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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?

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  • $\begingroup$ 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

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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 $v(p_1,p_2,M)=\frac{\max\left\{M-p_2\left(\frac{\beta p_1}{p_2}\right)^{\frac{1}{1-\beta}},0\right\}}{p_1}+\left(\min\left\{\left(\frac{\beta p_1}{p_2}\right)^{\frac{1}{1-\beta}},\frac{M}{p_2}\right\}\right)^\beta$

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

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