# Jensen inequality first order differentiaton

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.

• You will need to work harder to reach equality. – mathreadler May 2 '17 at 13:10
• 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. – Macavity May 2 '17 at 13:13
• 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 $f(x) = f(x)$. – mlc May 2 '17 at 13:20
• @Macavity Yes i stated that f is strictly convex? – aceminer May 2 '17 at 14:06
• 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. – Macavity May 2 '17 at 16:37

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.