1
$\begingroup$

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) $?

  • 0
    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] $2011-10-19
  • 3
    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.2011-10-19
  • 1
    I will do as soon as 8 hours has passed.2011-10-19

1 Answers 1