If $g''(x)>0$ prove that if $E(g(X)) = g(E(X))$ then $X$ is a constant random variable.

Question

I know that when $g''(x)>0$ it means that $g$ is strictly convex. By Taylor's theorem we have:

$$ g(x) = g(E(x)) + [x - E(x)]g'(E(x)) + \frac{[x-E(x)]^2}{2}g''(\epsilon_x) $$ for $\epsilon_x\in(x,E(x))$. So for $g''(x)>0$ we have $$ g(x) > g(E(x)) + [x - E(x)]g'(E(x)) $$

But I have no idea how to prove that if $E(g(x))=g(E(x))$ then $X$ is a constant random variable. Any help or hint would be appreciated.

See https://math.stackexchange.com/questions/4013864/equality-in-jensens-inequality-for-strictly-convex-functions — Andrew Shedlock, May 31 '22 at 03:33

score 2 · Answer 1 · answered May 31 '22 at 03:25

2

Hint:

Define the linear functional $l(x) = g(EX) + g'(EX)(x-EX)$, and note that $g(x) > l(x)$ for $x \neq EX$.

Now consider $E[g(X)-l(X)]$ and show that if $X \ne EX$ on a set of positive measure, then $E[g(X)-l(X)] >0$.

answered May 31 '22 at 03:25

copper.hat

178,207

1

Thanks for answering, but I don't get the second part – Mina May 31 '22 at 03:42
What is $E[l(X)]$? Remember that expectation is linear. – copper.hat May 31 '22 at 03:45

If $g''(x)>0$ prove that if $E(g(X)) = g(E(X))$ then $X$ is a constant random variable.

1 Answers1