<p>Here is the utility maximisation problem:
<span class="math-container">\begin{eqnarray*}\max_{(x_1,x_2)\in\mathbb{R}^2_+} &amp; x_1+x_2^\beta \\ \text{s.t. } &amp; p_1x_1+p_2x_2\leq M\end{eqnarray*}</span>
where <span class="math-container">$p_1&gt;0, p_2&gt;0, M\geq 0, \beta\in (0,1)$</span> are given.
Solving this problem, we get demand as:
<span class="math-container">\begin{eqnarray*}(x_1^d,x_2^d)(p_1,p_2,M)=\begin{cases} \left(0,\frac{M}{p_2}\right) &amp; \text{if } M&lt;p_2\left(\frac{\beta p_1}{p_2}\right)^{\frac{1}{1-\beta}} \\ \left(\frac{M-p_2\left(\frac{\beta p_1}{p_2}\right)^{\frac{1}{1-\beta}}}{p_1},\left(\frac{\beta p_1}{p_2}\right)^{\frac{1}{1-\beta}}\right) &amp; \text{otherwise}\end{cases}\end{eqnarray*}</span>
Indirect Utility function is
<span class="math-container">$v(p_1,p_2,M)=\frac{\max\left\{M-p_2\left(\frac{\beta p_1}{p_2}\right)^{\frac{1}{1-\beta}},0\right\}}{p_1}+\left(\min\left\{\left(\frac{\beta p_1}{p_2}\right)^{\frac{1}{1-\beta}},\frac{M}{p_2}\right\}\right)^\beta$</span></p>

CC BY-SA 4.0
2024-05-18T10:02:59.210Z
Thanks for your response but the indirect utility function is still function of $p_2$
giudale
CC BY-SA 4.0
2024-05-18T10:17:47.227Z
It is what it is.
Amit
CC BY-SA 4.0
2024-05-18T04:48:00.777Z
2024-05-18T04:48:00.777Z
Amit
<p>Hey guys I need your help to solve the following problem. Here's my attempt.</p>
<p>Given the function: <span class="math-container">$$u(x_1,x_2)=x_1+x_2^\beta, \ \ \beta \in (0,1).$$</span></p>
<ul>
<li>Write down the UMP and solve for the Walrasian Demands of <span class="math-container">$x_1, x_2$</span>.</li>
<li>Is good 1 a Normal good? And good 2?</li>
<li>Compute the indirect utility function for this problem and show that it can be written as <span class="math-container">$v(w,\mathbf{p})=\tilde{v}(w,p_1)$</span>.</li>
</ul>
<p>My attempt:
<span class="math-container">$$\mathcal{L}(\mathbf{x},\lambda)=x_1+x_2^\beta-\lambda(p_1x_1+p_2x_2-w).$$</span>
<span class="math-container">$$\frac{\partial\mathcal{L}}{\partial x_1}=1-\lambda p_1 = 0.$$</span>
<span class="math-container">$$\frac{\partial\mathcal{L}}{\partial x_2}=\beta x_2^{(\beta - 1)}-\lambda p_2 = 0$$</span>
<span class="math-container">$$\frac{\partial\mathcal{L}}{\partial \lambda}=p_1x_1+p_2x_2=w$$</span>
By solving the system I end up with:
<span class="math-container">$$x_1^*(\mathbf{p},w)=\frac{w}{p_1}-(\frac{p_2}{p_1})^{\frac{\beta}{\beta -1}}\cdot \frac{1}{\beta^{\frac{1}{\beta - 1}}} \\ x_2^*(\mathbf{p},w)=(\frac{p_2}{\beta p_1})^{\frac{1}{\beta -1}}$$</span>
A good is defined normal if <span class="math-container">$\frac{\partial x_i^*}{\partial w} &gt; 0.$</span>
<span class="math-container">$$\frac{\partial x_1^*}{\partial w} = \frac{1}{p_1} &gt; 0.$$</span> Good 1 is a normal good.
Good 2 is not a function of <span class="math-container">$w$</span>, implying that <span class="math-container">$x_2$</span> is a neutral good.</p>
<p>Indirect utility function:
<span class="math-container">$$v(\mathbf{p},w) \equiv u(x_1^*,x_2^*) $$</span>
<span class="math-container">$$v(\mathbf{p},w)= \frac{w}{p_1}-(\frac{p_2}{p_1})^{\frac{\beta}{\beta -1}}\cdot \frac{1}{\beta^{\frac{1}{\beta - 1}}}+((\frac{p_2}{\beta p_1})^{\frac{1}{\beta -1}})^\beta .$$</span>
The indirect utility function is function of both <span class="math-container">$p_1,p_2$</span>, how can I show that is function of <span class="math-container">$p_1$</span> only?</p>

