assume X1 and X2 are independent and identically distributed r.v. with mean E[X] =1 let Y = max(X1,X2) how to show that E[Y] <= 2E[X] Thanks!
s*q
2 楼
I tried for a while. What I believe is that you missed some condition, such as X1 is positive definite. Reason: counter-example: P(X1=-2)=1/2 P(X1=4)=1/2 X1, X2 iid, E(X1)=(-2)*1/2+4/2=-1+2=1 Then P(Y=-2)=1/4 P(Y=4)=3/4 E(Y)=-1/2+3=5/2>2 Do you think so?
【在 e****s 的大作中提到】 : assume X1 and X2 are independent and identically distributed r.v. : with mean E[X] =1 : let Y = max(X1,X2) : how to show that E[Y] <= 2E[X] : Thanks!
s*q
3 楼
Let's re by myself, hehe. If X1 is non-negative. The statement is true by using the equation: E(X)=integer(1-F_X(x)){x from 0 to infitie} You can find it is easy to show and the condition E(X1)=1 is useless.
as
【在 s*q 的大作中提到】 : I tried for a while. What I believe is that you missed some condition, such as : X1 is positive definite. : Reason: counter-example: : P(X1=-2)=1/2 : P(X1=4)=1/2 : X1, X2 iid, : E(X1)=(-2)*1/2+4/2=-1+2=1 : Then : P(Y=-2)=1/4 : P(Y=4)=3/4
e*s
4 楼
Yeah, sorry, i forget a condition, X>0, actually, X = k*a, a is a small number and a>0, k is positive integer.
such
【在 s*q 的大作中提到】 : Let's re by myself, hehe. : If X1 is non-negative. The statement is true by using the equation: : E(X)=integer(1-F_X(x)){x from 0 to infitie} : You can find it is easy to show and the condition E(X1)=1 is useless. : : as