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I want to proof that the function $\phi:\mathbb R\rightarrow\mathbb R: \phi(\lambda)=\mathbb{E}U(w+\lambda X)$, which is everywhere finite-valued, is concave.

$U:\mathbb R\rightarrow\mathbb R$ is concave and the random variable $X$ has zero mean.

My first idea was to use Taylor expansion to show that the second derivative is smaller than $0$. When I use Taylor expansion I get approximately $U(w)+U'(w)\mathbb{E}(\lambda X)+U''(w)\mathbb{E}\lambda^2X^2=U(w)+\lambda^2U''(w)\mathbb E{X^2}$ and I now that $U''(w)\le0$ because $U$ is concave, but how can I continue?

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Hint: $U$ is concave iff $$ U(\eta w_1+(1-\eta) w_2) \ge \eta U(w_1)+(1-\eta)U(w_2),\;\;\forall w_1, w_2 \mbox{ and } \eta \in [0,1] $$ Now $$ EU(\eta w_1+(1-\eta) w_2 + \lambda X)=EU\big(\eta (w_1+ \lambda X)+(1-\eta) (w_2 + \lambda X)\big)\ge\ldots $$

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  • $\begingroup$ I am very sorry, I defined the function in a wrong way, $\phi$ depends on $\lambda$, not on $w$ $\endgroup$ – Alexander Oct 12 '13 at 19:00
  • $\begingroup$ Don't problem. Same idea. $EU(w+(1-\eta)\lambda_1X+\eta\lambda_2X)...$ $\endgroup$ – Pocho la pantera Oct 12 '13 at 19:19
  • $\begingroup$ Ok I got the first. Is there a simple argument why $\phi$ is decreasing in $[0,\infty)$ and inccreasing in $(-\infty,0]$? $\endgroup$ – Alexander Oct 12 '13 at 19:37

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