验证总方差公式 v a r ( X ) = E [ v a r ( X ∣ Y ) ] + v a r ( E [ X ∣ Y ] ) var(X) = E[var(X|Y)] + var(E[X|Y])var(X)=E[var(X∣Y)]+var(E[X∣Y]).
主要要知道基本的公式 v a r ( X ) = E [ X 2 ] − ( E [ X ] ) 2 var(X) = E[X^2]-(E[X])^2var(X)=E[X2]−(E[X])2 以及 E [ E [ X ∣ Y ] ] = E [ X ] E[E[X|Y]] = E[X]E[E[X∣Y]]=E[X]. 然后可以完成证明.
证明: 先看等号右边第一项, 因为 v a r ( X ∣ Y ) = E [ X 2 ∣ Y ] − ( E [ X ∣ Y ] ) 2 var(X|Y) = E[X^2|Y] - (E[X|Y])^2var(X∣Y)=E[X2∣Y]−(E[X∣Y])2,
所以 E [ v a r ( X ∣ Y ) ] = E [ X 2 ] − E [ ( E [ X ∣ Y ] ) 2 ] E[var(X|Y)] = E[X^2] - E[(E[X|Y])^2]E[var(X∣Y)]=E[X2]−E[(E[X∣Y])2]
之后看等号右边第二项, 因为
v a r ( E [ X ∣ Y ] ) = E [ ( E [ X ∣ Y ] ) 2 ] − ( E [ E [ X ∣ Y ] ] ) 2 = E [ ( E [ X ∣ Y ] ) 2 ] − ( E [ X ] ) 2 \begin{aligned} var(E[X|Y]) &= E[(E[X|Y])^2] - (E[E[X|Y]])^2 \\ &=E[(E[X|Y])^2] - (E[X])^2 \end{aligned}var(E[X∣Y])=E[(E[X∣Y])2]−(E[E[X∣Y]])2=E[(E[X∣Y])2]−(E[X])2
所以,
E [ v a r ( X ∣ Y ) ] + v a r ( E [ X ∣ Y ] ) = E [ X 2 ] − ( E [ X ] ) 2 = v a r ( X ) \begin{aligned} &E[var(X|Y)] + var(E[X|Y])\\ =&E[X^2]-(E[X])^2 \\ =&var(X) \end{aligned}==E[var(X∣Y)]+var(E[X∣Y])E[X2]−(E[X])2var(X)