# Showing concavity of a function defined in terms of expectancy of another function

Suppose $U : \mathbb{R} \to \mathbb{R}$ is concave, and that the random variable $\epsilon$ has zero mean. Assuming that the function $\phi : \mathbb{R} \to \mathbb{R}$, defined by $\phi(\lambda) = \mathbb{E} U(\mu + \lambda \epsilon)$ is everywhere finite-valued, prove that $\phi$ is concave.

I've tried a few different things, including Jensen's inequality, but I can't get it to work. Any help would be greatly appreciated. Thanks

EDIT: I'll show some of my working

$p \phi(\lambda_1) + (1-p)\phi(\lambda_2) = p \mathbb{E} U (\mu + \lambda_1 \epsilon) + (1-p) \mathbb{E} U(\mu + \lambda_2 \epsilon)$

$\leq pU(\mathbb{E}(\mu+\lambda_1 \epsilon)) + (1-p)(U(\mathbb{E}(\mu + \lambda_2 \epsilon)) = pU(\mu) + (1-p)U(\mu) = U(\mu)$

The inequality comes from Jensen and the fact that $U$ is concave. But I'm not sure where to go from here. Am I right in thinking $U(\mu) = \phi(0)$?

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I think I might have done it. If I use linearity of expectation, concavity of $U$ and the fact that $X \leq Y \implies \mathbb{E}[X] \leq \mathbb{E}[Y]$ – MartinP Oct 19 '11 at 14:32
If you've found the answer to your question, you can write it up as an answer and accept it so that the question doesn't remain unanswered. – joriki Oct 19 '11 at 14:35
I will do as soon as 8 hours has passed. – MartinP Oct 19 '11 at 16:16

$p \phi(\lambda_1) + (1-p)\phi(\lambda)2) = p \mathbb{E}U(\mu + \lambda_1 \epsilon) + (1-p) \mathbb{E}U(\mu + \lambda_2 \epsilon)$

$= \mathbb{E}[ pU(\mu + \lambda_1 \epsilon) + (1-p)U(\mu + \lambda_2 \epsilon)]$, by linearity of expectation.

By concavity of $U$, we have that

$pU(\mu + \lambda_1 \epsilon) + (1-p)U(\mu + \lambda_2 \epsilon) \leq U [ p(\mu + \lambda_1 \epsilon) + (1-p)(\mu + \lambda_2 \epsilon) ] = U [ \mu + p\lambda_1 \epsilon + (1-p) \lambda_2 \epsilon]$

Now $X \leq Y \implies \mathbb{E}[X] \leq \mathbb{E}[Y]$ gives us that

$\mathbb{E}[ pU(\mu + \lambda_1 \epsilon) + (1-p)U(\mu + \lambda_2 \epsilon)] \leq \mathbb{E}U [ \mu + p\lambda_1 \epsilon + (1-p) \lambda_2 \epsilon] = \phi(p\lambda_1 + (1-p)\lambda_2)$

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