$V (a X + b Y)$ $V (a X + b Y)$ $= a^{2} V (X) + 2 a b C (X, Y) + b^{2} V (X)$ $= a^{2} V (X) + 2 a b C (X, Y) + b^{2} V (X)$

2つの確率変数 $X$ $X$ ， $Y$ $Y$ について

$V (a X + b Y)$ $V (a X + b Y)$ $= a^{2} V (X) + 2 a b C (X, Y) + b^{2} V (X)$ $= a^{2} V (X) + 2 a b C (X, Y) + b^{2} V (X)$ 　･･････(1)

が成り立つ．

■証明

$E (X) = μ_{x}$ $E (X) = μ_{x}$ 　･･････(2)

$E (Y) = μ_{y}$ $E (Y) = μ_{y}$ 　･･････(3)

とする．

分散の定義より

$V (a X + b Y)$ $V (a X + b Y)$ $= E ({(a X + b Y) - (a μ_{x} + b μ_{y})}^{2})$ $= E ({\{(a X + b Y) - (a μ_{x} + b μ_{y})\}}^{2})$

$= E ({(a X + b Y)}^{2}) - {E (a μ_{x} + b μ_{y})}^{2}$ $= E ({(a X + b Y)}^{2}) - {\{E (a μ_{x} + b μ_{y})\}}^{2}$

∵ $V (X) = E (X^{2}) - {E (X)}^{2}$ $V (X) = E (X^{2}) - {\{E (X)\}}^{2}$ (分散を参照)

$= E (a^{2} X^{2} + 2 a b X Y + b^{2} Y^{2}) - {(a μ_{x} + b μ_{y})}^{2}$ $= E (a^{2} X^{2} + 2 a b X Y + b^{2} Y^{2}) - {(a μ_{x} + b μ_{y})}^{2}$

$= E (a^{2} X^{2}) + E (2 a b X Y) + E (b^{2} Y^{2})$ $= E (a^{2} X^{2}) + E (2 a b X Y) + E (b^{2} Y^{2})$ $- a^{2} μ_{x}^{2}$ $- a^{2} μ_{x}^{2}$ $- 2 a b μ_{x} μ_{y}$ $- 2 a b μ_{x} μ_{y}$ $- b^{2} μ_{y}^{2}$ $- b^{2} μ_{y}^{2}$

$= a^{2} E (X^{2}) - a^{2} μ_{x}^{2} + 2 a b E (X Y)$ $= a^{2} E (X^{2}) - a^{2} μ_{x}^{2} + 2 a b E (X Y)$ $- 2 a b μ_{x} μ_{y}$ $- 2 a b μ_{x} μ_{y}$ $+ b^{2} E (Y^{2})$ $+ b^{2} E (Y^{2})$ $- b^{2} μ_{y}^{2}$ $- b^{2} μ_{y}^{2}$

$= a^{2} {E (X^{2}) - μ_{x}^{2}}$ $= a^{2} \{E (X^{2}) - μ_{x}^{2}\}$ $+ 2 a b {E (X Y) - μ_{x} μ_{y}}$ $+ 2 a b \{E (X Y) - μ_{x} μ_{y}\}$ $+ 2 a b {E (X Y) - μ_{x} μ_{y}}$ $+ 2 a b \{E (X Y) - μ_{x} μ_{y}\}$

$= a^{2} V (X) + 2 a b C (X, Y) + b^{2} V (X)$ $= a^{2} V (X) + 2 a b C (X, Y) + b^{2} V (X)$

∵ $V (X) = E (X^{2}) - {E (X)}^{2}$ $V (X) = E (X^{2}) - {\{E (X)\}}^{2}$ ， $C (X, Y) = E (X Y) - E (X) E (Y)$ $C (X, Y) = E (X Y) - E (X) E (Y)$

ホーム>>カテゴリー分類>>確率統計>> $E (X)$ $E (X)$ ， $V (X)$ $V (X)$ ， $C (X, Y)$ $C (X, Y)$ の計算則>> $C (X, Y) = 0$ $C (X, Y) = 0$

最終更新日： 2024年2月23日