Consider two functions, $g(x)=\left(1-\mathbb{E}_y (f(x,y))\right)^n$, $h(x)=\mathbb{E}_y\left(\left(1- f(x,y)\right)^n\right)$, $n=1,2,\cdots$, where $\mathbb{E}_y(f(x,y))$ denotes the expectation of function $f(x,y)\in[0,1]$, $x\in[0,1]$, with respect to (w.r.t.) $y$, where $y\in(-\infty.\infty)$.
I have a conjecture which is stated as follows. If $\mathbb{E}_y(f(x,y))$ is concave w.r.t. $x$, then both the functions $g(x)$ and $h(x)$ are convex w.r.t. $x$.
However, I cannot prove my conjecture above. Can anyone give me some hints to prove it, or take a counter example to tell me that it is wrong.
I appreciate any comment on this post.
Yes to the $g$ part, as you have probably checked.
But no to the $h$ part. Let $n=2$, let $y\sim N(0,1)$ and let $f(x,y)=1/2+x(1-x)$ if $y\ge0$ and $f(x,y)=1/2-x(1-x)$ if $y<0$. (Think of this as $1-f(x,y) = 1/2 \pm x(1-x)$ where the sign depends on $y$.) The expectation $\mathbb E_y(1-f(x,y)) = 1/2$, a convex function. But $$h(x)= \mathbb E_y(1-f(x,y))^2 = \mathbb E_y(1/4 \pm x(1-x)) + (x(1-x))^2) = 1/4 + (x(1-x))^2,$$ which is neither concave nor convex.
The problem is that requiring $\mathbb E_y(f(x,y))$ to be concave in $x$ does not put much of a constraint on the slices you get by holding $y$ fixed. As above, the average of non-concave functions can be concave. This does not hurt you in the $g$ case, but in the $h$ case it can.