F(x)& =& G(x)\end{array}$$ が成り立つなら(特定の積分範囲の定積分結果が等しいだけではないところに注意)， $$\begin{array}{rcl}{F}^{\prime }(x)& =& G^{\prime }(x)\end{array}$$ つまり $$\begin{array}{rcl}f(x)& =& g(x)\end{array}$$ が成り立つ．
よって $$\begin{array}{rcl}{\int }_a^{x}f(t)\mathrm{d}t& =& {\int }_a^{x}g(t)\mathrm{d}t\end{array}$$ ならば $$\begin{array}{r}\\ f(x)& =& g(x)\end{array}$$ であり，被積分凾数同士も等しい．

カイ二乗分布の導出

カイ二乗${\chi }^2$分布の導出

$k=1$のXの確率密度凾数

$$\begin{array}{rcl}X& =& Z_1^2\cdots Z_1:\mathrm{標}\mathrm{準}\mathrm{正}\mathrm{規}\mathrm{分}\mathrm{布}\mathrm{に}\mathrm{従}\mathrm{う}\mathrm{確}\mathrm{率}\mathrm{変}\mathrm{数}\\ F_1(X=t)& =& {\int }_0^t{f}_1(x)\mathrm{d}x\cdots k=1\mathrm{の}X\mathrm{の}\mathrm{累}\mathrm{積}\mathrm{分}\mathrm{布}\mathrm{凾}\mathrm{数}{F}_1,\mathrm{確}\mathrm{率}\mathrm{密}\mathrm{度}\mathrm{凾}\mathrm{数}{f}_1\\ F_1(X=t)& =& {\int }_{-\sqrt{t}}^{+\sqrt{t}}g(z_1)\mathrm{d}{z}_1\cdots Z_1=\pm \sqrt{X},x:0\to t,z_1:0\to \pm \sqrt{t}\mathrm{標}\mathrm{準}\mathrm{正}\mathrm{規}\mathrm{分}\mathrm{布}\mathrm{の}\mathrm{確}\mathrm{率}\mathrm{密}\mathrm{度}\mathrm{凾}\mathrm{数}g\\ & =& 2{\int }_0^{\sqrt{t}}g(z_1)\mathrm{d}{z}_1\cdots \mathrm{標}\mathrm{準}\mathrm{正}\mathrm{規}\mathrm{分}\mathrm{布}\mathrm{は}\mathrm{偶}\mathrm{凾}\mathrm{数}(y\mathrm{軸}(x＝0)\mathrm{で}\mathrm{左}\mathrm{右}\mathrm{対}\mathrm{称}),Z_1\ge 0\mathrm{の}\mathrm{み}\mathrm{考}\mathrm{え}\mathrm{れ}\mathrm{ば}\mathrm{よ}\mathrm{く}\mathrm{す}\mathrm{る}\\ & =& 2{\int }_0^{t}g(\sqrt{x})\frac{1}{2}{x}^{-\frac{1}{2}}\mathrm{d}x\\ & & \cdots Z_1=\sqrt{X}(Z_1\ge 0)\mathrm{よ}\mathrm{り}\frac{\mathrm{d}{z}_1}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}\sqrt{x}=\frac{\mathrm{d}}{\mathrm{d}x}{x}^{\frac{1}{2}}=\frac{1}{2}{x}^{-\frac{1}{2}},\mathrm{d}{z}_1=\frac{1}{2}{x}^{-\frac{1}{2}}\mathrm{d}x\\ & & \cdots z_1:0\to \sqrt{t},x:0\to t\\ & =& {\int }_0^{t}g(\sqrt{x})x^{-\frac{1}{2}}\mathrm{d}x\\ {\int }_0^t{f}_1(x)\mathrm{d}x& =& {\int }_0^{t}g(\sqrt{x})x^{-\frac{1}{2}}\mathrm{d}x\cdots \mathrm{累}\mathrm{積}\mathrm{分}\mathrm{布}\mathrm{凾}\mathrm{数}\mathrm{同}\mathrm{士}\mathrm{な}\mathrm{の}\mathrm{で}\mathrm{等}\mathrm{し}\mathrm{い}(0\mathrm{か}\mathrm{ら}\mathrm{任}\mathrm{意}\mathrm{の}t(>0)\mathrm{ま}\mathrm{で}\mathrm{の}\mathrm{定}\mathrm{積}\mathrm{分}\mathrm{が}\mathrm{等}\mathrm{し}\mathrm{い})\\ f_1(x)& =& g(\sqrt{x})x^{-\frac{1}{2}}\cdots 0\mathrm{か}\mathrm{ら}\mathrm{任}\mathrm{意}\mathrm{の}t(>0)\mathrm{ま}\mathrm{で}\mathrm{の}\mathrm{定}\mathrm{積}\mathrm{分}\mathrm{が}\mathrm{等}\mathrm{し}\mathrm{い}\to 0\mathrm{か}\mathrm{ら}t\mathrm{の}\mathrm{範}\mathrm{囲}\mathrm{で}\mathrm{等}\mathrm{し}\mathrm{い}\to \mathrm{各}t\mathrm{で}\mathrm{の}\mathrm{微}\mathrm{分}\mathrm{が}\mathrm{等}\mathrm{し}\mathrm{い}\to \mathrm{被}\mathrm{積}\mathrm{分}\mathrm{凾}\mathrm{数}\mathrm{同}\mathrm{士}\mathrm{も}\mathrm{等}\mathrm{し}\mathrm{い}\\ & =& \frac{1}{\sqrt{2\pi }}{e}^{-\frac{{\sqrt{x}}^2}{2}}{x}^{-\frac{1}{2}}\cdots \mathrm{標}\mathrm{準}\mathrm{正}\mathrm{規}\mathrm{分}\mathrm{布}\mathrm{の}\mathrm{確}\mathrm{率}\mathrm{密}\mathrm{度}\mathrm{凾}\mathrm{数}:g(x)=\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}\\ & =& \frac{1}{\sqrt{2\pi }}{e}^{-\frac{x}{2}}{x}^{-\frac{1}{2}}\end{array}$$

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$k=2$のXの確率密度凾数

$k=2$ $$\begin{array}{rcl}X& =& Z_1^2+Z_2^2\cdots Z_1,Z_2:\mathrm{互}\mathrm{い}\mathrm{に}\mathrm{独}\mathrm{立}\mathrm{に}\mathrm{標}\mathrm{準}\mathrm{正}\mathrm{規}\mathrm{分}\mathrm{布}\mathrm{に}\mathrm{従}\mathrm{う}\mathrm{確}\mathrm{率}\mathrm{変}\mathrm{数}\\ Y& =& Z_1^2\\ X-Y& =& Z_2^2\\ f_2(x)& =& {\int }_0^{\mathrm{\infty }}{f}_1(y)f_1(x-y)\mathrm{d}y\cdots k=2\mathrm{の}X\mathrm{の}\mathrm{確}\mathrm{率}\mathrm{密}\mathrm{度}\mathrm{凾}\mathrm{数}{f}_2,Z_1\mathrm{と}{Z}_2\mathrm{は}\mathrm{独}\mathrm{立}\mathrm{な}\mathrm{の}\mathrm{で}\mathrm{同}\mathrm{時}\mathrm{確}\mathrm{率}\mathrm{は}{f}_1(Z_1)\mathrm{と}{f}_1(Z_2)\mathrm{の}\mathrm{積}\\ & =& {\int }_0^x\left\{\frac{1}{\sqrt{2\pi }}{e}^{-\frac{y}{2}}{y}^{-\frac{1}{2}}\right\}\cdot \{\frac{1}{\sqrt{2\pi }}{e}^{-\frac{(x-y)}{2}}(x-y{)}^{-\frac{1}{2}}\}\mathrm{d}y\\ & =& \frac{1}{\sqrt{2\pi }}\frac{1}{\sqrt{2\pi }}{\int }_0^x{e}^{-\frac{y}{2}}{e}^{-\frac{(x-y)}{2}}{y}^{-\frac{1}{2}}(x-y{)}^{-\frac{1}{2}}\mathrm{d}y\\ & =& \frac{1}{2\pi }{\int }_0^x{e}^{-\frac{y}{2}-\frac{(x-y)}{2}}{y}^{-\frac{1}{2}}(x-y{)}^{-\frac{1}{2}}\mathrm{d}y\\ & =& \frac{1}{2\pi }{e}^{-\frac{x}{2}}{\int }_0^x{y}^{-\frac{1}{2}}(x-y{)}^{-\frac{1}{2}}\mathrm{d}y\\ & =& \frac{1}{2\pi }{e}^{-\frac{x}{2}}\pi \cdots {\int }_0^a\frac{1}{\sqrt{x}\sqrt{a-x}}\mathrm{d}x=\pi \\ & =& \frac{1}{2}{e}^{-\frac{x}{2}}\end{array}$$

