s*****n
发帖数: 2174 | 1 从本科同学的微信群上看来的, 蛮有意思的一道题
Let X and Y be two positive random variables with identical distribution (
not necessarily independent), prove E(X/Y) >= 1. | e********2
发帖数: 495 | 2 e(x/y) + e(y/x) >= 2
and
e(x/y) == e(y/x)?
【在 s*****n 的大作中提到】
: 从本科同学的微信群上看来的, 蛮有意思的一道题
: Let X and Y be two positive random variables with identical distribution (
: not necessarily independent), prove E(X/Y) >= 1.
| s*****n
发帖数: 2174 | 3 E(X/Y) 不一定等于 E(Y/X)
反例如下: X, Y有下面的联合分布
P(X=1, Y=2) = 1/3
P(X=2, Y=4) = 1/3
P(X=4, Y=1) = 1/3
X和Y的分布都是(1,2,4)的均匀分布, 但是
P(X/Y=0.5) = 2/3, P(X/Y=4) = 1/3, E(X/Y)=5/3
P(Y/X=2)=2/3, P(Y/X=1/4)=1/3, E(Y/X)=17/12
【在 e********2 的大作中提到】
: e(x/y) + e(y/x) >= 2
: and
: e(x/y) == e(y/x)?
| e********2
发帖数: 495 | 4 Jensen's inequality
E(e^(ln(Y/X)))>=e^(E(ln(Y/X)))=1
【在 s*****n 的大作中提到】
: E(X/Y) 不一定等于 E(Y/X)
: 反例如下: X, Y有下面的联合分布
: P(X=1, Y=2) = 1/3
: P(X=2, Y=4) = 1/3
: P(X=4, Y=1) = 1/3
: X和Y的分布都是(1,2,4)的均匀分布, 但是
: P(X/Y=0.5) = 2/3, P(X/Y=4) = 1/3, E(X/Y)=5/3
: P(Y/X=2)=2/3, P(Y/X=1/4)=1/3, E(Y/X)=17/12
| s*****n
发帖数: 2174 | 5 但是你不等式右边展开的时候有可能出现无穷减无穷的情况。所以这个证明只对E(log(
X)) 和 E(log(Y)) 存在这个情况成立。
【在 e********2 的大作中提到】
: Jensen's inequality
: E(e^(ln(Y/X)))>=e^(E(ln(Y/X)))=1
| P****i
发帖数: 1362 | 6 强，我想到用Jensen's inequality，没想出来这么变换
【在 e********2 的大作中提到】
: Jensen's inequality
: E(e^(ln(Y/X)))>=e^(E(ln(Y/X)))=1
| I*****a
发帖数: 5425 | 7 E(X/Y) = exp( ln ( E(X/Y) ) )
ln( E(X/Y) ) >= E( ln(X/Y) ) = 0
so E(X/Y) >= exp(0) = 1
【在 e********2 的大作中提到】
: Jensen's inequality
: E(e^(ln(Y/X)))>=e^(E(ln(Y/X)))=1
| s*****n
发帖数: 2174 | 8 这个证明和上面有同样的问题
这题的精华在于如何证明一般的情况
X,Y>0
【在 I*****a 的大作中提到】
: E(X/Y) = exp( ln ( E(X/Y) ) )
: ln( E(X/Y) ) >= E( ln(X/Y) ) = 0
: so E(X/Y) >= exp(0) = 1
| L****o
发帖数: 1 | 9 民科’s guess:
Since X>0, E(X) is always defined. Two cases to be considered.
(1)E(X)0, ln(X) is defined and hence Eln(X)<00 because of E(X)<
oo. Then by Jensen inequality, we see that E(X/Y)=Eexp{ln(X/Y)}>=exp{Eln(X/Y
)}=exp{Eln(X)-ElnY}=0, where the last equality is obtained since lnX and LnY
have the same marginal distribution and their first moments are finite.
