a********a
发帖数: 346 | 1 if X，Y are idd , E(X/Y)>=E(X) / E(Y) given E(Y) 不等于0成立吗？需要怎么证
明？谢谢。 | h*******e
发帖数: 4 | 2 不能除以E(X),因为E(X)正负未定呀。你这个例子里，E(X)<0的，所以原不等式还是成
立的。 | g******n
发帖数: 339 | 3 sorry, I wasn't very careful.
The inequality you try to prove is essentially E(X)*E(1/X)>=1, since X and Y
are IID.
It seems that if the support of X includes 0, then you would have to define
E(1/X) carefully, as E(1/X) could be infinity. In special cases when the
support of X is positive, i.e, when X is defined (or a subset) on (0,+
infinity), the proof is simply Jensen's inequality:
Jensen's: if g(.) is convex and defined on the support of X, then E[g(X)]>=g
(E(X))
let g(w) be 1/w (w in (0, +i
【在 h*******e 的大作中提到】
: 不能除以E(X),因为E(X)正负未定呀。你这个例子里，E(X)<0的，所以原不等式还是成
: 立的。
| h*******e
发帖数: 4 | 4 Yeah, this time, it makes sense. Thank you.
Y
define
=g
(0
【在 g******n 的大作中提到】
: sorry, I wasn't very careful.
: The inequality you try to prove is essentially E(X)*E(1/X)>=1, since X and Y
: are IID.
: It seems that if the support of X includes 0, then you would have to define
: E(1/X) carefully, as E(1/X) could be infinity. In special cases when the
: support of X is positive, i.e, when X is defined (or a subset) on (0,+
: infinity), the proof is simply Jensen's inequality:
: Jensen's: if g(.) is convex and defined on the support of X, then E[g(X)]>=g
: (E(X))
: let g(w) be 1/w (w in (0, +i
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