# Reverse of Jensen's Inequality

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 $$v(x) = x^2$$.

Fix $$\lambda\in[0,1]$$ and $$a.
Let $$X$$ be the random variable equal to $$a$$ with probability $$\lambda$$ and to $$b$$ with probability $$1-\lambda$$.
$$v(\lambda a + (1-\lambda) b) =v(E[X]) \le E[ v(X)]= \lambda v(a) + (1-\lambda) v(b)$$
• My answer (hidden part) uses the given assumption to conclude that $v$ is convex according to the standard definition, a definition not involving Taylor's expansion. See en.wikipedia.org/wiki/Convex_function for more details. I will keep Taylor's expansion out of picture here because in general convex functions don't have a Taylor's expansion around some points (e.g. $|x|$ around $0$). If you look around, I'm sure you'll find a connection between the general definition and derivatives, when they exist. This, however, is a topic for a separate question. – Fnacool Mar 21 at 16:19