(2)E(X)=oo. Since X>0, we can always find a large positive integer m such
that E{X^(1/m)}={E[(X/Y)^(1/m)]}^m by
virtue of Lyapunov's Inequality. It remains to establish E[(X/Y)^(1/m)]>=1
. But E[(X/Y)^(1/m)]=E{[X^(1/m)]/[Y^(1/m)]). Then the desired result
follows by using the same procedure in (1) with X^(1/m) and Y^(1/m)
replacing X and Y respectively.
【在 s*****n 的大作中提到】
: 从本科同学的微信群上看来的, 蛮有意思的一道题
: Let X and Y be two positive random variables with identical distribution (
: not necessarily independent), prove E(X/Y) >= 1.
| l******t
发帖数: 96 | 10 in (1) E log(X) could be - infty.
)<
/Y
LnY
by
=1
【在 L****o 的大作中提到】
: 民科’s guess:
: Since X>0, E(X) is always defined. Two cases to be considered.
: (1)E(X)0, ln(X) is defined and hence Eln(X)<00 because of E(X)<
: oo. Then by Jensen inequality, we see that E(X/Y)=Eexp{ln(X/Y)}>=exp{Eln(X/Y
: )}=exp{Eln(X)-ElnY}=0, where the last equality is obtained since lnX and LnY
: have the same marginal distribution and their first moments are finite.
: (2)E(X)=oo. Since X>0, we can always find a large positive integer m such
: that E{X^(1/m)}={E[(X/Y)^(1/m)]}^m by
: virtue of Lyapunov's Inequality. It remains to establish E[(X/Y)^(1/m)]>=1
: . But E[(X/Y)^(1/m)]=E{[X^(1/m)]/[Y^(1/m)]). Then the desired result
| | | D**u
发帖数: 288 | 11 let me give a try:
Expectation is basically an integral, when we write E(X/Y), we are assuming
that X/Y is integrable. Then Y should not have continuous density
approaching and at zero, otherwise the integral goes to infinity. And, since
we are told "X and Y be two positive(should be non-negative?) random
variables", then X, Y > 0. And then the Jensen's inequality follows. | l******t
发帖数: 96 | 12 we are not assuming X/Y is integrable, if it's not, then expectation is
infity, again >= 1.
Thus, Y could have continuous density approaching 0.
assuming
since
【在 D**u 的大作中提到】
: let me give a try:
: Expectation is basically an integral, when we write E(X/Y), we are assuming
: that X/Y is integrable. Then Y should not have continuous density
: approaching and at zero, otherwise the integral goes to infinity. And, since
: we are told "X and Y be two positive(should be non-negative?) random
: variables", then X, Y > 0. And then the Jensen's inequality follows.
| D**u
发帖数: 288 | 13 if the integral is infinity, then it is not well defined, and it does not
exist, and you can't say it is >=1.
【在 l******t 的大作中提到】
: we are not assuming X/Y is integrable, if it's not, then expectation is
: infity, again >= 1.
: Thus, Y could have continuous density approaching 0.
:
: assuming
: since
| l******t
发帖数: 96 | 14 that's usually the case, but if the RV is positive, we could still say that
it's infinity, and of course it's > 1.
The reason we usually ignore infinity is that expectation of both positive
and negative part are infinity, and we could not get an clear answer for
infinity - infinity.
【在 D**u 的大作中提到】
: if the integral is infinity, then it is not well defined, and it does not
: exist, and you can't say it is >=1.
| e********2
发帖数: 495 | 15 你可以把积分区间分成小块，然后两小块，两小块相加。
or
[1/r, r] x [1/r, r]区间积分然后对r->inf取极限。
log(
【在 s*****n 的大作中提到】
: 但是你不等式右边展开的时候有可能出现无穷减无穷的情况。所以这个证明只对E(log(
: X)) 和 E(log(Y)) 存在这个情况成立。
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