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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.

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  • $\begingroup$ You will need to work harder to reach equality. $\endgroup$ – mathreadler May 2 '17 at 13:10
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    $\begingroup$ 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
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    $\begingroup$ 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
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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.

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