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I have an indirect utility function - $P(a,b ;\theta)$ - where $a$ and $b$ are positive, deterministic parameters and $\theta$ is a random variable. I would like to study the properties of the Expected Indirect Utility (EIU), namely:

$$V(a,b) = \int_{\Theta} P(a,b ;\theta) \ f(\theta)d\theta$$

In particular, I would like to know under which conditions - both on the indirect utility function and on the density function of $\theta$ - the resulting $V$ is (strictly-quasi) convex in $a$ and $b$.

There are two cases that I considered.

Case One

$$P(a,b ;\theta) = \max \{\theta a, (1-\theta)b\}~, \quad \text{with}~~\theta \sim U[0,1]$$

Then:

\begin{align} V(a,b) &= \int_{0}^1 \max \{\theta a, (1-\theta)b\} \ d\theta \\ &= \int_{0}^{\frac{b}{a+b}} (1-\theta) b \ d\theta + \int_{\frac{b}{a+b}}^{1} \theta a \ d\theta = \frac{a^2 + ab +b^2}{2(a+b)} \end{align}

This function is indeed strictly convex in $a$ and $b$ (the level curves are ellipses).

Case Two

$$P(a,b ;\theta) = \max \{\theta a, (1-\theta)b\}~, \quad \text{with}~~\theta \sim \mathrm{Triangular}[0, \frac{b}{a+b}, 1] \\ \begin{aligned} V(a,b) &= \int_{0}^{1} \max \{\theta a, (1-\theta)b\} \ f(\theta) d\theta \\ & =\int_{0}^{\frac{b}{a+b}} (1-\theta)b \cdot \frac{2\theta}{\frac{b}{a+b}} \ d\theta + \int_{\frac{b}{a+b}}^{1} \theta a \cdot \frac{2(1-\theta)}{(1-\frac{b}{a+b})} \ d\theta \\ &= \frac{a+b}3 \end{aligned}$$

In this case, $V$ is linear in $a$ and $b$ and the level curves are linear as a result.

I would like to understand if there is a way to disentangle the effect of the functional form of $P(a,b ;\theta)$ from the distribution of $\theta$ in shaping the convexity of the resulting $V$ function.

Notice that in the two cases presented, the problem is even more complicated given that $\theta$ and the parameters $a$ and $b$ are related through the extremes of integration.

All I know to begin with is that $P$ is increasing in $a$ and $b$ and it is continuous in $\theta$.

Does anyone know how to tackle the problem? Any reference?

Thank you very much.

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Jancsi
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  • Convexity readily holds if $P(a,b,\theta)$ is convex with respect to $(a,b)$ for almost every $\theta$ and the distribution of $\theta$ does not depend on $a$ and $b$. A more in-depth analysis is presented in Theorem 7.46 here. The case when the distribution of $\theta$ depends on $a$ and $b$ is more complicated. Also, you cannot expected that if $P$ is quasi-convex, then $V$ is also quasi-convex. This is because quasi-convexity is not closed under addition (and expectation is like weighted sum) – madnessweasley Feb 19 '20 at 19:22
  • Thanks a lot for your comment and reference. May I ask you to elaborate further on the last two points - namely: a) if theta depends on the parameters it gets more complicated; b) quasi-convexity is not closed under addition? – Jancsi Feb 19 '20 at 21:54
  • See this link for (b). For (a), I simply meant that it seems likely that convexity of $V$ will only hold for very special cases, i.e., you should not expect a general result – madnessweasley Feb 19 '20 at 21:58
  • Thank you again. Any idea/reference about a) ? Or just by intuition? – Jancsi Feb 19 '20 at 22:08
  • Just intuition for (a) – madnessweasley Feb 19 '20 at 23:05

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