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I'm trying to follow the in-text examples from Mas-Colell. I can confirm I have the correct first order-conditions and hence the Marshallian demand functions for Example 3.D.1: $u(x_1,x_2) = x_1^\alpha x_2^{1-\alpha}$ subject to the budget constraint, $p_1x_1 + p_2x_2 =w$. By going through the example and taking Lagrange, I found the Marshallian demands to be $x_1(p,w)=\frac{\alpha w}{p_1}$ and $x_2(p,w)=\frac{(1-\alpha) w}{p_2}$. Fine.

The part I'm having difficulty with is finding the Hicksian demand, we know from 3.E.4 that $x(p,w)=h(p,v(p,w))$, where $v(p,w)$ is indirect utility, it's made clear that we can find indirect utility by substituting our Marshallian demand functions into $u(p,w)$.

I attempt to find v(p,w) by substituting the Marshallian demand for $x_2$ into the $u(x_1,x_2)$ : V(p,w)= $(\frac{\alpha w}{p_1})^\alpha [ \frac{(1-\alpha)m}{p_2} ] ^{1-\alpha}$

To quote the notes given to me by my tutor, 'Marshallian is Hicksian if we substitute utility by indirect tility function:

$x^*(p,w)=h^*(p,V(p,w))$

I've reached a bit of a dead end here, I have the indirect utility function, but don't know how to get the Hicksian demand.

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  • $\begingroup$ When you have the value function $V(p,w) = k \cdot w$ - which in this case is simply a constant $k(p,\alpha)$ multiplied by income - invert to solve for income $w = V(p,I) /k$ then insert this expression in Marshall demand to get $h(p,V) = x(p,V/k)$. $\endgroup$ – Jesper Hybel Dec 18 '20 at 15:16
  • $\begingroup$ TYPO: the »I« as argument in V should be w. $\endgroup$ – Jesper Hybel Dec 18 '20 at 16:23
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    $\begingroup$ You could consider accepting the answer or entering a dialogue about what you think is missing. $\endgroup$ – Jesper Hybel Dec 19 '20 at 14:02
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You have found the value function $V(p,I)$ which in your case is given as

$$V(p,I) = \alpha_1^{\alpha_1}\alpha_2^{\alpha_2}\frac{I}{p_1^{\alpha_1}p_2^{\alpha_2}},$$

where $\alpha_2 = 1- \alpha_1$ and $I$ is income.

The Hicks demand is a function of the utility level denoted $u$. An agent facing prices $p$ and income $I$ gets utility $V(p,I) = u$. Therefore you can set

$$u = \alpha_1^{\alpha_1}\alpha_2^{\alpha_2}\frac{I}{p_1^{\alpha_1}p_2^{\alpha_2}},$$

and invert to solve for $I$ in order to get

$$\frac{p_1^{\alpha_1}p_2^{\alpha_2}}{\alpha_1^{\alpha_1}\alpha_2^{\alpha_2}}u = I,$$ which is the expenditure function. You can insert this in your Marshall demand

$$\frac{\alpha_j I}{p_j} = \left(\frac{\alpha_j}{p_j}\right)\frac{p_1^{\alpha_1}p_2^{\alpha_2}} {\alpha_1^{\alpha_1}\alpha_2^{\alpha_2}}u = h^\star(p,u).$$

Or in the case of $J$ goods: You have CD preferences with well known Marshall demand

$$(1)\ \ x^\star_j(p,I) = \frac{\alpha_j I}{p_j} \Leftrightarrow \frac{x_j}{\alpha_j} = \frac{I}{p_j},$$

and the utility function is

$$(2) \ \ U(x) = \prod_j x_j^{\alpha_j} = A \prod_j \left(\frac{x_j}{\alpha_j} \right)^{\alpha_j},$$ where $A := \prod_j \alpha_j^{\alpha_j}$ the constant $A$ serves the purpose of making the algebra somewhat easier. Inserting from (1) in (2) gives you value function

$$V(p,I) = A \prod_j \left(\frac{I}{p_j} \right)^{\alpha_j} = \frac{AI}{\bar p},$$ where $\bar p:= \prod_j p_j^{\alpha_j}$ and $I$ is constant across goods hence not bound by index so when $\sum_j \alpha_j = 1$ it $\prod_j I^{\alpha_j} = I$. From here you isolate income to get

$$\frac{\bar p}{A}V(p,I) = I$$

you then insert this in Marshall demand to get

$$x^\star(p,I) = x^\star\left(p,\frac{\bar p V(p,I)}{A}\right) = h^\star(p,V(p,I)).$$