CC BY-SA 4.0
2024-05-26T14:19:28.500Z
Why would it only be a function of p1? P2 affects the indirect utility as well...
ChinG
CC BY-SA 4.0
2024-05-31T00:19:10.590Z
It's what the exercise asked. It may be an error of the professor.
giudale
<microeconomics>
Quasi-linear microeconomics problem
giudale
# Quasi-linear microeconomics problem

(Asked by: username_0 on 2024-05-17)

Hey guys I need your help to solve the following problem. Here's my attempt.

Given the function: $$u(x\_1,x\_2)=x\_1+x\_2^\beta, \ \ \beta \in (0,1).$$

* Write down the UMP and solve for the Walrasian Demands of $x\_1, x\_2$.
* Is good 1 a Normal good? And good 2?
* Compute the indirect utility function for this problem and show that it can be written as $v(w,\mathbf{p})=\tilde{v}(w,p\_1)$.

My attempt:
$$\mathcal{L}(\mathbf{x},\lambda)=x\_1+x\_2^\beta-\lambda(p\_1x\_1+p\_2x\_2-w).$$
$$\frac{\partial\mathcal{L}}{\partial x\_1}=1-\lambda p\_1 = 0.$$
$$\frac{\partial\mathcal{L}}{\partial x\_2}=\beta x\_2^{(\beta - 1)}-\lambda p\_2 = 0$$
$$\frac{\partial\mathcal{L}}{\partial \lambda}=p\_1x\_1+p\_2x\_2=w$$
By solving the system I end up with:
$$x\_1^\*(\mathbf{p},w)=\frac{w}{p\_1}-(\frac{p\_2}{p\_1})^{\frac{\beta}{\beta -1}}\cdot \frac{1}{\beta^{\frac{1}{\beta - 1}}} \\ x\_2^\*(\mathbf{p},w)=(\frac{p\_2}{\beta p\_1})^{\frac{1}{\beta -1}}$$
A good is defined normal if $\frac{\partial x\_i^\*}{\partial w} > 0.$
$$\frac{\partial x\_1^\*}{\partial w} = \frac{1}{p\_1} > 0.$$ Good 1 is a normal good.
Good 2 is not a function of $w$, implying that $x\_2$ is a neutral good.

Indirect utility function:
$$v(\mathbf{p},w) \equiv u(x\_1^\*,x\_2^\*) $$
$$v(\mathbf{p},w)= \frac{w}{p\_1}-(\frac{p\_2}{p\_1})^{\frac{\beta}{\beta -1}}\cdot \frac{1}{\beta^{\frac{1}{\beta - 1}}}+((\frac{p\_2}{\beta p\_1})^{\frac{1}{\beta -1}})^\beta .$$
The indirect utility function is function of both $p\_1,p\_2$, how can I show that is function of $p\_1$ only?

**Tags:** microeconomics
> **Comments:**
> **username_1**: Why would it only be a function of p1? P2 affects the indirect utility as well...  
> **username_0**: It's what the exercise asked. It may be an error of the professor.  

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## 1 answers

### Answer by username_2 on 2024-05-18. Score: 5

Here is the utility maximisation problem:
\begin{eqnarray\*}\max\_{(x\_1,x\_2)\in\mathbb{R}^2\_+} & x\_1+x\_2^\beta \\ \text{s.t. } & p\_1x\_1+p\_2x\_2\leq M\end{eqnarray\*}
where $p\_1>0, p\_2>0, M\geq 0, \beta\in (0,1)$ are given.
Solving this problem, we get demand as:
\begin{eqnarray\*}(x\_1^d,x\_2^d)(p\_1,p\_2,M)=\begin{cases} \left(0,\frac{M}{p\_2}\right) & \text{if } M
Indirect Utility function is
$v(p\_1,p\_2,M)=\frac{\max\left\{M-p\_2\left(\frac{\beta p\_1}{p\_2}\right)^{\frac{1}{1-\beta}},0\right\}}{p\_1}+\left(\min\left\{\left(\frac{\beta p\_1}{p\_2}\right)^{\frac{1}{1-\beta}},\frac{M}{p\_2}\right\}\right)^\beta$

> **Comments:**
> **username_0**: Thanks for your response but the indirect utility function is still function of $p_2$  
> **username_2**: It is what it is.  

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