I have an indirect utility function - $P(a,b ;\theta)=max\{\theta a, (1-\theta)b\}$ - where $a$ and $b$ are positive, deterministic parameters and $\theta$ is a random variable. I would like to study the property of symmetry 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 symmetric to the line $a=b$ in the $(a,b)$ space.
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 symmetric wrt the $a=b$ line (the level curves are symmetric 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 symmetric to the 45° line again.
I would like to understand whether there is a mathematical reason behind the symmetry in these cases.
Intuitively, for the mean value of $\theta$, we have $\theta = 1-\theta$, so that $max\{\theta a, (1-\theta)b\}$ is symmetric to the $a=b$ line. But I would like to know whether there is a deeper reason behind this property.
Also, does convexity play any role in this?
Does anyone know how to tackle the problem? Any reference?
Thank you very much.