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

<p>Set up the bordered Hessian:</p>

<p><span class="math-container">$$\bar{H}=\begin{bmatrix}0 &amp; \frac{\partial v}{\partial p_1} &amp; \frac{\partial v}{\partial p_2} &amp; \frac{\partial v}{\partial p_3} &amp; \frac{\partial v}{\partial m}\\ 
\frac{\partial v}{\partial p_1} &amp; \frac{\partial^2 v}{\partial p_1^2} &amp; \frac{\partial^2 v}{\partial p_1\partial p_2} &amp; \frac{\partial^2 v}{\partial p_1\partial p_3} &amp; \frac{\partial^2 v}{\partial p_1\partial m}\\
\frac{\partial v}{\partial p_2} &amp; \frac{\partial^2 v}{\partial p_2 \partial p_1} &amp; \frac{\partial^2 v}{\partial p_2^2} &amp; \frac{\partial^2 v}{\partial p_2\partial p_3} &amp; \frac{\partial^2 v}{\partial p_2\partial m}\\
\frac{\partial v}{\partial p_3} &amp; \frac{\partial^2 v}{\partial p_3 \partial p_1} &amp; \frac{\partial^2 v}{\partial p_3 \partial p_2} &amp; \frac{\partial^2 v}{\partial p_3^2} &amp; \frac{\partial^2 v}{\partial p_3\partial m}\\
\frac{\partial v}{\partial m} &amp; \frac{\partial^2 v}{\partial m \partial p_1} &amp; \frac{\partial^2 v}{\partial m \partial p_2} &amp; \frac{\partial^2 v}{\partial m \partial p_3} &amp; \frac{\partial^2 v}{\partial m^2}\\
\end{bmatrix}$$</span></p>

<p>Define <span class="math-container">$\bar{H}_k$</span> as the <span class="math-container">$k_{th}$</span> order leading principal submatrix. We can apply the following theorem:</p>

<p>If the <span class="math-container">$\text{Det}\left(\bar{H}_k\right) &lt;0$</span> for all <span class="math-container">$k=2,3,4,5$</span>, then <span class="math-container">$v$</span> is quasiconvex in <span class="math-container">$(\textbf{p},m)$</span>. </p>

<p>This is an application of the more general result:</p>

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

CC BY-SA 4.0
2019-11-23T18:47:13.627Z
2019-11-23T18:47:13.627Z
lunar_props
<p>The indirect utility function is as follows:</p>

<p><span class="math-container">$$ v(m,p) = \frac{m}{p_{1}^{1/2} p_{2}^{1/4} p_{3}^{1/4}} $$</span></p>

<p>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.</p>

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

CC BY-SA 4.0
2019-11-22T19:55:30.003Z
Could you please type in what you tried with the definition of quasiconvexity, so that we can see where you run into problems?
Giskard
<microeconomics><utility>
Prove this indirect utility function is quasi-convex
dunguyen
# Prove this indirect utility function is quasi-convex

(Asked by: username_0 on 2019-11-22)

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.

**Tags:** microeconomics, utility
> **Comments:**
> **username_1**: Could you please type in what you tried with the definition of quasiconvexity, so that we can see where you run into problems?  

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## 1 answers

### Answer by username_2 on 2019-11-23. Score: 1 | Accepted answer

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 $X$.


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