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The indirect utility function looks like this:

$v(p_x, p_y, I) = \frac{I^2}{p_x*p_y}$

where: $I$ - income, $p_i$ - price for good $i ∈ x,y$.

Find the utility function $U(x,y)$

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  • $\begingroup$ What have you tried? $\endgroup$
    – Rumi
    Oct 1, 2022 at 15:40
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    $\begingroup$ I will help you get started. See if you can figure it out. Isolate Income in terms of V. Use partial derivatives with respect to each goods price to get Hicksian (Shephards Lemma). Get the price ratio in one of the Hicksians and plug it in the other Hicksian. Solve for V. $\endgroup$
    – Rumi
    Oct 1, 2022 at 15:48
  • $\begingroup$ Also see this answer for an alternative approach: economics.stackexchange.com/a/16745/11824 $\endgroup$
    – Amit
    Oct 2, 2022 at 10:20

1 Answer 1

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Suppose utility function is $u(x, y) = kxy$ where $k>0$. Consider the following problem

$\displaystyle \max_{x\geq 0, y\geq 0} \ \ kxy \\ \text{s.t.} \ p_Xx + p_Yy \leq I $

where $p_X >0$, $p_Y > 0$, and $I\geq 0$.

Solving the above utility maximization problem we get the following solution (also known as demand):

\begin{eqnarray*} (x^d,y^d)(p_X, p_Y, I) = \left(\frac{I}{2p_X},\frac{I}{2p_Y} \right) \end{eqnarray*}

and the optimal value of utility or the indirect utility function as:

\begin{eqnarray*} V(p_X, p_Y, I) = u\left(\frac{I}{2p_X},\frac{I}{2p_Y} \right) = \frac{kI^2}{4p_Xp_Y} \end{eqnarray*}

Therefore, when $k = 4$ we get the desired indirect utility function. So, $u(x, y) = 4xy$.

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  • $\begingroup$ (-1) This is a guess and validate type of answer (ideal for cramming) that does not impart knowledge, while a general answer exists; see Rumi's comment. $\endgroup$
    – Giskard
    Oct 2, 2022 at 9:36
  • $\begingroup$ That is funny. According to you memorising Shepherd's Lemma or Roy's identity is not cramming. I think a randomly selected Economics student is more likely to know demand for Cobb Douglas than these Lemmas :) $\endgroup$
    – Amit
    Oct 2, 2022 at 9:53
  • $\begingroup$ You are right about the random student for sure! Is that a good thing though? $\endgroup$
    – Giskard
    Oct 2, 2022 at 14:28

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