4
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ Why would it only be a function of p1? P2 affects the indirect utility as well... $\endgroup$
    – ChinG
    Commented May 26 at 14:19
  • $\begingroup$ It's what the exercise asked. It may be an error of the professor. $\endgroup$
    – giudale
    Commented May 31 at 0:19

1 Answer 1

4
$\begingroup$

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<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) & \text{otherwise}\end{cases}\end{eqnarray*} Indirect Utility function is $v(p_1,p_2,M)=\frac{\max\left\{M-p_2\left(\frac{\beta p_1}{p_2}\right)^{\frac{1}{1-\beta}},0\right\}}{p_1}+\left(\min\left\{\left(\frac{\beta p_1}{p_2}\right)^{\frac{1}{1-\beta}},\frac{M}{p_2}\right\}\right)^\beta$

$\endgroup$
2
  • $\begingroup$ Thanks for your response but the indirect utility function is still function of $p_2$ $\endgroup$
    – giudale
    Commented May 18 at 10:02
  • 4
    $\begingroup$ It is what it is. $\endgroup$
    – Amit
    Commented May 18 at 10:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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