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I am trying to proof that:

Let f $\in$ $C^2(R,R)$,$f''>0$, and $E[f(X)]=f(E[X])$. I am trying to proof that X should be a constant.

I know that $f(E[X])\leq E[f(X)]$, this holds because of the Jensen inequality. Which states

f($\sum\nolimits_{k=1}^n \lambda_kx_k$)$\leq$ $\sum\nolimits_{k=1}^n \lambda_kf(x_k)$ If equality holds, this meanes that f is linear. My problem now is, that if f is two times coutinously differentiable, that would mean that $f''$ should equal 0. (i am assuming X is defined on finitely many points)

WanyM
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    what do you mean by $C^2(R,R)$ here? Jensen's inequality works only for convex functions. – Suman Chakraborty Apr 25 '21 at 17:17
  • two times coutinously differentiable, and $f''$>0 – WanyM Apr 25 '21 at 17:19
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    If you walk through the proof of Jensen's inequality and examine which steps have inequalities, I think you can arrive to the conclusion. This answer essentially does this. – angryavian Apr 25 '21 at 17:28
  • @angryavian i used the condition that f is convex i.e. $f(\lambda x+(1-\lambda)y)=\lambda f(x)+(1-\lambda)f(y)$, i got so far that this only holds if f is linear or if x=y. Is this right? – WanyM Apr 25 '21 at 18:16

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Hint: Observe that if $f''>0$ then $f'$ is strictly increasing. This then implies $f$ is strictly convex.

grand_chat
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