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I need to prove that for a given continuous non-decreasing distribution $F_X(x)$, and a concave non-decreasing distortion function $g(.)$ defined on $[0,1]$, the following holds: $$g(1-F_X(x)) \ge 1-F_X(x)$$ We know that $g(0) = 0$ and $g(1) = 1$. I was thinking to use the fact that if $g$ is concave then for $t$ in $[0,1]: g(tx + (1-t)y) \ge tg(x) + (1-t)g(y)$ , for any $x,y$ in the domain of the distribution but can't seem to find a solution. It's probably pretty obvious, but I'm stuck so any help would be truly appreciated.

Best, tb

blah_crusader
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If you're not familiar with it already, Jensen's Inequality is applicable here:

$E[g(1-\mathbf{1}_{\leq x}(X))]\leq g(1-E[\mathbf{1}_{\leq x}(X)])=g(1-F_X(x))$ [for $g$ concave].

Since $g(0)=0$ and $g(1)=1$ then

$E[g(1-\mathbf{1}_{\leq x}(X))]=E[\mathbf{1}_{\geq x}(X)]=1-F_X(x)\implies g(1-F_x)\geq 1-F_X$

$\square$...