# Does this variation of Jensen's inequality hold?

The original Jensen's inequality in probability theory is generally stated in the following form: if $X$ is a random variable and $f$ is a convex function, then $f \left(\mathbb{E}[X]\right) \leq \mathbb{E}\left[f(X)\right]$ holds, where $\mathbb{E}$ means the expectation of $X$.

Then I have another function $g(\cdot)$ and I do not know if it is convex or not. Does $f \left(\mathbb{E}[g(X)]\right) \leq \mathbb{E}\left[f(g(X))\right]$ hold? Why?

• Please, check if your Jensen's inequality is correct... – Olivier Oloa Dec 27 '14 at 14:55

It follows from an application of Jensen's inequality to the random variable $g(X)$ (provided that $g$ is Borel measurable) instead of $X$.
• Thanks for your kind help. But in my question $f \left(\mathbb{E}[g(X)]\right) \leq \mathbb{E}\left[f(g(X))\right]$, the expectation is on $X$ not $g(X)$, so it seems this is not a direct application of Jensen's inequality to the random variable $g(X)$. Can you help me? – olivia Jan 18 '15 at 12:28
• What do you mean by "the expectation is on $X$"? – Davide Giraudo Jan 18 '15 at 13:38
• @ Davide Giraudo $f \left(\mathbb{E}[X]\right) \leq \mathbb{E}\left[f(X)\right]$ holds, where f is a convex function and $\mathbb{E}$ means the expectation of $X$. So if you directly use Jensen's inequality and regard $g(X)$ as a random variable, then accordingly $\mathbb{E}$ will mean the expectation of $g(X)$. However, in the final equation of my question, the inequality holds with $\mathbb{E}$ denoting the expectation of $X$ not $g(X)$. – olivia Jan 18 '15 at 13:53