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$k=3$のXの確率密度凾数

$$\begin{array}{rcl}X& =& Z_1^2+Z_2^2+Z_3^2\cdots Z_1,Z_2,Z_3:\mathrm{互}\mathrm{い}\mathrm{に}\mathrm{独}\mathrm{立}\mathrm{に}\mathrm{標}\mathrm{準}\mathrm{正}\mathrm{規}\mathrm{分}\mathrm{布}\mathrm{に}\mathrm{従}\mathrm{う}\mathrm{確}\mathrm{率}\mathrm{変}\mathrm{数}\\ Y& =& Z_1^2+Z_2^2\\ X-Y& =& Z_3^2\\ f_3(x)& =& {\int }_0^{\mathrm{\infty }}{f}_2(y)f_1(x-y)\mathrm{d}y\cdots k=3\mathrm{の}X\mathrm{の}\mathrm{確}\mathrm{率}\mathrm{密}\mathrm{度}\mathrm{凾}\mathrm{数}{f}_3,Y\mathrm{と}{Z}_3\mathrm{は}\mathrm{独}\mathrm{立}\mathrm{な}\mathrm{の}\mathrm{で}\mathrm{同}\mathrm{時}\mathrm{確}\mathrm{率}\mathrm{は}{f}_2(Y)\mathrm{と}{f}_1(Z_3)\mathrm{の}\mathrm{積}\\ & =& {\int }_0^x\left\{\frac{1}{2}{e}^{-\frac{y}{2}}\right\}\cdot \{\frac{1}{\sqrt{2\pi }}{e}^{-\frac{(x-y)}{2}}(x-y{)}^{-\frac{1}{2}}\}\mathrm{d}y\\ & =& \frac{1}{2}\frac{1}{\sqrt{2\pi }}{\int }_0^x{e}^{-\frac{y}{2}}{e}^{-\frac{(x-y)}{2}}(x-y{)}^{-\frac{1}{2}}\mathrm{d}y\\ & =& \frac{1}{2\sqrt{2\pi }}{\int }_0^x{e}^{-\frac{y}{2}-\frac{(x-y)}{2}}(x-y{)}^{-\frac{1}{2}}\mathrm{d}y\\ & =& \frac{1}{2\sqrt{2\pi }}{e}^{-\frac{x}{2}}{\int }_0^x(x-y{)}^{-\frac{1}{2}}\mathrm{d}y\\ & =& \frac{1}{2\sqrt{2\pi }}{e}^{-\frac{x}{2}}2x^{\frac{1}{2}}\cdots {\int }_0^a\frac{1}{\sqrt{a-x}}\mathrm{d}x=2a^{\frac{1}{2}}\\ & =& \frac{1}{\sqrt{2\pi }}{e}^{-\frac{x}{2}}{x}^{\frac{1}{2}}\end{array}$$

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$k=4$のXの確率密度凾数

$$\begin{array}{rcl}X& =& Z_1^2+Z_2^2+Z_3^2+Z_4^2\cdots Z_1,Z_2,Z_3,Z_4:\mathrm{互}\mathrm{い}\mathrm{に}\mathrm{独}\mathrm{立}\mathrm{に}\mathrm{標}\mathrm{準}\mathrm{正}\mathrm{規}\mathrm{分}\mathrm{布}\mathrm{に}\mathrm{従}\mathrm{う}\mathrm{確}\mathrm{率}\mathrm{変}\mathrm{数}\\ Y& =& Z_1^2+Z_2^2+Z_3^2\\ X-Y& =& Z_4^2\\ f_4(x)& =& {\int }_0^{\mathrm{\infty }}{f}_3(y)f_1(x-y)\mathrm{d}y\cdots k=4\mathrm{の}X\mathrm{の}\mathrm{確}\mathrm{率}\mathrm{密}\mathrm{度}\mathrm{凾}\mathrm{数}{f}_4,Y\mathrm{と}{Z}_4\mathrm{は}\mathrm{独}\mathrm{立}\mathrm{な}\mathrm{の}\mathrm{で}\mathrm{同}\mathrm{時}\mathrm{確}\mathrm{率}\mathrm{は}{f}_3(Y)\mathrm{と}{f}_1(Z_4)\mathrm{の}\mathrm{積}\\ & =& {\int }_0^x\left\{\frac{1}{\sqrt{2\pi }}{e}^{-\frac{y}{2}}{y}^{\frac{1}{2}}\right\}\cdot \{\frac{1}{\sqrt{2\pi }}{e}^{-\frac{(x-y)}{2}}(x-y{)}^{-\frac{1}{2}}\}\mathrm{d}y\\ & =& \frac{1}{\sqrt{2\pi }}\frac{1}{\sqrt{2\pi }}{\int }_0^x{e}^{-\frac{y}{2}}{e}^{-\frac{(x-y)}{2}}{y}^{\frac{1}{2}}(x-y{)}^{-\frac{1}{2}}\mathrm{d}y\\ & =& \frac{1}{2\pi }{\int }_0^x{e}^{-\frac{y}{2}-\frac{(x-y)}{2}}{y}^{\frac{1}{2}}(x-y{)}^{-\frac{1}{2}}\mathrm{d}y\\ & =& \frac{1}{2\pi }{e}^{-\frac{x}{2}}{\int }_0^x{y}^{\frac{1}{2}}(x-y{)}^{-\frac{1}{2}}\mathrm{d}y\\ & =& \frac{1}{2\pi }{e}^{-\frac{x}{2}}\frac{x\pi }{2}\cdots {\int }_0^a\frac{\sqrt{x}}{\sqrt{a-x}}\mathrm{d}x=\frac{a\pi }{2}\\ & =& \frac{1}{4}{e}^{-\frac{x}{2}}x\end{array}$$

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$k=5$のXの確率密度凾数

$$\begin{array}{rcl}X& =& Z_1^2+Z_2^2+Z_3^2+Z_4^2+Z_5^2\cdots Z_1,Z_2,Z_3,Z_4,Z_5:\mathrm{互}\mathrm{い}\mathrm{に}\mathrm{独}\mathrm{立}\mathrm{に}\mathrm{標}\mathrm{準}\mathrm{正}\mathrm{規}\mathrm{分}\mathrm{布}\mathrm{に}\mathrm{従}\mathrm{う}\mathrm{確}\mathrm{率}\mathrm{変}\mathrm{数}\\ Y& =& Z_1^2+Z_2^2+Z_3^2+Z_4^2\\ X-Y& =& Z_5^2\\ f_5(x)& =& {\int }_0^{\mathrm{\infty }}{f}_4(y)f_1(x-y)\mathrm{d}y\cdots k=5\mathrm{の}X\mathrm{の}\mathrm{確}\mathrm{率}\mathrm{密}\mathrm{度}\mathrm{凾}\mathrm{数}{f}_5,Y\mathrm{と}{Z}_4\mathrm{は}\mathrm{独}\mathrm{立}\mathrm{な}\mathrm{の}\mathrm{で}\mathrm{同}\mathrm{時}\mathrm{確}\mathrm{率}\mathrm{は}{f}_4(Y)\mathrm{と}{f}_1(Z_5)\mathrm{の}\mathrm{積}\\ & =& {\int }_0^x\left\{\frac{1}{4}{e}^{-\frac{y}{2}}y\right\}\cdot \{\frac{1}{\sqrt{2\pi }}{e}^{-\frac{(x-y)}{2}}(x-y{)}^{-\frac{1}{2}}\}\mathrm{d}y\\ & =& \frac{1}{4}\frac{1}{\sqrt{2\pi }}{\int }_0^x{e}^{-\frac{y}{2}}{e}^{-\frac{(x-y)}{2}}y(x-y{)}^{-\frac{1}{2}}\mathrm{d}y\\ & =& \frac{1}{4\sqrt{2\pi }}{\int }_0^x{e}^{-\frac{y}{2}-\frac{(x-y)}{2}}y(x-y{)}^{-\frac{1}{2}}\mathrm{d}y\\ & =& \frac{1}{4\sqrt{2\pi }}{e}^{-\frac{x}{2}}{\int }_0^{x}y(x-y{)}^{-\frac{1}{2}}\mathrm{d}y\\ & =& \frac{1}{4\sqrt{2\pi }}{e}^{-\frac{x}{2}}\frac{4}{3}{x}^{\frac{3}{2}}\cdots {\int }_0^a\frac{x}{\sqrt{a-x}}\mathrm{d}x=\frac{4}{3}{a}^{\frac{3}{2}}\\ & =& \frac{1}{3\sqrt{2\pi }}{e}^{-\frac{x}{2}}{x}^{\frac{3}{2}}\end{array}$$

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$k=n$の$X$の確率密度凾数の類推

$$\begin{array}{}k=1& :& f_1(x)=& \frac{1}{\sqrt{2\pi }}& e^{-\frac{x}{2}}& x^{-\frac{1}{2}}\\ k=2& :& f_2(x)=& \frac{1}{2}& e^{-\frac{x}{2}}& x^0\\ k=3& :& f_3(x)=& \frac{1}{\sqrt{2\pi }}& e^{-\frac{x}{2}}& x^{\frac{1}{2}}\\ k=4& :& f_4(x)=& \frac{1}{4}& e^{-\frac{x}{2}}& x^1\\ k=5& :& f_5(x)=& \frac{1}{3\sqrt{2\pi }}& e^{-\frac{x}{2}}& x^{\frac{3}{2}}\\ k=n& :& f_n(x)=& {\alpha }_n& e^{-\frac{x}{2}}& x^{\frac{n}{2}-1}\end{array}$$

