TITLE: Prove that expected value of X is greater than Y, if given that $P(X\ge Y)=1$
QUESTION [6 upvotes]: I have to prove that $E(X)$ (Expected Value of a random variable X), is greater than $E(Y)$, if given that $P(X\ge Y)=1$.
my thoughts so far:
I know from the $P(X\ge Y)=1$ statement, that the values that X "receives" are always greater than the values that Y "receives", and because the expected value of a random variable is always a weighted mean of its valid values, than of course $E(X) \ge E(Y)$. But how do I prove it formally?

REPLY [3 votes]: Take a new random variable, say Z, where Z = X - Y. Then you have $P(Z \geq 0) = 1$. Now try to show $E(Z) \geq 0$.