Can anyone explain why this is true.

If a function f is strictly convex, then $$f(E(x)) = E(f(x))$$ which means $$x = E(x)$$

I do not seem to be able to prove it.

  • $\begingroup$ You will need to work harder to reach equality. $\endgroup$ – mathreadler May 2 '17 at 13:10
  • 1
    $\begingroup$ The existence of convex functions can't obviouslyforce random variables to become constant. I suspect what you mean is if $f$ is strictly convex and your first equality holds then a.s. it implies the second or something similar. Please recheck the wording. $\endgroup$ – Macavity May 2 '17 at 13:13
  • $\begingroup$ As stated, your claim does not hold. Jensen's inequality states only that $f(E(X)) \le E(f(X))$. Note that $X$ is a r.v.. If $X$ is degenerate, you get the trivial identity $f(x) = f(x)$. $\endgroup$ – mlc May 2 '17 at 13:20
  • $\begingroup$ @Macavity Yes i stated that f is strictly convex? $\endgroup$ – aceminer May 2 '17 at 14:06
  • 1
    $\begingroup$ Possibly this is a duplicate of math.stackexchange.com/questions/513951/… Though you may need to read up more on convexity and probability before tackling it. $\endgroup$ – Macavity May 2 '17 at 16:37

As stated, your claim is false.

Jensen's inequality states that $f(E(X)) \le E(f(X))$ for any (integrable) r.v. $X$.

If $X$ is degenerate (that is, $P(X=x) =1$), then $E(X) = x$ and the trivial identity $$f(E(X)) = f(x) = E(f(x))$$ holds.


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.