Knowing the Marshall demand $\alpha_j I/p_j$ insert $\bar p V(p,I)/A$ with $V(p,I)$to get

$$h^\star(p,V(p,I)) = \frac{\alpha_j}{p_j} \frac{\bar p V(p,I)}{A} ,$$

let $V(p,I) = u$ to get

$$h^\star(p,u) = \frac{\alpha_j}{p_j} \frac{\bar p u}{A} = \frac{\alpha_j}{p_j} E(p,u),$$

where $E(p,u)= \frac{\bar p u}{A}$ is expenditure function (this follows from the fact that we arrived at $\bar p u/A$ as an expression for income by inverting the value function).

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  • $\begingroup$ @CorporateNationalism you should accept this answer so that other people looking for this same answer can more easily find it, it also helps get the site off of beta :) $\endgroup$ – Brennan Dec 27 '20 at 20:58
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I prefer another approach. I use duality to find the cost function and then use Shepard's Lemma to derive the Hicksian demands. I will demonstrate below for the simple 2 good case, rather than generally, for simplicity.

Begin with the indirect utility function: $$V(p,I) = \alpha_1^{\alpha_1}\alpha_2^{\alpha_2}\frac{I}{p_1^{\alpha_1}p_2^{\alpha_2}} = \left(\frac{\alpha_1}{p_1}\right)^{\alpha_1}\left(\frac{\alpha_2}{p_2}\right)^{\alpha_2}I$$ and we use duality to impose that $V(p,I)=u$ and $C(p,u) = I$ at the optimal value of these functions (the solution to the utility maximization problem is the same as that of the cost minimization problem, you can explore this concept on your own time).

Then we can solve for $C(p,u)$ by substituting the above relationships in: $$u=\left(\frac{\alpha_1}{p_1}\right)^{\alpha_1}\left(\frac{\alpha_2}{p_2}\right)^{\alpha_2}C(p,u)$$ then let $A:=\left(\frac{\alpha_1}{p_1}\right)^{\alpha_1}\left(\frac{\alpha_2}{p_2}\right)^{\alpha_2}$ for simplicity, the above can be represented as $u=AC(p,u)$ which implies that the cost function is $C(p,u) = A^{-1}u$, therefore $$C(p,u) = \left(\frac{p_1}{\alpha_1}\right)^{\alpha_1}\left(\frac{p_2}{\alpha_2}\right)^{\alpha_2}u$$

Now that we have the form of the cost function, we can invoke Shepard's Lemma which states that the derivative of the cost function with respect to the price of a good gives us the Hicksian demand for said good. Think of it in terms of the budget constraint: $p_{1}x_1 + p_{2}x_2 \leq I$, where $I$ would be our expenditure should this constraint be binding. If we have optimal values for $x_i$, then taking a derivative of the constraint/expenditure with respect to, say, $p_1$ should give us $x_{1}^*$ right?

Shepard's Lemma can be written succinctly as $$\frac{\partial c(p,u)}{\partial p_i} = h_{i}(p,u)$$

So, for the Hicksian demand of good 1, for example, it would be $$h_{1}(p,u) = \frac{\partial C(p,u)}{\partial p_{1}} = \frac{1}{\alpha_1}\alpha_{1}\left(\frac{p_1}{\alpha_1}\right)^{\alpha_1-1}\left(\frac{p_2}{\alpha_2}\right)^{\alpha_2}u = \left(\frac{p_1}{\alpha_1}\right)^{\alpha_1-1}\left(\frac{p_2}{\alpha_2}\right)^{\alpha_2}u$$

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