Import definitions and concepts in “Probabilistic Machine Learning: An Introduction“
作者：互联网
E [ V [ X ∣ Y ] ] = P ( Y = 1 ) V [ X ∣ Y = 1 ] + P ( Y = 2 ) V [ X ∣ Y = 2 ] = 0.5 × 0.5 + 0.5 × 0.5 = 0.5 (2.48) E[V[X|Y]] \\= P(Y=1)V[X|Y=1] + P(Y=2)V[X|Y=2] \\= 0.5 \times 0.5 + 0.5 \times 0.5 = 0.5 \tag{2.48} E[V[X∣Y]]=P(Y=1)V[X∣Y=1]+P(Y=2)V[X∣Y=2]=0.5×0.5+0.5×0.5=0.5(2.48)
V [ E [ X ∣ Y ] ] = E [ ( E [ X ∣ Y ] ) 2 ] − ( E [ E [ X ∣ Y ] ] ) 2 = E [ ( E [ X ∣ Y ] ) 2 ] − ( E [ X ] ) 2 = P ( Y = 1 ) ( E [ X ∣ Y = 1 ] ) 2 + P ( Y = 2 ) ( E [ X ∣ Y = 2 ] ) 2 − ( 2 + 0 2 ) 2 = 0.5 ∗ 0 + 0.5 ∗ 2 2 − 1 = 1 (2.49) V[E[X|Y]] \\= E[(E[X|Y])^2] - (E[E[X|Y]])^2 \\= E[(E[X|Y])^2] - (E[X])^2 \\= P(Y=1)(E[X|Y=1])^2 + P(Y=2)(E[X|Y=2])^2 - (\frac{2+0}{2})^2 \\= 0.5 * 0 + 0.5 * 2^2 - 1 = 1 \tag{2.49} V[E[X∣Y]]=E[(E[X∣Y])2]−(E[E[X∣Y]])2=E[(E[X∣Y])2]−(E[X])2=P(Y=1)(E[X∣Y=1])2+P(Y=2)(E[X∣Y=2])2−(22+0)2=0.5∗0+0.5∗22−1=1(2.49)
标签：22,Introduction,2.49,0.5,times,Machine,2.48,tag,definitions 来源： https://blog.csdn.net/qq_27039891/article/details/118875689