[問題] X1,X2~N(0,1) Y=min(X1,X2) 求E(Y)
X1, X2 iid~ N(0,1) Y=min(X1,X2)之期望值?
我猜也許可以這樣解?
pdf of N(0,1) = f(x)
cdf of N(0,1) = Φ(x)=P(X<=x)
P(Y<=y)=P(min(X1,X2)<=y)=1-P(min(X1,X2)>y)
=1-P(X1>y)P(X2>y)=1-(1-P(X1<=y))(1-P(X2<=y)
=1-(1-Φ(y))(1-Φ(y))
=1-(1-2Φ(y)+(Φ(y))^2)
=2Φ(y)-(Φ(y))^2
接下來對y微分，求pdf of Y，得到
f_Y(y)=2f(y)-2f(y)Φ(y)
取期望值E(Y)=
∞
∫2yf(y)-2yf(y)Φ(y) dy
-∞
∞ y
=0-2∫ yf(y)(∫f(x)dx) dy
-∞ -∞
∞ y
=-(1/π)∫ y∫exp[-(x^2+y^2)/2] dx dy
-∞ -∞
令y=rcosθ, x=rsinθ, Jacobian=r
π+arctan(1) ∞
=-(1/π)∫ ∫rcosθexp(-r^2 /2)rdrdθ
arctan(1) 0
接下來卡住了....
這有比較容易的解法嗎?
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※ 編輯: anovachen 來自: 111.255.238.16 (02/20 08:40)