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Assume that $ u(\cdot) $ is concave, $ u'(x_1) > 0 $ and $ u''(x_1) < 0 $. Assume that $\theta_1, \theta_2$ are i.i.d. uniformly distributed on the square $\theta_1, \theta_2 \in [0,1]^2$.

Denote

\begin{align} U = & \int_0^{\frac{1}{u(x_1)-u(x_2)}} \int_0^1 \left( \theta_1 u(x_1)+ \theta_2 u(x_2) \right) \,d\theta_1 \,d\theta_2 \\[6pt] {}+{} & \int_{\frac{1}{u(x_1)-u(x_2)}}^1 \int_{\theta_2 - \frac{1}{u(x_1)-u(x_2)}}^1 \left(\theta_1 u(x_1)+\theta_2 u(x_2)\right) \, d\theta_1\,d\theta_2 \\[6pt] {}+{} & \int_{\frac{1}{u(x_1)-u(x_2)}}^1 \int_0^{\theta_2-\frac{1}{u(x_1)-u(x_2)}} \left(\theta_1 u(x_2)+\theta_2 u(x_1)\right) \,d\theta_1 \,d\theta_2 \end{align}

We restrict the case that $ x_2 = \overline{x} - x_1 $, and $\overline{x}$ is fixed. $u$ is increasing and concave in $x_1$, so is $u(\overline{x}-x_1)$ concave in $x_1$. And suppose $0 <\frac{1}{u(x_1) - u(x_2)} < 1$, so the line $\theta_1 = \theta_2 - \frac{1}{u(x_1) - u(x_2)}$ split the square $[0,1]^2$ into two regions.

Is $U$ still concave in $x_1$ locally? Maybe I can check the sign of the second order derivatives. But is there any properties to indicate the concavity of things like this?

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  • $\begingroup$ What do you take to be gained by writing about the "probability density function $f(\cdot)$" rather than about the "probability density function $f$"? $\qquad$ $\endgroup$ – Michael Hardy Jan 2 '18 at 2:07
  • $\begingroup$ Maybe I didn't consider that so far. What I have done is to consider the simplest case, the uniform distribution while $f(\theta_1,\theta_2) = 1 $. $\endgroup$ – Randy Jan 2 '18 at 3:00
  • $\begingroup$ Sorry maybe I missed something, are those $x_1, x_2$ just deterministic constants? and $u$ is a deterministic function? If yes then you just pull them out from the expectation? If $x_1, x_2$ are also random variables, then why it does not appear in the integral when you compute the expectation? $\endgroup$ – BGM Jan 2 '18 at 11:32
  • $\begingroup$ @BGM yes $ x_1 $ and $ x_2 $ are controls of the optimization and $u$ is assumed to be concave in $x$, and only $\theta_1$ and $\theta_2$ are random variables here. Yes I can compute the expectation and check the second order derivatives of the outcome. The problem is the integral limit also contains $u$, so I am not sure whether concavity can be preserved unless directly computing the integral. $\endgroup$ – Randy Jan 2 '18 at 22:50
  • $\begingroup$ I am quite confused by the current formulation of the question; the expression $\mathbb{E}[\theta_1 u(x_1) + \theta_2 u(x_2)$ does not have an x in it; what concavity are you looking for? Also, a stylistic remark (not sure if this is the norm but this is what I am used to): when you write your expectation, you shouldn't write your domain of integration as a conditional; $E[X \mid Y]$ is not a constant but rather a random variable... I know there is a convention to write an event in the conditioning, in which case you should write $\theta_1, \theta_2 \in \Omega$ to indicate it is an event. $\endgroup$ – E-A Jan 2 '18 at 23:15

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