When is Jensen's Inequality strict?

Question

For a homework problem, I have to prove that for a random sample $X_1, \ldots, X_n$, drawn from a population with finite variance $\sigma^2$, with sample mean $\bar{x}$ and sample variance $s^2$, that $E(S) \leq \sigma$ and if $\sigma^2 > 0$, $E(S) < \sigma$.

The first part of the statement is a straightforward application of Jensen's Inequality: since $g(x) = x^2$ is convex everywhere, $E(S^2) = \sigma^2 \geq (E(S))^2 \implies \sigma \geq E(S)$.

However, I am not sure how to prove the second part of the statement. I got that since $\sigma^2 > 0$, then $0 \geq E(S)$, but I'm not sure if that helps. Are there any conditions that I just don't know about for Jensen's Inequality to be strict?

Answer 1

Jensen's inequality is strict if the function is strictly convex and the distribution is non-degenerate.

If the function is twice differentiable there's an explicit lower bound on the difference described here