$V\left(aX\right)=a^{2}V\left(X\right)$

確率変数 ${\displaystyle X}$ について

$V\left(aX\right)=a^{2}V\left(X\right)$

が成り立つ．

■証明

まず

${\displaystyle \mu =E\left(X\right)}$　（ここでは，平均(期待値)を $\overline{x}$ ではなく $\mu$ を使うことにする．）

${\displaystyle \eta =E\left(aX\right)=aE\left(X\right)=a\mu }$

とする．

●離散型確率変数の場合

分散の定義

${\displaystyle V\left(X\right)={\displaystyle \sum _{i=1}^n{\left(x_i-\mu \right)}^{2}f\left(x_i\right)}}$

より

${\displaystyle V\left(aX\right)={\displaystyle \sum _{i=1}^n{\left(a{x}_i-\eta \right)}^{2}f\left(x_i\right)}}$

${\displaystyle ={\displaystyle \sum _{i=1}^n{\left(a{x}_i-a\mu \right)}^{2}f\left(x_i\right)}}$

${\displaystyle ={\displaystyle \sum _{i=1}^n{\left\{a\left(x_i-\mu \right)\right\}}^{2}f\left(x_i\right)}}$

${\displaystyle ={\displaystyle \sum _{i=1}^n{a}^2{\left(x_i-\mu \right)}^{2}f\left(x_i\right)}}$

${\displaystyle =a^2{\displaystyle \sum _{i=1}^n{\left(x_i-\mu \right)}^{2}f\left(x_i\right)}}$

${\displaystyle =a^{2}V\left(X\right)}$

●連続型確率変数の場合

分散の定義

${\displaystyle V\left(X\right)={\displaystyle {\int }_{-\infty }^{\infty }{\left(x_i-\mu \right)}^{2}f\left(x_i\right)dx}}$

より

${\displaystyle V\left(X\right)={\displaystyle {\int }_{-\infty }^{\infty }{\left(a{x}_i-\eta \right)}^{2}f\left(x_i\right)f\left(x_i\right)dx}}$

${\displaystyle ={\displaystyle {\int }_{-\infty }^{\infty }{\left(a{x}_i-a\mu \right)}^{2}f\left(x_i\right)f\left(x_i\right)dx}}$

${\displaystyle ={\displaystyle {\int }_{-\infty }^{\infty }{\left\{a\left(x_i-\mu \right)\right\}}^{2}f\left(x_i\right)f\left(x_i\right)dx}}$

${\displaystyle ={\displaystyle {\int }_{-\infty }^{\infty }{a}^2{\left(x_i-\mu \right)}^{2}f\left(x_i\right)f\left(x_i\right)dx}}$

${\displaystyle =a^2{\displaystyle {\int }_{-\infty }^{\infty }{\left(x_i-\mu \right)}^{2}f\left(x_i\right)f\left(x_i\right)dx}}$

${\displaystyle =a^{2}V\left(X\right)}$

　

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最終更新日： 2024年2月23日