1. ## Expectation inequality probelems

If E(X) < E(X')
and E(Y) < E(Y')
Can we prove that E(XY) < E(X'Y') ?

2. No, in general E(XY) < E(X'Y') is not true.

First, if the variables are not independent, here is a counter example:
X = 0, 1 with equal probability
Y = X (not independent of X)
X' = 0, 100 with equal probability
Y' = 100 - X' (not independent of X')
E(X) = E(Y) = 0.5
E(X') = E(Y') = 50
E(XY) = 0.5
E(X'Y') = 0

Second if they were independent E(XY) = E(X)E(Y), and E(X'Y') = E(X')E(Y')
The expectations still have to be positive.
Example:
E(X) = E(Y) = -1
E(X') = E(Y') = 0
E(XY) = E(X)E(Y) = 1
E(X'Y') = E(X')E(Y') = 0

3. ## Case where X and X', Y and Y' are related

@snowtea : Nice counter-example !

Can we still come up with a counter-example if X & X' are related, and Y & Y' are related ?

4. If X, X' are related and Y, Y' are related. Lets assume that X,X',Y,Y' can only take nonnegative values, and X' = f(X), Y' = g(Y). Otherwise, the counter example will be quite easy to come up with (try it).

There is still a counter example for the case described.
X = 0, 10 with equal probability
X' = 4 when X = 0 and X' = 8 when X = 10
Y=X and Y'=X'
E(X) = E(Y) = 5
E(X') = E(Y') = 6
E(XY) = E(X^2) = 50
E(X'Y') = E(X'^2) = 40