${\displaystyle E\left(aX+b\right)=aE\left(X\right)+b}$

確率変数 ${\displaystyle X}$ について

${\displaystyle E\left(aX+b\right)=aE\left(X\right)+b}$

が成り立つ．

■証明

●離散型確率変数の場合

期待値の定義

${\displaystyle E\left(X\right)={\displaystyle \sum _{i=1}^n{x}_{i}f\left(x_i\right)}}$

より

${\displaystyle E\left(aX+b\right)={\displaystyle \sum }_{i=1}^n\left(a{x}_i+b\right)f\left(x_i\right)}$

${\displaystyle ={\displaystyle \sum }_{i=1}^n\left\{a{x}_{i}f\left(x_i\right)+bf\left(x_i\right)\right\}}$

${\displaystyle ={\displaystyle \sum }_{i=1}^{n}a{x}_{i}f\left(x_i\right)+{\displaystyle \sum }_{i=1}^{n}bf\left(x_i\right)}$

${\displaystyle =a{\displaystyle \sum }_{i=1}^n{x}_{i}f\left(x_i\right)+b{\displaystyle \sum }_{i=1}^{n}f\left(x_i\right)}$

${\displaystyle =aE\left(X\right)+b}$

●連続型確率変数の場合

期待値の定義

${\displaystyle E\left(X\right)={\displaystyle {\int }_{-\infty }^{\infty }{x}_{i}f\left(x_i\right)dx}}$

より

${\displaystyle E\left(aX\right)={{\displaystyle \int }}_{-\infty }^{\infty }\left(a{x}_i+b\right)f\left(x_i\right)dx}$

${\displaystyle ={{\displaystyle \int }}_{-\infty }^{\infty }\left\{a{x}_{i}f\left(x_i\right)+bf\left(x_i\right)\right\}dx}$

${\displaystyle ={{\displaystyle \int }}_{-\infty }^{\infty }a{x}_{i}f\left(x_i\right)dx+{{\displaystyle \int }}_{-\infty }^{\infty }bf\left(x_i\right)dx}$

${\displaystyle =a{{\displaystyle \int }}_{-\infty }^{\infty }{x}_{i}f\left(x_i\right)dx+b{{\displaystyle \int }}_{-\infty }^{\infty }f\left(x_i\right)dx}$

${\displaystyle =aE\left(X\right)+b}$

　

ホーム>>カテゴリー分類>>確率・統計>> $E\left(X\right)$ ， $V\left(X\right)$ ， $C\left(X,Y\right)$ の計算則>> ${\displaystyle E\left(aX+b\right)=aE\left(X\right)+b}$

最終更新日： 2026年3月30日