One form of Jensen's inequality is
If $X$ is a random variable and $g$ is a convex function, then $\mathbb{E}(g(X))\geq g(\mathbb{E}(X))$.
Just out of curiosity, when do we have equality? If and only if $g$ is constant?
Asked
Active
Viewed 2.2k times
39
RobPratt
- 50,938
eeeeeeeeee
- 2,311
-
12Also if $g$ is affine. – hardmath Jan 05 '14 at 19:40
-
10And/or $X$ constant almost surely. – Did Jan 05 '14 at 19:45
-
6I think an if and only if condition would be: there is equality iff there exists an interval A such that g is affine on A, and X belongs to A almost surely. – D. Thomine Nov 20 '14 at 22:01
-
2https://math.stackexchange.com/questions/1160095/convexity-and-equality-in-jensen-inequality – StubbornAtom May 15 '20 at 14:39
-
An example of X being constant almost surely is KL(p||q)=0 iff p=q almost surely. Here g is -ln() so can't be affine. The constant must be one (almost surely) due to normalization. – user5280911 Mar 15 '22 at 07:18
2 Answers
16
Just for the sake of having an "answered" question (thanks to @hardmath and @Did), Jensen's inequality is equality when $g$ is affine or $X$ is constant almost surely.
Jacob A
- 579
eeeeeeeeee
- 2,311
1
I have come across this old question and I have noticed that while it states the result correctly it does not prove it. Let me prove it here for the benefit of the future readers.
If there is no set $A$ such that $\mathbb{P}(X\in A)=1$ and $g$ is affine on $A$, it must be the case that for all fixed $x\in \mathbb{R},b\in\partial g(x)$ (the set of subgradients of $g$ at $x$) $\mathbb{P}(g(X)>g(x)+b(X-x))>0$. Now, taking $x=\mathbb{E}(X)$ and any $b\in\partial g(x)$, it follows that $g(X)>g(\mathbb{E}(X))+b(X-\mathbb{E}(X))$ on a set of positive measure, so hence taking the expectations the inequality remains strict (since it holds weakly everywhere by convexity). Then we get $\mathbb{E}(g(X))>g(\mathbb{E}(X))+b(\mathbb{E}(X)-\mathbb{E}(X))=g(\mathbb{E}(X))$ and the result is proved.
No-one
- 776