[Math] Reverse of Jensen’s Inequality

convex-analysisjensen-inequalityproof-verification

If $E[v(x)] \geq v(E[X])$ for every random variable $X$, then $v$ is convex. I know that a function $v(x)$ is convex iff for every $x_0$ a line, we have $l_0(x) = a_0x + b_0$ exists such that $l_0(x_0) = v(x_0)$ and moreover $v(x) \geq l_0(x)$ for all $x$.

How can I prove the reverse of Jensen's Inequality for a convex function $v(x)$? In the question, if $v(x)$ is convex, then we have $E[v(x)] \geq v(E[X])$ . However, I did not understand if we have $E[v(x)] \geq v(E[X])$ then $v(x)$ is convex, considering especially the case $v(x) = x^2$.

Best Answer

Let's just use the definition of convexity with corresponding RVs.

Fix $\lambda\in[0,1]$ and $a<b\in{\mathbb R}$.

Let $X$ be the random variable equal to $a$ with probability $\lambda$ and to $b$ with probability $1-\lambda$.

Try to continue from here.

$$ v(\lambda a + (1-\lambda) b) =v(E[X]) \le E[ v(X)]= \lambda v(a) + (1-\lambda) v(b) $$