e****s
发帖数: 46 | 1 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
发帖数: 31 | 2 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
发帖数: 31 | 3 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
发帖数: 46 | 4 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
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