$E(X) \geq E(Y) \implies E(f(X)) \geq E(f(Y))$ for non-decreasing $f$?

Question

Is the following statement true?

Let $X,Y$ be random variables such that $E(X) \geq E(Y)$. Then $$E(f(X)) \geq E(f(Y)) $$ for all non-decreasing functions $f$.

I feel like it should be, but I can't prove it. Am I missing something obvious?

Answer 1

Statement is false by the following counterexample: Let $X = 1$ w.p. 1, $$ Y = \begin{cases} 0, \quad \text{w.p. } 1/2 \\ 2, \quad \text{w.p. } 1/2 \end{cases} $$ and let $f(x) = x^2$. Then $E(X) = 1 = E(Y)$ and $f$ is non-decreasing, but $$1 = E(f(X)) < 2 = E(f(Y)) $$ $\square$