Upper bound of $\mathbb{E}(\max(X,Y))$ with known mean and variance

Question

I’m looking for any idea to go to the end of this... Let us consider random variables $X$ and $Y$, positive and independent, with finite mean value ($\mu_{X}$ and $\mu_{Y}$) and finite variance ($\sigma_{\!X}^{2}$ and $\sigma_{Y}^{2}$). No other assumptions. We’re looking for the best upper bound of $\mathbb{E}(\max (X,Y))$ with these conditions.

My feeling is that best upper bound is reached when taking $X$ and $Y$ as discrete distributions, with $2$ points. Assuming this, I can show that the best upper bound is given by: $$ M = \max\left( \mu_X^{\phantom.} + \frac{\mu_Y^{\phantom.}\,\sigma_X^2}{\mu_X^{2} + \sigma_X^2},\, \mu_Y^{\phantom.} + \frac{\mu_X^{\phantom.}\,\sigma_Y^2}{\mu_Y^{2} + \sigma_Y^2} \right). $$

“Best upper bound” meaning that if we note $D_{+}$ the space of positive and independent distributions with finite mean value and variance, $$ M = \inf\,\left\{ m \ \middle\vert\ \forall X, Y \in D_{+},\, \mathbb{E}\bigl(\max (X,Y)\bigr) \leq m(\mu_{1},\sigma_{1},\mu_{2},\sigma_{2}) \right\}. $$

Numerically with various distributions with only given mean and variance, this upper bound is okay. And $M$ is exactly reached for well tuned $2$-points discrete distributions.

But I can’t find an approach to demonstrate and justify that discrete distributions (with $2$ points) are the ones that are minimising upper bound of $\mathbb{E}\bigl(\max(X,Y)\bigr)$ with these constraints. It looks like a problem of optimization with constraints on the space of positive and independent distribution, but not sure how to formulate/solve this.

Any idea to proceed? Thank you in advance for help and advice.

Answer 1

Just an upper bound, not the best.

Note that $\max(X, Y) = \frac{X+Y+|X-Y|}{2}$, therefore \begin{align*} \mathbb E\big[ \max(X, Y) \big] &= \frac{\mu_X+\mu_Y}{2} + \frac{1}{2} \mathbb E\big[ |X-Y| \big] \end{align*} So you are looking for an upper bound on $\mathbb E\big[ |X-Y| \big]$. By the Cauchy–Schwarz inequality, we have \begin{align*} \mathbb E\big[ |X-Y| \big]^2&\leq \mathbb E\big[ (X-Y)^2 \big]\\ &=\mathbb E\big[ X^2 - 2 XY + Y^2 \big]\\ &=\sigma_X+\mu_X^2 -2 \mu_X \mu_Y + \sigma_Y+\mu_Y^2\\ &=\sigma_X+\sigma_Y + (\mu_X-\mu_Y)^2 \end{align*} We get the following upper bound: \begin{align*} \mathbb E\big[ \max(X, Y) \big] &\leq \frac{\sigma_X+\mu_X+\sigma_Y+\mu_Y + (\mu_X-\mu_Y)^2}{2} \end{align*} Note that this bound is in general not tight unless $|X-Y|$ is constant, or in other words, both $X$ and $Y$ are constant (because of their independence).