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係数${\alpha }_n$を求める

$$\begin{array}{rcl}{F}_n(t)& =& {\int }_0^t{f}_n(x)\mathrm{d}x\cdots \mathrm{任}\mathrm{意}\mathrm{の}n\mathrm{に}\mathrm{お}\mathrm{け}\mathrm{る}X\mathrm{の}\mathrm{累}\mathrm{積}\mathrm{分}\mathrm{布}\mathrm{凾}\mathrm{数}{F}_n,\mathrm{確}\mathrm{率}\mathrm{密}\mathrm{度}\mathrm{凾}\mathrm{数}{f}_n\\ F_n(t)& =& {\int }_0^t{\alpha }_n{e}^{-\frac{x}{2}}{x}^{\frac{n}{2}-1}\mathrm{d}x\cdots \mathrm{前}\mathrm{述}\mathrm{の}\mathrm{類}\mathrm{推}\mathrm{よ}\mathrm{り}\\ & =& {\alpha }_n{\int }_0^t{e}^{-\frac{x}{2}}{x}^{\frac{n}{2}-1}\mathrm{d}x\cdots {\alpha }_n\mathrm{は}x\mathrm{に}\mathrm{よ}\mathrm{ら}\mathrm{な}\mathrm{い}\mathrm{定}\mathrm{数},\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x\\ & =& {\alpha }_n{\int }_0^t{e}^{-t}(2u{)}^{\frac{n}{2}-1}2\mathrm{d}u\cdots u=\frac{x}{2},x=2u,\frac{dx}{du}=2,x:0\to \mathrm{\infty },u:0\to \mathrm{\infty }\\ & =& {\alpha }_n{\int }_0^t{e}^{-t}{2}^{\frac{n}{2}-1}{t}^{\frac{n}{2}-1}2\mathrm{d}t\\ & =& {\alpha }_n{\int }_0^t{e}^{-t}{2}^{\frac{n}{2}}{t}^{\frac{n}{2}-1}\mathrm{d}t\\ & =& {\alpha }_n{2}^{\frac{n}{2}}{\int }_0^t{e}^{-t}{t}^{\frac{n}{2}-1}\mathrm{d}t\\ F_n(t=\mathrm{\infty })& =& {\int }_0^{\mathrm{\infty }}{f}_n(x)\mathrm{d}x\\ & =& {\alpha }_n{2}^{\frac{n}{2}}{\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{\frac{n}{2}-1}\mathrm{d}t\\ & =& {\alpha }_n{2}^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)\cdots \mathrm{\Gamma }\left(s\right)={\int }_0^{\mathrm{\infty }}{e}^{-t}{t}^{s-1}\mathrm{d}t\\ & =& {\alpha }_n{2}^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)\\ & =& 1\cdots \mathrm{全}\mathrm{事}\mathrm{象}\mathrm{は}1\\ {\alpha }_n& =& \frac{F_n(t=\mathrm{\infty })}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}\\ & =& \frac{1}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}\end{array}$$

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係数${\alpha }_k$の確認

$$\begin{array}{}{\alpha }_1& =& \frac{1}{2^{\frac{1}{2}}\mathrm{\Gamma }\left(\frac{1}{2}\right)}& =& \frac{1}{2^{\frac{1}{2}}\sqrt{\pi }}& =& \frac{1}{\sqrt{2\pi }}\\ {\alpha }_2& =& \frac{1}{2^{\frac{2}{2}}\mathrm{\Gamma }\left(\frac{2}{2}\right)}& =& \frac{1}{2^{1}1}& =& \frac{1}{2}\\ {\alpha }_3& =& \frac{1}{2^{\frac{3}{2}}\mathrm{\Gamma }\left(\frac{3}{2}\right)}& =& \frac{1}{2^{\frac{3}{2}}\frac{\sqrt{\pi }}{2}}& =& \frac{1}{\sqrt{2\pi }}\\ {\alpha }_4& =& \frac{1}{2^{\frac{4}{2}}\mathrm{\Gamma }\left(\frac{4}{2}\right)}& =& \frac{1}{2^{2}1}& =& \frac{1}{4}\\ {\alpha }_5& =& \frac{1}{2^{\frac{5}{2}}\mathrm{\Gamma }\left(\frac{5}{2}\right)}& =& \frac{1}{2^{\frac{5}{2}}\frac{3\sqrt{\pi }}{4}}& =& \frac{1}{3\sqrt{2\pi }}\end{array}$$

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累積分布凾数の等式より$k=n$の$X$の確率密度凾数を求める

$$\begin{array}{r}\\ {\int }_0^t{f}_n(x)\mathrm{d}x& =& {\int }_0^t\frac{1}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{e}^{-\frac{x}{2}}{x}^{\frac{n}{2}-1}\mathrm{d}x\cdots \mathrm{累}\mathrm{積}\mathrm{分}\mathrm{布}\mathrm{凾}\mathrm{数}\mathrm{同}\mathrm{士}\mathrm{な}\mathrm{の}\mathrm{で}\mathrm{等}\mathrm{し}\mathrm{い}(0\mathrm{か}\mathrm{ら}\mathrm{任}\mathrm{意}\mathrm{の}t(>0)\mathrm{ま}\mathrm{で}\mathrm{の}\mathrm{定}\mathrm{積}\mathrm{分}\mathrm{が}\mathrm{等}\mathrm{し}\mathrm{い})\\ f_n(x)& =& \frac{1}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{e}^{-\frac{x}{2}}{x}^{\frac{n}{2}-1}\cdots 0\mathrm{か}\mathrm{ら}\mathrm{任}\mathrm{意}\mathrm{の}t(>0)\mathrm{ま}\mathrm{で}\mathrm{の}\mathrm{定}\mathrm{積}\mathrm{分}\mathrm{が}\mathrm{等}\mathrm{し}\mathrm{い}\to 0\mathrm{か}\mathrm{ら}t\mathrm{の}\mathrm{範}\mathrm{囲}\mathrm{で}\mathrm{等}\mathrm{し}\mathrm{い}\to \mathrm{各}t\mathrm{で}\mathrm{の}\mathrm{微}\mathrm{分}\mathrm{が}\mathrm{等}\mathrm{し}\mathrm{い}\to \mathrm{被}\mathrm{積}\mathrm{分}\mathrm{凾}\mathrm{数}\mathrm{同}\mathrm{士}\mathrm{も}\mathrm{等}\mathrm{し}\mathrm{い}\end{array}$$

