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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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