Comparing between $(\mathbb{E}[X] \cdot \mathbb{E}[Y])^{1/2}$ and $\mathbb{E}[(XY)^{1/2}]$

Question

I have the following problem: let $X$ and $Y$ be non-negative random variables. I know that both $(\mathbb{E}[X] \cdot \mathbb{E}[Y])^{1/2}$ and $\mathbb{E}[(XY)^{1/2}]$ are at least $\mathbb{E}[X^{1/2}] \cdot \mathbb{E}[Y^{1/2}]$ by concavity of $\sqrt{x}$ and the non-negativity of covariance (respectively). However, it is unclear to me whether the two terms are comparable. Would appreciate to hear comments on this.

Thanks!

Answer 1

The short answer is: no, we can't say anything about the relationship between these.

• Consider the case where $Y = X$; here, the left and right terms are both $\mathbb E[X]$.

• Consider the case where $X, Y$ are independent Bernoulli variables; then the left term is $0.5$ and the right term is $0.25$.

• Consider the case where $X$ is a Bernoulli variable and $Y = 1-X$. Then the left term is $0.5$ and the right term is $0$.

But this last example illustrates an important point -- the original claim isn't correct! Covariance can be negative.