7
$\begingroup$

I want to prove (or find a counterexample for) the following variant of Jensen's inequality.

Let $f$ and $g$ be convex functions (then $f(g(x))$ and $g(f(x))$ are convex functions). From the standard Jensen's inequality, we have

$$ \mathbb{E_{\sim i}}[f(g(x_i))] \geq f(g(\mathbb{E_{\sim i}}[x_i])) $$ or alternatively $$ \mathbb{E_{\sim i}}[f(g(x_i))] \geq f(\mathbb{E_{\sim i}}[g(x_i)]) $$

where in the second case we are only "extracting" the first function, but we can take the first as well since the composition of $f,g$ is convex.

I would like to know what necessary assumptions on $f,g$ are required such that the following holds:

$$ \mathbb{E_{\sim i}}[f(g(x_i))] \geq g(\mathbb{E_{\sim i}}[f(x_i)]) $$ A sufficient condition, of course is that $f\circ g \geq g \circ f$: $$ \mathbb{E_{\sim i}}[f(g(x_i))] \geq \mathbb{E_{\sim i}}[g(f(x_i))] \geq g(\mathbb{E_{\sim i}}[f(x_i)]) $$

but this is not very interesting and I was hoping for something more general.

Edit: Some additional constraints of interest to consider: $f$ is monotonic, $g$ is sublinear.

$\endgroup$

1 Answer 1

3
+50
$\begingroup$

Unless I'm misunderstand your question, your sufficient condition for the inequality $$ \mathbb{E}_{\sim\mathbb{i}}\left[f\big(g(x_i)\big)\right]\ge g\left(\mathbb{E}_{\sim\mathbb{i}}\left[f\big(x_i\big)\right]\right) $$ to hold is also necessary. I'm presuming $\ f\ $ and $\ g\ $ are defined (and therefore finite and continuous) on the whole of some Euclidean space $\ \mathbb{R}^m\ $, and you require the inequality to hold for all distributions $\ \mathbb{i}\ $. If you choose the distribution $\ \mathbb{i}\ $ to have its entire weight concentrated at the single point $\ x\ $, then \begin{align} \mathbb{E}_{\sim\mathbb{i}}\left[f\big(g(x_i)\big)\right]&=f\big(g(x)\big)\ \ \text{and}\\ g\left(\mathbb{E}_{\sim\mathbb{i}}\left[f\big(x_i\big)\right]\right)&=g\big(f(x)\big)\ , \end{align} so the inequality implies that $\ f\big(g(x)\big)\ge g\big(f(x)\big)\ $. Even if you restrict the inequality to holding only for distributions $\ \mathbb{i}\ $ with strictly positive variance, you can still get the same result by taking a sequence $\ \mathbb{i}_n\ $ of distributions with mean $\ x\ $ and variances which tend to $\ 0\ $ as $\ n\rightarrow\infty\ $.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.