Sufficient condition on $f$ to make $g(x) = \int_{-\infty}^{\infty} \max(y-x, 0)f(y)dy$ strictly convex?

Question

Let $Y$ be a random variable and $$g(x) = \mathrm{E}[\text{max}(Y-x, 0)] = \int_{-\infty}^{\infty} \text{max}(y-x, 0)f(y)dy$$ I know that $g$ is convex since for any fixed $y$, $\text{max}(y-x, 0)$ is convex and $f(y) \geq 0$.

My question is that what is the most general condition on $f$ (if any), such that $g$ becomes strictly convex?

Thanks!

marty cohen · Accepted Answer · 2016-11-14T19:38:39.797

2

Here's my take.

$\begin{array}\\ g(x) &=\int_{-\infty}^{\infty} \text{max}(y-x, 0)f(y)dy\\ &=\int_{x}^{\infty} (y-x)f(y)dy\\ &=\int_{x}^{\infty} yf(y)dy -x\int_{x}^{\infty} f(y)dy\\ \end{array} $

Differentiating, $g'(x) =-xf(x) -(x(-f(x)+\int_{x}^{\infty} f(y)dy) =-\int_{x}^{\infty} f(y)dy $.

Differentiating again, $g''(x) =-(-f(x)) =f(x) $.

So, if $f(x) > 0$, then $g(x)$ is strictly convex.

edited Nov 14 '16 at 19:38

answered Nov 14 '16 at 07:41

marty cohen

107,799

This implies that $g''(x) = -f(x) \leq 0$. It seems that this result must not be true, since it proves that $g$ is not even convex. – Mehdi Jafarnia Jahromi Nov 14 '16 at 19:27
Or maybe my primary justification that $g$ is convex is not true. – Mehdi Jafarnia Jahromi Nov 14 '16 at 19:34
It turns out that $g''(x) = f(x)$, as my corrected answer shows. – marty cohen Nov 14 '16 at 19:39

Sufficient condition on $f$ to make $g(x) = \int_{-\infty}^{\infty} \max(y-x, 0)f(y)dy$ strictly convex?

1 Answers1