Is Expectation of Maximum of two Convex Random Variables Convex?

Question

Assume that $\xi_1(x)$ and $\xi_2(y)$ are two non-negative independent random variables with pdf $f(x,\tau)$, $f(y,\tau)$ and cdf $F(x,\tau)$, $F(y,\tau)$, where $x$, $y \in \mathbb{R}^n$ are parameters.

We know, that their expected values $$ E[\xi_1(x)] = \mu(x), $$ $$ E[\xi_2(y)] = \mu(y), $$

where $\mu$ is a convex function.

It is known, that the expected value $ M(x, y) = E[max\{\xi_1(x), \xi_2(y)\}]$ is

$$ M(x,y) = \int_0^\infty F(x,\tau)f(y, \tau) \tau d\tau + \int_0^\infty F(y,\tau)f(x, \tau) \tau d \tau. $$

Is $M(x,y)$ convex?

Thanks a lot!

Answer 1

It isn’t. For instance, for $n=1$, take $\xi_1(x)=0$ and $\xi_2(y)=\pm\sqrt{|y|}$ with probability $\frac12$ each. Then $\mu(x)=\mu(y)=0$ is convex, but $M(x,y)=\frac12\sqrt{|y|}$ isn’t.

Answer 2

Just a slight modification of the example already proposed. Let's have $\xi_1(x)=\pm \sqrt{|x|}$ and $\xi_2(y)=\pm \sqrt{|y|}$ each value with probability $1/2$.

If we call $f_1(x,*)$ and $f_2(y,*)$ the associated parametrized pdfs, than there is in this case a common parameterized pdf $f(z,*)$ such that, $f_1(x,*)=f(x,*)$ and $f_2(y,*)=f(y,*)$.

We now have $E[\xi_1(x)]=E[\xi_2(y)]=0$ and we can define $Z(x,y)=max(\xi_1(x),\xi_2(y))$.

From an explicit computation (there are just 4 combinations to consider because we consider the variables independent):

$E[Z(x,y)]=\sqrt{max(|x|,|y|)}/2$

which is not convex in $x,y$.