I want to find out whether the following expectation is concave or convex in $y$. $$E[g(f(x)−y)]$$ Given $g$ and $f$ are both concave functions, and $x\sim U(i,i+y)$ where $i>0$ is the only constant and $y\geq 0$. This is a follow-up from this thread.
I think I would start by substituting $h(x)\equiv g(f(x)-y)$, through which I get $$E[h(x)]=y^{-1}\int^{i+y}_{i}h(z)\;dz$$
I do not know how to proceed from this, however. The above form also does not give much information regarding the concavity of the expectation. Is it possible to actually know whether it is concave or convex, or do we need a more specific form of either $g$ or $f$?
Let's $y_1,\,y_2\ge 0$ and $t\in[0,1]$. As $g$ in concave, we have $$\begin{align*} g\left(f(x)+ty_1+(1-t)y_2\right) &= g\left(tf(x) + (1-t)f(x) +ty_1+(1-t)y_2\right)\\ &= g\left(t(f(x)+y_1) + (1-t)(f(x)+y_2)\right)\\ &\ge tg(f(x)+y_1) + (1-t)g(f(x)+y_2) \end{align*}$$
Then $$\begin{align*} E\left[g\left(f(x)+ty_1+(1-t)y_2\right)\right] &\ge E\left[tg(f(x)+y_1) + (1-t)g(f(x)+y_2)\right]\\ E\left[g\left(f(x)+ty_1+(1-t)y_2\right)\right]&\ge tE\left[g(f(x)+y_1)\right] + (1-t)E\left[g(f(x)+y_2)\right]\\ \end{align*}$$
Conclusion : The expression is concave in $y$.