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 $v(x) = x^2$.


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.

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

  • $\begingroup$ But how do I use Taylor's approximation to get v(x) is a convex function. The teacher told me to continue from there and I am stuck with this part. If we assume that E[v(x)]≥v(E[X]) then I did not get how to conclude that the function v(x) is convex by using Taylor's formula? $\endgroup$ – yorukobasi Mar 16 at 11:16
  • $\begingroup$ 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. $\endgroup$ – Fnacool Mar 21 at 16:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.