Question

if X and Y are two random variables and f ,g two increasing functions how to prove that E(f(X)g(Y))>E(f(X))E(g(Y)) , E is the expectation

Answer 1

So, this? \( E[f(X)g(Y)]>E[f(X)] E[g(Y)]\)

Answer 2

yes with the conditions mentionned in the question I want to prove this inequality E[f(X)g(Y)]>E[f(X)]E[g(Y)]

Answer 3

Hm let me see if I can think of something

Answer 4

okay thanks

Answer 5

for example take E(x)=x/2, then
E(f(x)g(y))=[ f(x)g(y) ]/2 where
E(f(x))E(g(y))=[ f(x)g(y) ]/4
=> E(f(X)g(Y))>E(f(X))E(g(Y))

Answer 6

but what about the general case !!

Answer 7

Gosh I am sorry. I have tried doing this in multiple ways but haven't succeeded :(

Answer 8

it's okay me too I've tried everything I know !!

Answer 9

By definition,
$$
\large{
var[g(X)f(Y)]=E[g^2(X)]E[f^2(Y)]-E[g(X)]^2E[f(Y)]^2 \ge 0
}
$$
http://en.wikipedia.org/wiki/Variance#Product_of_independent_variables

By Cauchy-Schwartz,
$$
\large{
E[g(X)f(Y)]^2\le E[g^2(X)]E[f^2(Y)]
}
$$
http://en.wikipedia.org/wiki/Cauchy-Schwartz_inequality#Probability_theory

Combining these two facts we have
$$
E[g^2(X)]E[f^2(Y)]-E[g(X)]^2E[f(Y)]^2 \le E[g(X)f(Y)]^2-E[g(X)]^2E[f(Y)]^2 \ge 0\\
\implies E[g(X)f(Y)]^2\ge E[g(X)]^2E[f(Y)]^2\\
\implies E[g(X)f(Y)]\ge E[g(X)]E[f(Y)]
$$
Now how do we use the fact that f and g are increasing to make this relation strictly increasing?

Answer 10

*I just noticed that I interchanged g(X) and f(Y). It should be f(X) and g(Y) throughout instead. Sorry about that.

Answer 11

Because f(X) and g(Y) are increasing, the inverse mappings exist and are given by
$$
X=f^{-1}X\\
Y=g^{-1}Y
$$
This justified our use in applying the following:
http://en.wikipedia.org/wiki/Variance#Product_of_independent_variables

Since we assume that X and Y are independent (because the problem did not state otherwise) we can use this fact to indicate that f and g are independent and the relation is strictly increasing. Otherwise, equality would hold and we could write X = aY, for some constant \(a\), and therefore f(X)=bf(Y), for some \(b\), which we were not given.

Here is some info on increasing functions an their relationship to inverse functions - http://en.wikipedia.org/wiki/Monotonic_function
http://www.math.uh.edu/~jiwenhe/Math1432/lectures/lecture01_handout.pdf

Answer 12

thank for all this great effort, but when you said that the inverse map exists because the function is increasing , I think we need here the continuity as a condition. otherwise I was thinking about an other method but I need to proove this :
if we compose a bilinear form with two increasing functions do we get a positive biliear form ??
if I can proove that then as we know the Cov(X,Y) is a bilinear form and if we compose it with f and g ;Cov(f(x),g(y)) will be positive

Answer 13

@loser or @satellite may be able to provide some guidance on whether the composition of two increasing functions produce a positive bilinear form

Answer 14

I find the assumption of independence interesting... because whenever independence is not indicated (on our homework/tests), we always assume dependence :O!

Answer 15

@hidy

But if \(X,Y\) are independent (as we assumed), then \(f(X), g(Y)\) are independent, then \( Cov(X,Y)=Cov\left(f(X),g(Y) \right)=0\)

Answer 16

yes, you're right if the X and Y were independent then we have the equality !
but unfortunately in this question X and Y are not necessarily independents :( :(

Answer 17

I misstated, X and Y should NOT be assumed independent, otherwise equality would hold therefore, our relation IS strictly increasing.

$$
E[g(X)f(Y)]\gt E[g(X)]E[f(Y)]
$$