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数学的帰納法による証明

$f_1(x),f_n(x)$を認めた上で，$f_{n+1}(x)$が$f_n(x)$の$n$を$n+1$とした式になるかを確認する． $$\begin{array}{rcl}{f}_1(x)& =& \frac{1}{\sqrt{2\pi }}{e}^{-\frac{x}{2}}{x}^{-\frac{1}{2}}\\ f_n(x)& =& \frac{1}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{e}^{-\frac{x}{2}}{x}^{\frac{n}{2}-1}\\ f_{n+1}(x)& =& \frac{1}{2^{\frac{n+1}{2}}\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}{e}^{-\frac{x}{2}}{x}^{\frac{n+1}{2}-1}\cdots f_n(x)\mathrm{の}n\mathrm{を}n+1\mathrm{と}\mathrm{し}\mathrm{た}\mathrm{式},\mathrm{上}2\mathrm{式}\mathrm{を}\mathrm{認}\mathrm{め}\mathrm{た}\mathrm{上}\mathrm{で}\mathrm{こ}\mathrm{の}\mathrm{式}\mathrm{が}\mathrm{導}\mathrm{出}\mathrm{で}\mathrm{き}\mathrm{る}\mathrm{か}．\end{array}$$ $$\begin{array}{rcl}X& =& Z_1^2+\cdots +Z_n^2+Z_{n+1}^2\cdots Z_1,\cdots ,Z_n,Z_{n+1}:\mathrm{互}\mathrm{い}\mathrm{に}\mathrm{独}\mathrm{立}\mathrm{に}\mathrm{標}\mathrm{準}\mathrm{正}\mathrm{規}\mathrm{分}\mathrm{布}\mathrm{に}\mathrm{従}\mathrm{う}\mathrm{確}\mathrm{率}\mathrm{変}\mathrm{数}\\ Y& =& Z_1^2+\cdots +Z_n\\ X-Y& =& Z_{n+1}^2\\ f_{n+1}(x)& =& {\int }_0^{\mathrm{\infty }}{f}_n(y)f_1(x-y)\mathrm{d}y\\ & =& {\int }_0^{\mathrm{\infty }}\left\{\frac{1}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{e}^{-\frac{y}{2}}{y}^{\frac{n}{2}-1}\right\}\{\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x-y}{2}}(x-y{)}^{-\frac{1}{2}}\}\mathrm{d}y\\ & =& \frac{1}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}\frac{1}{\sqrt{2\pi }}{\int }_0^{\mathrm{\infty }}{e}^{-\frac{y}{2}}{e}^{-\frac{x-y}{2}}{y}^{\frac{n}{2}-1}(x-y{)}^{-\frac{1}{2}}\mathrm{d}y\cdots \int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x\\ & =& \frac{1}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}\frac{1}{\sqrt{2}\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}{e}^{-\frac{y}{2}-\frac{x-y}{2}}{y}^{\frac{n}{2}-1}(x-y{)}^{-\frac{1}{2}}\mathrm{d}y\cdots A^B{A}^C=A^{B+C}\\ & =& \frac{1}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}\frac{1}{2^{\frac{1}{2}}\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}{e}^{-\frac{x}{2}}{y}^{\frac{n}{2}-1}(x-y{)}^{-\frac{1}{2}}\mathrm{d}y\\ & =& \frac{1}{2^{\frac{n}{2}}{2}^{\frac{1}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi }}{e}^{-\frac{x}{2}}{\int }_0^{\mathrm{\infty }}{y}^{\frac{n}{2}-1}(x-y{)}^{-\frac{1}{2}}\mathrm{d}y\cdots \int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x\\ & =& \frac{1}{2^{\frac{n+1}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi }}{e}^{-\frac{x}{2}}{\int }_0^1(xu{)}^{\frac{n}{2}-1}(x-xu{)}^{-\frac{1}{2}}x\mathrm{d}u\cdots y=xu,\frac{\mathrm{d}y}{\mathrm{d}u}=x,x:0\to \mathrm{\infty },u:0\to 1\\ & =& \frac{1}{2^{\frac{n+1}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi }}{e}^{-\frac{x}{2}}{\int }_0^1(xu{)}^{\frac{n}{2}-1}{\{x(1-u)\}}^{-\frac{1}{2}}x\mathrm{d}u\\ & =& \frac{1}{2^{\frac{n+1}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi }}{e}^{-\frac{x}{2}}{\int }_0^1{x}^{\frac{n}{2}-1}{u}^{\frac{n}{2}-1}{x}^{-\frac{1}{2}}(1-u{)}^{-\frac{1}{2}}x\mathrm{d}u\cdots (AB{)}^C=A^C{B}^C\\ & =& \frac{1}{2^{\frac{n+1}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi }}{e}^{-\frac{x}{2}}{x}^{\frac{n}{2}-1}{x}^{-\frac{1}{2}}x{\int }_0^1{u}^{\frac{n}{2}-1}(1-u{)}^{-\frac{1}{2}}\mathrm{d}u\cdots \int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x\\ & =& \frac{1}{2^{\frac{n+1}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi }}{e}^{-\frac{x}{2}}{x}^{\frac{n}{2}-1-\frac{1}{2}+1}{\int }_0^1{u}^{\frac{n}{2}-1}(1-u{)}^{-\frac{1}{2}}\mathrm{d}u\cdots A^B{A}^C=A^{B+C}\\ & =& \frac{1}{2^{\frac{n+1}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi }}{e}^{-\frac{x}{2}}{x}^{\frac{n+1}{2}-1}{\int }_0^1{u}^{\frac{n}{2}-1}(1-u{)}^{-\frac{1}{2}}\mathrm{d}u\\ & =& \frac{1}{2^{\frac{n+1}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi }}{e}^{-\frac{x}{2}}{x}^{\frac{n+1}{2}-1}B(\frac{n}{2}-1,-\frac{1}{2})\cdots B(p,q)={\int }_0^1{x}^p(1-x{)}^q\mathrm{d}x\\ & =& \frac{1}{2^{\frac{n+1}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi }}{e}^{-\frac{x}{2}}{x}^{\frac{n+1}{2}-1}\frac{(\frac{n}{2}-1)!(-\frac{1}{2})!}{(\frac{n}{2}-1-\frac{1}{2}+1)!}\cdots B(p,q)=\frac{p!q!}{(p+q+1)!}\\ & =& \frac{1}{2^{\frac{n+1}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi }}{e}^{-\frac{x}{2}}{x}^{\frac{n+1}{2}-1}\frac{(\frac{n}{2}-1)!(-\frac{1}{2}+1-1)!}{(\frac{n}{2}+\frac{1}{2}-1)!}\\ & =& \frac{1}{2^{\frac{n+1}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi }}{e}^{-\frac{x}{2}}{x}^{\frac{n+1}{2}-1}\frac{(\frac{n}{2}-1)!(\frac{1}{2}-1)!}{(\frac{n+1}{2}-1)!}\\ & =& \frac{1}{2^{\frac{n+1}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi }}{e}^{-\frac{x}{2}}{x}^{\frac{n+1}{2}-1}\frac{\mathrm{\Gamma }\left(\frac{n}{2}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}\cdots \mathrm{\Gamma }(n)=(n-1)!\\ & =& \frac{1}{2^{\frac{n+1}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}\frac{1}{\sqrt{\pi }}{e}^{-\frac{x}{2}}{x}^{\frac{n+1}{2}-1}\frac{\mathrm{\Gamma }\left(\frac{n}{2}\right)\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}\\ & =& \frac{1}{2^{\frac{n+1}{2}}}\frac{1}{\sqrt{\pi }}{e}^{-\frac{x}{2}}{x}^{\frac{n+1}{2}-1}\frac{\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}\\ & =& \frac{1}{2^{\frac{n+1}{2}}}\frac{1}{\sqrt{\pi }}{e}^{-\frac{x}{2}}{x}^{\frac{n+1}{2}-1}\frac{\sqrt{\pi }}{\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}\cdots \mathrm{\Gamma }\left(\frac{1}{2}\right)=\sqrt{\pi }\\ & =& \frac{1}{2^{\frac{n+1}{2}}\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}{e}^{-\frac{x}{2}}{x}^{\frac{n+1}{2}-1}\end{array}$$ 任意の$n(n\ge 1)$において上記式が成り立つことを証明できた．
以上により${\chi }^2$分布の確率密度凾数が導出された．

x/√(a-x)の積分

x/√(a-x)の積分

不定積分

$$\begin{array}{rcl}\int \frac{x}{\sqrt{a-x}}\mathrm{d}x& =& \int x(a-x{)}^{-\frac{1}{2}}\mathrm{d}x\\ & =& \int x{\{-2(a-x{)}^{\frac{1}{2}}\}}^{\prime }\mathrm{d}x\cdots \int (a-x{)}^{-\frac{1}{2}}\mathrm{d}x=-2(a-x{)}^{\frac{1}{2}}+C\\ & =& -2x(a-x{)}^{\frac{1}{2}}-\int -2(a-x{)}^{\frac{1}{2}}\mathrm{d}x\cdots \int f^{\prime }(x)g(x)\mathrm{d}x=fg-\int f{g}^{\prime }\mathrm{d}x\\ & =& -2x(a-x{)}^{\frac{1}{2}}+2\int (a-x{)}^{\frac{1}{2}}\mathrm{d}x\\ & =& -2x(a-x{)}^{\frac{1}{2}}+2\{\frac{1}{\frac{1}{2}+1}(a-x{)}^{\frac{1}{2}+1}(-1)\}\\ & =& -2x(a-x{)}^{\frac{1}{2}}+2\{\frac{-1}{\frac{3}{2}}(a-x{)}^{\frac{3}{2}}\}\\ & =& -2x(a-x{)}^{\frac{1}{2}}+2\{\frac{-2}{3}(a-x{)}^{\frac{3}{2}}\}\\ & =& -2x(a-x{)}^{\frac{1}{2}}-\frac{4}{3}(a-x{)}^{\frac{3}{2}}\\ & =& -2x(a-x{)}^{\frac{1}{2}}-\frac{4}{3}(a-x)(a-x{)}^{\frac{1}{2}}\\ & =& (a-x{)}^{\frac{1}{2}}\{-2x-\frac{4}{3}(a-x)\}\\ & =& (a-x{)}^{\frac{1}{2}}(-2x-\frac{4}{3}a+\frac{4}{3}x)\\ & =& (a-x{)}^{\frac{1}{2}}(-\frac{6}{3}x-\frac{4}{3}a+\frac{4}{3}x)\\ & =& (a-x{)}^{\frac{1}{2}}(-\frac{2}{3}x-\frac{4}{3}a)\\ & =& -\frac{2}{3}(a-x{)}^{\frac{1}{2}}(x+2a)+C\cdots C:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数}\\ & =& -\frac{2}{3}\sqrt{a-x}(x+2a)+C\end{array}$$

定積分

$$\begin{array}{rcl}{\int }_0^a\frac{x}{\sqrt{a-x}}\mathrm{d}x& =& {[-\frac{2}{3}\sqrt{a-x}(x+2a)]}_0^a\\ & =& (-\frac{2}{3}\sqrt{a-a}(a+2a))-(-\frac{2}{3}\sqrt{a-0}(0+2a))\\ & =& 0-(-\frac{2}{3}\sqrt{a}\left(2a\right))\\ & =& 0-(-\frac{4}{3}{\sqrt{a}}^3)\\ & =& \frac{4}{3}{a}^{\frac{3}{2}}\end{array}$$

