When does $E[X] \leq E[Y]$ imply $E[X^2] \leq E[Y^2]$?

Question

Assume $X, Y$ are two non-negative random variables. When does $E[X] \leq E[Y]$ imply $E[X^2] \leq E[Y^2]$? Always?

I think the conclusion is not always true.

Answer 1

As a counterexample, let $X=B(p)/p$ and $Y=B'(q)/q$, where $B,B'$ are iid Bernoulli with probability $p,q$. Then $E[X]=E[Y]=1$, but $E[X^2]=1/p$ and $E[Y^2]=1/q$, which will violate your inequality when $p<q$.

There's a lot of conditions that would imply your inequality. For example in terms of second moments, $E[X^2]\leq E[Y]^2$ would imply $E[X^2]\leq E[Y^2]$.

Answer 2

Take $Y=1$ and let $X$ be uniformly distributed on $[0, M]$, with $M \le 2$ so that $EX \le EY$.

Then $EX^2 = {M^3 \over 3}, E Y^2 = 1$ and $EX^2 \le EY^2$ iff $M \le \sqrt[3]{3} \approx 1.44 < 2$.

Answer 3

Since $X,Y$ are non-negative, $E[X] \le E[Y] \iff E[X]^2 \le E[Y]^2$. Meanwhile, you want:

$$ 0 \le E[Y^2] - E[X^2] \\ = (E[Y]^2 + Var(Y)) - (E[X]^2 + Var(X)) \\ = (E[Y]^2 - E[X]^2) + (Var(Y) - Var(X))$$

So, as Alex R. said, there are many conditions which would lead to what you want. A particularly simple additional sufficient condition would be $Var(Y) \ge Var(X)$, but it is not necessary. Another way to say this is that any counterexample must have $Var(Y) < Var(X)$ -- the counterexample of copper.hat exploited this.