Let $B(x):[0,1] \rightarrow [0,1]$ be piecewise linear increasing convex function with $B(0)=0$ and $B(1)=1$. (Think of the power of the neyman-persron test). Let $E(x)=-log(B(1-e^{-x}))$ a logaritmic scaled version of $B(x)$ in both dimentions.
I need $E(x)$ to be convex. I also get this from simulations.
I derive as if there are no problems (i.e. between jumps of the pieces in $B$): Define for simplicity $F(x)=B(1-e^{-x})$ so $E(x)=-log(F(x))$.
$E'(x) = -\frac{F'(x)}{F(x)}$
$E''(x) = {\left( {\frac{{F'\left( R \right)}}{{F\left( R \right)}}} \right)^2} - \frac{{F''\left( R \right)}}{{F\left( R \right)}}$
Also using direct derivative of $F'(x)$ we can get: $F''(x)=B''(1-e^{-x})e^{-2x}-F'(x)$
and substiute back:
$E''(x) = {\left( {E'\left( R \right)} \right)^2} - E'\left( R \right) - \frac{{B''\left( {1 - {e^{ - R}}} \right)}}{{F\left( R \right)}}{\left( {{e^{ - R}}} \right)^2}$
now, $B$ is piecewise linear so $B''(x)$ is zero "almost" everywhere and the rest is positive as $E$ is decresing.
How do I handle the jump points? I didn't use yet the convexity of $B$ and I think this should be in the jump point,
Any ideas?