√(x)/√(a-x)の積分

√(x)/√(a-x)の積分

不定積分

$$\begin{array}{rcl}\int \frac{\sqrt{x}}{\sqrt{a-x}}\mathrm{d}x& =& \int x^{\frac{1}{2}}(a-x{)}^{-\frac{1}{2}}\mathrm{d}x\\ & =& \int u(a-u^2{)}^{-\frac{1}{2}}2u\mathrm{d}u\cdots u=\sqrt{x},\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{1}{2\sqrt{x}},\mathrm{d}x=2\sqrt{x}\mathrm{d}u=2u\mathrm{d}u\\ & =& 2\int u^2(a-u^2{)}^{-\frac{1}{2}}\mathrm{d}u\cdots \int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x\\ & =& 2\int {\{\sqrt{a}\sin \left(\theta \right)\}}^2{[a-{\{\sqrt{a}\sin \left(\theta \right)\}}^2]}^{-\frac{1}{2}}\sqrt{a}\cos \left(\theta \right)\mathrm{d}\theta \cdots u=\sqrt{a}\sin \left(\theta \right),\frac{\mathrm{d}u}{\mathrm{d}\theta }=\sqrt{a}\cos \left(\theta \right),\mathrm{d}u=\sqrt{a}\cos \left(\theta \right)\mathrm{d}\theta \\ & =& 2\int a{\sin }^2\left(\theta \right){\{a-a{\sin }^2\left(\theta \right)\}}^{-\frac{1}{2}}\sqrt{a}\cos \left(\theta \right)\mathrm{d}\theta \\ & =& 2a\sqrt{a}\int {\sin }^2\left(\theta \right){\left[a\{1-{\sin }^2\left(\theta \right)\}\right]}^{-\frac{1}{2}}\cos \left(\theta \right)\mathrm{d}\theta \\ & =& 2a\sqrt{a}\int \sqrt{a}{\sin }^2\left(\theta \right)a^{-\frac{1}{2}}{\{1-{\sin }^2\left(\theta \right)\}}^{-\frac{1}{2}}\cos \left(\theta \right)\mathrm{d}\theta \\ & =& 2a\sqrt{a}{a}^{-\frac{1}{2}}\int {\sin }^2\left(\theta \right){\{1-{\sin }^2\left(\theta \right)\}}^{-\frac{1}{2}}\cos \left(\theta \right)\mathrm{d}\theta \\ & =& 2a\int {\sin }^2\left(\theta \right){\{1-{\sin }^2\left(\theta \right)\}}^{-\frac{1}{2}}\cos \left(\theta \right)\mathrm{d}\theta \\ & =& 2a\int {\sin }^2\left(\theta \right){\{{\cos }^2\left(\theta \right)\}}^{-\frac{1}{2}}\cos \left(\theta \right)\mathrm{d}\theta \\ & =& 2a\int {\sin }^2\left(\theta \right){\{\cos \left(\theta \right)\}}^{-1}\cos \left(\theta \right)\mathrm{d}\theta \\ & =& 2a\int {\sin }^2\left(\theta \right)\mathrm{d}\theta \\ & =& 2a\{\frac{1}{2}\theta -\frac{1}{2}\sin \left(\theta \right)\cos \left(\theta \right)\}\cdots \int {\sin }^2\left(\theta \right)\mathrm{d}\theta =\frac{1}{2}\{\theta -\sin \left(\theta \right)\cos \left(\theta \right)\}+C(C:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数})\\ & =& 2a\{\frac{1}{2}{\sin }^{-1}\left(\frac{u}{\sqrt{a}}\right)-\frac{1}{2}\sin ({\sin }^{-1}\left(\frac{u}{\sqrt{a}}\right))\cos ({\sin }^{-1}\left(\frac{u}{\sqrt{a}}\right))\}\cdots u=\sqrt{a}\sin \left(\theta \right),\frac{u}{\sqrt{a}}=\sin \left(\theta \right),\theta ={\sin }^{-1}\left(\frac{u}{\sqrt{a}}\right)\\ & =& 2a\{\frac{1}{2}{\sin }^{-1}\left(\frac{u}{\sqrt{a}}\right)-\frac{1}{2}\left(\frac{u}{\sqrt{a}}\right)\sqrt{1-{\left(\frac{u}{\sqrt{a}}\right)}^2}\}\cdots \sin ({\sin }^{-1}\left(x\right))=x,\cos ({\sin }^{-1}\left(x\right))=\sqrt{1-x^2}\\ & =& 2a\frac{1}{2}\{{\sin }^{-1}\left(\frac{u}{\sqrt{a}}\right)-\left(\frac{u}{\sqrt{a}}\right)\sqrt{1-{\left(\frac{u}{\sqrt{a}}\right)}^2}\}\\ & =& a\{{\sin }^{-1}\left(\frac{u}{\sqrt{a}}\right)-\left(\frac{u}{\sqrt{a}}\right)\sqrt{1-{\left(\frac{u}{\sqrt{a}}\right)}^2}\}\\ & =& a\{{\sin }^{-1}\left(\frac{\sqrt{x}}{\sqrt{a}}\right)-\left(\frac{\sqrt{x}}{\sqrt{a}}\right)\sqrt{1-{\left(\frac{\sqrt{x}}{\sqrt{a}}\right)}^2}\}\cdots u=\sqrt{x}\\ & =& a\{{\sin }^{-1}\left(\sqrt{\frac{x}{a}}\right)-\frac{\sqrt{x}}{\sqrt{a}}\sqrt{1-\frac{x}{a}}\}\\ & =& a\{{\sin }^{-1}\left(\sqrt{\frac{x}{a}}\right)-\frac{\sqrt{x}}{\sqrt{a}}\sqrt{\frac{a-x}{a}}\}\\ & =& a\{{\sin }^{-1}\left(\sqrt{\frac{x}{a}}\right)-\frac{\sqrt{x}}{\sqrt{a}}\frac{\sqrt{a-x}}{\sqrt{a}}\}\cdots \sqrt{\frac{A}{B}}={\left(\frac{A}{B}\right)}^{\frac{1}{2}}={\left(A\frac{1}{B}\right)}^{\frac{1}{2}}=A^{\frac{1}{2}}{\left(\frac{1}{B}\right)}^{\frac{1}{2}}=A^{\frac{1}{2}}{\left(B^{-1}\right)}^{\frac{1}{2}}=A^{\frac{1}{2}}{B}^{-\frac{1}{2}}=\sqrt{A}\frac{1}{\sqrt{B}}=\frac{\sqrt{A}}{\sqrt{B}}\\ & =& a\{{\sin }^{-1}\left(\sqrt{\frac{x}{a}}\right)-\frac{\sqrt{x}\sqrt{a-x}}{a}\}\\ & =& a{\sin }^{-1}\left(\sqrt{\frac{x}{a}}\right)-a\frac{\sqrt{x}\sqrt{a-x}}{a}\\ & =& a{\sin }^{-1}\left(\sqrt{\frac{x}{a}}\right)-\sqrt{x}\sqrt{a-x}+C\cdots C:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数}\\ & =& a{\sin }^{-1}\left({\left(\frac{x}{a}\right)}^{\frac{1}{2}}\right)-x^{\frac{1}{2}}{(a-x)}^{\frac{1}{2}}+C\end{array}$$

定積分

$$\begin{array}{rcl}{\int }_0^a\frac{\sqrt{x}}{\sqrt{a-x}}\mathrm{d}x& =& {[a{\sin }^{-1}\left({\left(\frac{x}{a}\right)}^{\frac{1}{2}}\right)-x^{\frac{1}{2}}{(a-x)}^{\frac{1}{2}}]}_0^a\\ & =& (a{\sin }^{-1}\left({\left(\frac{a}{a}\right)}^{\frac{1}{2}}\right)-a^{\frac{1}{2}}{(a-a)}^{\frac{1}{2}})-(a{\sin }^{-1}\left({\left(\frac{0}{a}\right)}^{\frac{1}{2}}\right)-0^{\frac{1}{2}}{(a-0)}^{\frac{1}{2}})\\ & =& (a{\sin }^{-1}\left(1\right)-0)-(0-0)\cdots {\sin }^{-1}\left(0\right)=0,0^A=0\\ & =& a\frac{\pi }{2}\cdots {\sin }^{-1}\left(1\right)=\frac{\pi }{2}\\ & =& \frac{a\pi }{2}\end{array}$$

1/√(a-x)の積分

1/√(a-x)の積分

不定積分

$$\begin{array}{rcl}\int \frac{1}{\sqrt{a-x}}\mathrm{d}x& =& \int (a-x{)}^{-\frac{1}{2}}\mathrm{d}x\\ & =& \int u^{-\frac{1}{2}}(-1)\mathrm{d}u\cdots u=a-x,\frac{\mathrm{d}u}{\mathrm{d}x}=-1,\mathrm{d}x=-\mathrm{d}u\\ & =& -\int u^{-\frac{1}{2}}\mathrm{d}u\cdots \int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x\\ & =& -\frac{1}{-\frac{1}{2}+1}{u}^{-\frac{1}{2}+1}\cdots \int x^a\mathrm{d}x=\frac{1}{a+1}{x}^{a+1}+C(C:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数})\\ & =& -\frac{1}{\frac{1}{2}}{u}^{\frac{1}{2}}\\ & =& -2u^{\frac{1}{2}}\\ & =& -2(a-x{)}^{\frac{1}{2}}\cdots u=a-x\\ & =& -2(a-x{)}^{\frac{1}{2}}+C\cdots C:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数}\\ & =& -2\sqrt{a-x}+C\end{array}$$

