# Analog of Jensen Inequality

Jensen's inequality states that if $$\phi$$ is a convex function, then we have that: $$\phi(\mathbb{E}[X])\leq \mathbb{E}[\phi(X)]$$ Where in this case X is a discrete random variable. Is there any bound to the other side? I mean, $$\mathbb{E}[\phi(X)] \leq ?$$

In particular, I've been trying to bound $$E[-\log(X)]$$. I also have the variance of $$X$$, let's just call the mean and the variance $$\mu$$ and $$\sigma$$.

• First I think you mean $$\mathbb{E}[\phi(X)] \leq ?$$ Second, do you need an upper bound for $\mathbb{E}[\phi(X)]$ only based on $\mathbb{E}(X)$?
– Amir
Commented Apr 29 at 19:06
• My bad, edited the question with the updated information Commented Apr 29 at 23:29

When $$\phi$$ is twice differentiable, this paper shows that
$$\sigma^2\inf \frac{\phi''(x)}{2} \leq E\left[\phi \left(X\right)\right]-\phi\left(E[X]\right)\le \sigma^2\sup \frac{\phi''(x)}{2},$$
which yields the Jensen inequality if $$\phi''(x)\ge 0$$. Moreover, it gives the upper bound $$\phi\left(\mu\right) + \lambda \sigma^2$$ for $$E\left[\phi \left(X\right)\right]$$ when $$\lambda=\sup \frac{\phi''(x)}{2}$$ is finite.
When $$\phi''(x)$$ is not finite over the support of $$X$$, $$\lambda$$ becomes infinite, and so does the upper bound, which is the case for $$\phi(x)=-\log x$$. However, if the support of $$X$$ is $$(l, u)$$ for some $$l>0$$, you can use the above bound with $$\lambda=\frac{1}{2l^2}$$.
You can also use the approximation given here for $$E\left(\log X\right)$$.