The indirect utility function is as follows:

$$ v(m,p) = \frac{m}{p_{1}^{1/2} p_{2}^{1/4} p_{3}^{1/4}} $$

I need to prove that it is quasi-convex. I tried both definition of a quasiconvex function and the property of a convex lower contour set but couldn't get it.

This is a question in my homework so you may just want to give hint.

  • 3
    $\begingroup$ Could you please type in what you tried with the definition of quasiconvexity, so that we can see where you run into problems? $\endgroup$ – Giskard Nov 22 '19 at 19:55

If the other two methods did not work out for you, another possibility, albeit a more difficult one:

Set up the bordered Hessian:

$$\bar{H}=\begin{bmatrix}0 & \frac{\partial v}{\partial p_1} & \frac{\partial v}{\partial p_2} & \frac{\partial v}{\partial p_3} & \frac{\partial v}{\partial m}\\ \frac{\partial v}{\partial p_1} & \frac{\partial^2 v}{\partial p_1^2} & \frac{\partial^2 v}{\partial p_1\partial p_2} & \frac{\partial^2 v}{\partial p_1\partial p_3} & \frac{\partial^2 v}{\partial p_1\partial m}\\ \frac{\partial v}{\partial p_2} & \frac{\partial^2 v}{\partial p_2 \partial p_1} & \frac{\partial^2 v}{\partial p_2^2} & \frac{\partial^2 v}{\partial p_2\partial p_3} & \frac{\partial^2 v}{\partial p_2\partial m}\\ \frac{\partial v}{\partial p_3} & \frac{\partial^2 v}{\partial p_3 \partial p_1} & \frac{\partial^2 v}{\partial p_3 \partial p_2} & \frac{\partial^2 v}{\partial p_3^2} & \frac{\partial^2 v}{\partial p_3\partial m}\\ \frac{\partial v}{\partial m} & \frac{\partial^2 v}{\partial m \partial p_1} & \frac{\partial^2 v}{\partial m \partial p_2} & \frac{\partial^2 v}{\partial m \partial p_3} & \frac{\partial^2 v}{\partial m^2}\\ \end{bmatrix}$$

Define $\bar{H}_k$ as the $k_{th}$ order leading principal submatrix. We can apply the following theorem:

If the $\text{Det}\left(\bar{H}_k\right) <0$ for all $k=2,3,4,5$, then $v$ is quasiconvex in $(\textbf{p},m)$.

This is an application of the more general result:

Let $X$ be an open convex subset of $\mathbb{R}^n$ and let $f:X \rightarrow \mathbb{R}$ be a twice continuously differentiable function $\left(f\in C^2\right)$. Let $\bar{H}$ be the bordered Hessian and let $\bar{H}_k$ be the $k_{th}$ order leading principal submatrix. If the $\text{Det}\left(\bar{H}_k\right) <0$ for all $k=2,3,\dots,n+1$, then $f$ is quasiconvex on $X$.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.