定積分

$$\begin{array}{rcl}{\int }_0^a\frac{1}{\sqrt{a-x}}\mathrm{d}x& =& {[-2\sqrt{a-x}]}_0^a\\ & =& (-2\sqrt{a-a})-(-2\sqrt{a-0})\\ & =& 0-(-2\sqrt{a})\\ & =& 2\sqrt{a}\\ & =& 2a^{\frac{1}{2}}\end{array}$$

1/√(x(a-x))の積分

1/√(x(a-x))の積分

不定積分

$$\begin{array}{rcl}\int \frac{1}{\sqrt{x(a-x)}}\mathrm{d}x& =& \int {\{x(a-x)\}}^{-\frac{1}{2}}\mathrm{d}x\\ & =& \int x^{-\frac{1}{2}}{(a-x)}^{-\frac{1}{2}}\mathrm{d}x\\ & =& \int u^{-1}(a-u^2{)}^{-\frac{1}{2}}2u\mathrm{d}u\cdots u=\sqrt{x},\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{1}{2\sqrt{x}},\mathrm{d}x=2\sqrt{x}\mathrm{d}u=2u\mathrm{d}u\\ & =& 2\int (a-u^2{)}^{-\frac{1}{2}}\mathrm{d}u\cdots u^{-1}u=1,\int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x\\ & =& 2\int {[a-{\{\sqrt{a}\sin \left(\theta \right)\}}^2]}^{-\frac{1}{2}}\sqrt{a}\cos \left(\theta \right)\mathrm{d}\theta \cdots u=\sqrt{a}\sin \left(\theta \right),\frac{\mathrm{d}u}{\mathrm{d}\theta }=\sqrt{a}\cos \left(\theta \right),\mathrm{d}u=\sqrt{a}\cos \left(\theta \right)\mathrm{d}\theta \\ & =& 2\sqrt{a}\int {[a-\{{\sqrt{a}}^2{\sin }^2\left(\theta \right)\}]}^{-\frac{1}{2}}\cos \left(\theta \right)\mathrm{d}\theta \cdots (AB{)}^C=A^C{B}^C\\ & =& 2\sqrt{a}\int {\{a-a{\sin }^2\left(\theta \right)\}}^{-\frac{1}{2}}\cos \left(\theta \right)\mathrm{d}\theta \\ & =& 2\sqrt{a}\int {\left[a\{1-{\sin }^2\left(\theta \right)\}\right]}^{-\frac{1}{2}}\cos \left(\theta \right)\mathrm{d}\theta \\ & =& 2\sqrt{a}\int a^{-\frac{1}{2}}{\{1-{\sin }^2\left(\theta \right)\}}^{-\frac{1}{2}}\cos \left(\theta \right)\mathrm{d}\theta \\ & =& 2a\sqrt{a}{a}^{-\frac{1}{2}}\int {\{1-{\sin }^2\left(\theta \right)\}}^{-\frac{1}{2}}\cos \left(\theta \right)\mathrm{d}\theta \\ & =& 2\int {\{1-{\sin }^2\left(\theta \right)\}}^{-\frac{1}{2}}\cos \left(\theta \right)\mathrm{d}\theta \cdots \sqrt{A}{A}^{-\frac{1}{2}}=A^{\frac{1}{2}}{A}^{-\frac{1}{2}}=A^{-\frac{1}{2}+\frac{1}{2}}=A^0=1\\ & =& 2\int {\{{\cos }^2\left(\theta \right)\}}^{-\frac{1}{2}}\cos \left(\theta \right)\mathrm{d}\theta \cdots {\cos }^2\left(\theta \right)+{\sin }^2\left(\theta \right)=1,{\cos }^2\left(\theta \right)=1-{\sin }^2\left(\theta \right)\\ & =& 2\int {\{\cos \left(\theta \right)\}}^{2(-\frac{1}{2})}\cos \left(\theta \right)\mathrm{d}\theta \cdots {\left(A^B\right)}^C=A^{BC}\\ & =& 2\int {\{\cos \left(\theta \right)\}}^{-1}\cos \left(\theta \right)\mathrm{d}\theta \\ & =& 2\int \mathrm{d}\theta \cdots A^{-1}A=1\\ & =& 2\theta \cdots \int \mathrm{d}x=x+C(C:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数})\\ & =& 2{\sin }^{-1}\left(\frac{u}{\sqrt{a}}\right)\cdots \theta ={\sin }^{-1}\left(\frac{u}{\sqrt{a}}\right)\\ & =& 2{\sin }^{-1}\left(\frac{\sqrt{x}}{\sqrt{a}}\right)\cdots u=\sqrt{x}\\ & =& 2{\sin }^{-1}\left(\sqrt{\frac{x}{a}}\right)+C\cdots C:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数}\end{array}$$

定積分

$$\begin{array}{rcl}{\int }_0^a\frac{1}{\sqrt{x(a-x)}}\mathrm{d}x& =& {[2{\sin }^{-1}\left(\sqrt{\frac{x}{a}}\right)]}_0^a\\ & =& \{2{\sin }^{-1}\left(\sqrt{\frac{a}{a}}\right)\}-\{2{\sin }^{-1}\left(\sqrt{\frac{0}{a}}\right)\}\\ & =& \{2{\sin }^{-1}\left(1\right)\}-\{2{\sin }^{-1}\left(0\right)\}\\ & =& \left(2\frac{\pi }{2}\right)-(2\cdot 0)\\ & =& \pi -0\\ & =& \pi \end{array}$$

sinの二乗の不定積分

sinの二乗の不定積分

$$\begin{array}{rcl}\int {\sin }^2\left(\theta \right)\mathrm{d}\theta & =& \int \frac{1}{2}-\frac{1}{2}\cos \left(2\theta \right)\mathrm{d}\theta \\ & & \cdots \cos \left(2\theta \right)={\cos }^2\left(\theta \right)-{\sin }^2\left(\theta \right)=(1-{\sin }^2\left(\theta \right))-{\sin }^2\left(\theta \right)=1-2{\sin }^2\left(\theta \right)\\ & & \cdots {\sin }^2\left(\theta \right)=\frac{1}{2}-\frac{1}{2}\cos \left(2\theta \right)\\ & =& \frac{1}{2}\int \mathrm{d}\theta -\frac{1}{2}\int \cos \left(2\theta \right)\mathrm{d}\theta \\ & =& \frac{1}{2}\theta -\frac{1}{2}\int \cos \left(2\theta \right)\mathrm{d}\theta \\ & =& \frac{1}{2}\theta -\frac{1}{2}\int \cos \left(\varphi \right)\frac{1}{2}\mathrm{d}\varphi \cdots \varphi =2\theta ,\frac{\mathrm{d}\varphi }{\mathrm{d}\theta }=2,\mathrm{d}\theta =\frac{1}{2}\mathrm{d}\varphi \\ & =& \frac{1}{2}\theta -\frac{1}{4}\int \cos \left(\varphi \right)\mathrm{d}\varphi \cdots \int cf(x)\mathrm{d}x=c\int f(x)\mathrm{d}x\\ & =& \frac{1}{2}\theta -\frac{1}{2}(\frac{1}{2}\sin \left(\varphi \right))\cdots \int \cos \left(x\right)\mathrm{d}x=\sin \left(x\right)+C(C:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数})\\ & =& \frac{1}{2}\theta -\frac{1}{4}\sin \left(\varphi \right)\\ & =& \frac{1}{2}\theta -\frac{1}{4}\sin \left(2\theta \right)\\ & =& \frac{1}{2}\theta -\frac{1}{2}\sin \left(\theta \right)\cos \left(\theta \right)\\ & =& \frac{1}{2}\{\theta -\sin \left(\theta \right)\cos \left(\theta \right)\}+C\cdots C:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数}\end{array}$$

ベータ分布の期待値(平均)と分散

$$\begin{array}{rcl}f(x;m,n)& =& \frac{x^{(m-1)}(1-x{)}^{(n-1)}}{B(m,n)}\cdots \mathrm{ベ}\mathrm{ー}\mathrm{タ}\mathrm{分}\mathrm{布}\\ B(m,n)& =& {\int }_0^1{x}^{(m-1)}(1-x{)}^{(n-1)}\mathrm{d}x\cdots \mathrm{ベ}\mathrm{ー}\mathrm{タ}\mathrm{凾}\mathrm{数}\\ & =& \frac{(m-1)!(n-1)!}{\{(m-1)+(n-1)+1\}!}\cdots {\int }_{\alpha }^{\beta }(x-\alpha {)}^p(\beta -x{)}^q\mathrm{d}x=\frac{p!q!}{(p+q+1)!}(\beta -\alpha {)}^{(p+q+1)}(\mathrm{第}\mathrm{一}\mathrm{種}\mathrm{オ}\mathrm{イ}\mathrm{ラ}\mathrm{ー}\mathrm{積}\mathrm{分})\\ & =& \frac{(m-1)!(n-1)!}{(m+n-1)!}\end{array}$$

