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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 $v(x) = x^2$.

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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.

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

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  • $\begingroup$ 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

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