ベータ分布の期待値(一次モーメント・平均)

$$\begin{array}{rcl}\mathbf{E}[X]& =& {\int }_0^{1}xf(x)\mathrm{d}x\\ & =& {\int }_0^{1}x\frac{x^{(m-1)}(1-x{)}^{(n-1)}}{B(m,n)}\mathrm{d}x\\ & =& \frac{1}{B(m,n)}{\int }_0^1{x}^m(1-x{)}^{(n-1)}\mathrm{d}x\\ & =& \frac{(m+n-1)!}{(m-1)!(n-1)!}\frac{m!(n-1)!}{(m+n)!}\\ & =& \frac{(m+n-1)!}{(m+n)!}\frac{m!}{(m-1)!}\\ & =& \frac{1}{m+n}\frac{m}{1}\\ & =& \frac{m}{m+n}\cdots \mathrm{ベ}\mathrm{ー}\mathrm{タ}\mathrm{分}\mathrm{布}\mathrm{の}\mathrm{平}\mathrm{均}\end{array}$$

ベータ分布の二次モーメント

$$\begin{array}{rcl}\mathbf{E}[X^2]& =& {\int }_0^1{x}^{2}f(x)\mathrm{d}x\\ & =& {\int }_0^1{x}^2\frac{x^{(m-1)}(1-x{)}^{(n-1)}}{B(m,n)}\mathrm{d}x\\ & =& \frac{1}{B(m,n)}{\int }_0^1{x}^{(m+1)}(1-x{)}^{(n-1)}\mathrm{d}x\\ & =& \frac{(m+n-1)!}{(m-1)!(n-1)!}\frac{(m+1)!(n-1)!}{((m+1)+n)!}\\ & =& \frac{(m+n-1)!}{(m-1)!(n-1)!}\frac{(m+1)!(n-1)!}{(m+n+1)!}\\ & =& \frac{(m+n-1)!}{(m+n+1)!}\frac{(m+1)!}{(m-1)!}\\ & =& \frac{1}{(m+n+1)(m+n)}\frac{(m+1)m}{1}\\ & =& \frac{(m+1)m}{(m+n+1)(m+n)}\end{array}$$

ベータ分布の分散(二次の中心モーメント

$$\begin{array}{rcl}\mathbf{V}[X]& =& \mathbf{E}[X^2]-\mathbf{E}[X{]}^2\\ & =& \frac{(m+1)m}{(m+n+1)(m+n)}-{\left\{\frac{m}{m+n}\right\}}^2\\ & =& \frac{(m+1)m}{(m+n+1)(m+n)}-\frac{m^2}{{(m+n)}^2}\\ & =& \frac{(m+n)(m+1)m-(m+n+1)m^2}{(m+n+1)(m+n{)}^2}\\ & =& \frac{(m^2+m+mn+n)m-(m^3+m^{2}n+m^2)}{(m+n+1)(m+n{)}^2}\\ & =& \frac{m^3+m^2+m^{2}n+mn-m^3-m^{2}n-m^2}{(m+n+1)(m+n{)}^2}\\ & =& \frac{mn}{(m+n+1)(m+n{)}^2}\cdots \mathrm{ベ}\mathrm{ー}\mathrm{タ}\mathrm{分}\mathrm{布}\mathrm{の}\mathrm{分}\mathrm{散}\end{array}$$

KLダイバージェンスの下限(イェンセンの不等式による)

KLダイバージェンスの下限(イェンセンの不等式による)

$$\begin{array}{rcl}{p}_i& :& \mathrm{真}\mathrm{の}\mathrm{確}\mathrm{率}\mathrm{分}\mathrm{布}P(X)\mathrm{の}\mathrm{事}\mathrm{象}{x}_i\mathrm{の}\mathrm{確}\mathrm{率}\\ q_i& :& \mathrm{任}\mathrm{意}\mathrm{の}\mathrm{確}\mathrm{率}\mathrm{分}\mathrm{布}Q(X)\mathrm{の}\mathrm{事}\mathrm{象}{x}_i\mathrm{の}\mathrm{確}\mathrm{率}(Q:\mathrm{予}\mathrm{測}\mathrm{や}\mathrm{符}\mathrm{号}\mathrm{化}\mathrm{な}\mathrm{ど}「\mathrm{真}\mathrm{の}\mathrm{確}\mathrm{率}\mathrm{分}\mathrm{布}P\mathrm{に}\mathrm{対}\mathrm{す}\mathrm{る}\mathrm{モ}\mathrm{デ}\mathrm{ル}」\mathrm{が}\mathrm{示}\mathrm{す}\mathrm{確}\mathrm{率}\mathrm{分}\mathrm{布})\end{array}$$ $$\begin{array}{rcl}{D}_{KL}(P||Q)& =& \sum _{i=1}^k{p}_i\log \left(\frac{p_i}{q_i}\right)\\ & =& \sum _{i=1}^k{p}_i\{-\log \left(\frac{q_i}{p_i}\right)\}\cdots \log \left(\frac{A}{B}\right)=\log \left({\left(\frac{B}{A}\right)}^{-1}\right)=-\log \left(\frac{B}{A}\right)\\ & \ge & -\log \left(\sum _{i=1}^k{p}_i\frac{q_i}{p_i}\right)\\ & & \cdots -\log \left(x\right)\mathrm{は}\mathrm{凸}\mathrm{凾}\mathrm{数}(\mathrm{下}\mathrm{に}\mathrm{凸}\mathrm{の}\mathrm{凾}\mathrm{数})(\mathrm{判}\mathrm{定}\mathrm{に}\mathrm{つ}\mathrm{い}\mathrm{て}\mathrm{は}\mathrm{後}\mathrm{述})\\ & & \cdots \sum _{i=1}^n{\varphi }_{i}f(x_i)\ge f\left(\sum _{i=1}^n{\varphi }_i{x}_i\right)(\mathrm{イ}\mathrm{ェ}\mathrm{ン}\mathrm{セ}\mathrm{ン}\mathrm{の}\mathrm{不}\mathrm{等}\mathrm{式}(\mathrm{下}\mathrm{に}\mathrm{凸}\mathrm{の}\mathrm{凾}\mathrm{数}))\\ D_{KL}(P||Q)& \ge & -\log \left(\sum _{i=1}^k{p}_i\frac{q_i}{p_i}\right)\\ & \ge & -\log \left(\sum _{i=1}^k{q}_i\right)\\ & \ge & -\log \left(1\right)\cdots q_i\mathrm{は}\mathrm{確}\mathrm{率}\mathrm{な}\mathrm{の}\mathrm{で}\sum _{i=1}^k{q}_i=1\\ & \ge & 0\end{array}$$

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$-\log \left(x\right)$の凸凾数の判定

$$\begin{array}{rcl}\frac{{\mathrm{d}}^2}{\mathrm{d}{x}^2}\{-\log \left(x\right)\}& =& -\frac{{\mathrm{d}}^2}{\mathrm{d}{x}^2}\{\log \left(x\right)\}\cdots {\{cf(x)\}}^{\prime }=c{\{f(x)\}}^{\prime }\\ & =& -\frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{1}{x}\right\}\cdots {\{\log \left(x\right)\}}^{\prime }=x^{-1}=\frac{1}{x}\\ & =& -(-\frac{1}{x^2})\cdots {\left(x^{-1}\right)}^{\prime }=-x^{-2}=-\frac{1}{x^2}\\ & =& \frac{1}{x^2}>0\cdots x\mathrm{が}\mathrm{実}\mathrm{数}\mathrm{の}\mathrm{範}\mathrm{囲}\mathrm{で}\mathrm{は},\mathrm{二}\mathrm{階}\mathrm{微}\mathrm{分}\mathrm{が}\mathrm{常}\mathrm{に}\mathrm{正}\mathrm{な}\mathrm{の}\mathrm{で}\mathrm{凸}\mathrm{凾}\mathrm{数}(\mathrm{下}\mathrm{に}\mathrm{凸}\mathrm{の}\mathrm{凾}\mathrm{数})．\end{array}$$ 

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${\log }_{\mathrm{e}}x\le (x-1)$の関係を用いたKLダイバージェンスの下限

イェンセンの不等式の証明

イェンセンの不等式

${\varphi }_1,{\varphi }_2,\cdots {\varphi }_n$が$0<{\varphi }_i$かつ$\sum _{i=1}^n{\varphi }_i=1$を満たしまた，$x_1,x_2,\cdots x_n$を実数の列とするとき，凸凾数$f(x)$に対して以下のことが成り立つ. $$\begin{array}{r}\sum _{i=1}^n{\varphi }_{i}f(x_i)\ge f\left(\sum _{i=1}^n{\varphi }_i{x}_i\right)\cdots f(x)\mathrm{が}\mathrm{凸}\mathrm{凾}\mathrm{数}(\mathrm{下}\mathrm{に}\mathrm{凸}\mathrm{の}\mathrm{凾}\mathrm{数})\mathrm{の}\mathrm{と}\mathrm{き}.\\ (\sum _{i=1}^n{\varphi }_{i}f(x_i)\le f\left(\sum _{i=1}^n{\varphi }_i{x}_i\right)\cdots f(x)\mathrm{が}\mathrm{凹}\mathrm{凾}\mathrm{数}(\mathrm{上}\mathrm{に}\mathrm{凸}\mathrm{の}\mathrm{凾}\mathrm{数})\mathrm{の}\mathrm{と}\mathrm{き}.)\end{array}$$

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凸凾数(下に凸の凾数)

凸凾数(下に凸の凾数)の性質

任意の$x_1,x_2$に対して凸凾数(下に凸の凾数)$f(x)$では，$(x_1,f(x_1)),(x_2,f(x_2))$を結ぶ線分$g(x)$は凾数$f(x)$の上側にある.
逆に，$(x_1,f(x_1)),(x_2,f(x_2))$を結ぶ線分$g(x)$が凾数$f(x)$の上側にあるので，$f(x)$は凸凾数(下に凸の凾数)である． 

凸凾数(下に凸の凾数)にかかる線分$g(x)$の式

$$\begin{array}{rcl}{x}_t& =& (1-t)x_1+t{x}_2(\mathrm{た}\mathrm{だ}\mathrm{し}0\le t\le 1)\cdots x_1{x}_2\mathrm{間}\mathrm{を}\mathrm{単}\mathrm{位}\mathrm{と}\mathrm{し}{x}_1\mathrm{を}\mathrm{基}\mathrm{準}\mathrm{点}\mathrm{と}\mathrm{す}\mathrm{る}t\mathrm{を}\mathrm{用}\mathrm{い}\mathrm{た}x\mathrm{の}\mathrm{表}\mathrm{現}\mathrm{と}\mathrm{な}\mathrm{る}.\\ & =& \sum _{i=1}^2{\varphi }_i{x}_i\cdots {\varphi }_i=\{1-t,t\}\mathrm{で}，0\le {\varphi }_i\mathrm{か}\mathrm{つ}\sum _{i=1}^2{\varphi }_i=1\\ \\ \\ g(x)-f(x_1)& =& \frac{f(x_2)-f(x_1)}{x_2-x_1}(x-x_1)\cdots \mathrm{傾}\mathrm{き}\mathrm{が}\frac{f(x_2)-f(x_1)}{x_2-x_1}\mathrm{で}\mathrm{点}(x_1,f(x_1))\mathrm{を}\mathrm{通}\mathrm{る}\mathrm{直}\mathrm{線}\mathrm{の}\mathrm{方}\mathrm{程}\mathrm{式}\\ g(x)& =& \frac{f(x_2)-f(x_1)}{x_2-x_1}(x-x_1)+f(x_1)\\ & =& \frac{f(x_2)-f(x_1)}{x_2-x_1}(x_t-x_1)+f(x_1)\cdots x\mathrm{を}t\mathrm{で}\mathrm{表}\mathrm{現}\mathrm{す}\mathrm{る}{x}_t\mathrm{と}\mathrm{す}\mathrm{る}\\ & =& \frac{f(x_2)-f(x_1)}{x_2-x_1}\{(1-t)x_1+t{x}_2-x_1\}+f(x_1)\cdots x_t=(1-t)x_1+t{x}_2\\ & =& \frac{f(x_2)-f(x_1)}{x_2-x_1}(x_1-t{x}_1+t{x}_2-x_1)+f(x_1)\\ & =& \frac{f(x_2)-f(x_1)}{x_2-x_1}(-t{x}_1+t{x}_2)+f(x_1)\\ & =& \frac{f(x_2)-f(x_1)}{x_2-x_1}t(x_2-x_1)+f(x_1)\\ & =& \{f(x_2)-f(x_1)\}t+f(x_1)\\ & =& tf(x_2)-tf(x_1)+f(x_1)\\ & =& tf(x_2)+(1-t)f(x_1)\\ & =& (1-t)f(x_1)+tf(x_2)\\ & =& \sum _{i=1}^2{\varphi }_{i}f(x_i)\cdots {\varphi }_i=\{1-t,t\}\mathrm{で}，0\le {\varphi }_i\mathrm{か}\mathrm{つ}\sum _{i=1}^2{\varphi }_i=1\end{array}$$

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数学的帰納法によるイェンセンの不等式の証明

$n=2$の証明

$$\begin{array}{rcl}g(x_t)& \ge & f(x_t)\cdots \mathrm{凸}\mathrm{凾}\mathrm{数}(\mathrm{下}\mathrm{に}\mathrm{凸}\mathrm{の}\mathrm{凾}\mathrm{数})\mathrm{の}\mathrm{性}\mathrm{質}\mathrm{よ}\mathrm{り}g(x)\mathrm{が}f(x)\mathrm{よ}\mathrm{り}\mathrm{上}\mathrm{側}\mathrm{に}\mathrm{あ}\mathrm{る}.\\ (1-t)f(x_1)+tf(x_2)& \ge & f((1-t)x_1+t{x}_2)\cdots x_t=(1-t)x_1+t{x}_2\\ \sum _{i=1}^2{\varphi }_{i}f(x_i)& \ge & f\left(\sum _{i=1}^2{\varphi }_i{x}_i\right)\cdots n=2\mathrm{の}\mathrm{証}\mathrm{明}.\end{array}$$

$n=k$を認める状況で$n=k+1$を証明

$n=k+1$の式を変形していく． $$\begin{array}{rcl}\sum _{i=1}^{k+1}{\theta }_{i}f(x_i)& =& \sum _{i=1}^k{\theta }_{i}f(x_i)+{\theta }_{k+1}f(x_{k+1})\cdots 0\le {\theta }_i\mathrm{か}\mathrm{つ}\sum _{i=1}^{k+1}{\theta }_i=1\\ & =& {\mathrm{\Theta }}_k\sum _{i=1}^k\frac{{\theta }_i}{{\mathrm{\Theta }}_k}f(x_i)+{\theta }_{k+1}f(x_{k+1})\cdots {\mathrm{\Theta }}_k=\sum _{i=1}^k{\theta }_i=1-{\theta }_{k+1}<1,\sum _{i=1}^k\frac{{\theta }_i}{{\mathrm{\Theta }}_k}=\frac{1}{{\mathrm{\Theta }}_k}\sum _{i=1}^k{\theta }_i=1\\ & \ge & {\mathrm{\Theta }}_{k}f\left(\sum _{i=1}^k\frac{{\theta }_i}{{\mathrm{\Theta }}_k}{x}_i\right)+{\theta }_{k+1}f(x_{k+1})\cdots n=k\mathrm{は}\mathrm{認}\mathrm{め}\mathrm{る}\mathrm{の}\mathrm{で}(\sum _{i=1}^k{\varphi }_{i}f(x_i)\ge f\left(\sum _{i=1}^k{\varphi }_i{x}_i\right)),\sum _{i=1}^k\frac{{\theta }_i}{{\mathrm{\Theta }}_k}f(x_i)\ge f\left(\sum _{i=1}^k\frac{{\theta }_i}{{\mathrm{\Theta }}_k}{x}_i\right).\end{array}$$

$n=2$の式としてみる． $$\begin{array}{rcl}{\mathrm{\Theta }}_{k}f\left(\sum _{i=1}^k\frac{{\theta }_i}{{\mathrm{\Theta }}_k}{x}_i\right)+{\theta }_{k+1}f(x_{k+1})& =& {\lambda }_{1}f\left(a_1\right)+{\lambda }_{2}f(a_2)\cdots {\lambda }_j=\{{\mathrm{\Theta }}_k,{\theta }_{k+1}\},a_j=\{\sum _{i=1}^k\frac{{\theta }_i}{{\mathrm{\Theta }}_k}{x}_i,x_{k+1}\}\\ & =& \sum _{j=1}^2{\lambda }_{j}f(a_j)\\ & \ge & f\left(\sum _{j=1}^2{\lambda }_j{a}_j\right)\cdots n=2\mathrm{は}\mathrm{上}\mathrm{記}\mathrm{で}\mathrm{証}\mathrm{明}\mathrm{済}\mathrm{み}.\end{array}$$

$\sum _{j=1}^2{\lambda }_j{a}_j$を変形していく． $$\begin{array}{rcl}\sum _{j=1}^2{\lambda }_j{a}_j& =& {\mathrm{\Theta }}_k\sum _{i=1}^k\frac{{\theta }_i}{{\mathrm{\Theta }}_k}{x}_i+{\theta }_{k+1}{x}_{k+1}\\ & =& {\mathrm{\Theta }}_k\frac{1}{{\mathrm{\Theta }}_k}\sum _{i=1}^k{\theta }_i{x}_i+{\theta }_{k+1}{x}_{k+1}\cdots \sum _{i=1}^{n}C{x}_i=C\sum _{i=1}^n{x}_i\\ & =& \sum _{i=1}^k{\theta }_i{x}_i+{\theta }_{k+1}{x}_{k+1}\\ & =& \sum _{i=1}^{k+1}{\theta }_i{x}_i\end{array}$$

変形結果を戻す．

$$\begin{array}{rcl}f\left(\sum _{j=1}^2{\lambda }_j{a}_j\right)& =& f\left(\sum _{i=1}^{k+1}{\theta }_i{x}_i\right)\end{array}$$

以上より$n=k$を認めれば$n=k+1$でもイェンセンの不等式が成り立つ. $$\begin{array}{rcl}\sum _{i=1}^{k+1}{\theta }_{i}f(x_i)& \ge & f\left(\sum _{i=1}^{k+1}{\theta }_i{x}_i\right)\end{array}$$

$n=2$及び$n=k+1$の証明からの結論

数学的帰納法によりイェンセンの不等式が成り立つ.