1
$\begingroup$

Maximize $x_1^{\frac{1}{2}} + x_2^{\frac{1}{2}}$ subejct to the budget constraint $p_1x_1+p_2x_2=m$

Setting up the Lagrange and finding the first-order conditions:

$L(x_1, x_2, \lambda)=x_1^{\frac{1}{2}} + x_2^{\frac{1}{2}}+\lambda(p_1x_1+p_2x_2-m)$

$\frac{\partial L}{\partial x_1}=\frac{1}{2}x_1^\frac{-1}{2}+\lambda p_1=0$

$\frac{\partial L}{\partial x_2}=\frac{1}{2}x_2^\frac{-1}{2}+\lambda p_2=0$

Equating these,

$2p_2x_1^\frac{-1}{2}=2p_1x_2^\frac{-1}{2}$

Solving for $x_1$ and $x_2$, we get

$x_1=\frac{p_2^2x_2}{p_1^2}$

$x_2=\frac{p_1^2x_1}{p_2^2}$

Substituting these into the budget constraint and solving to find the Hicksian demand ($x^*(p,m)$),

$p_1x_1+p_2x_2=m$

$\frac{p_2^2x_2}{p_1}+p_2x_2=m$

Solving for $x_2$,

$x_2(\frac{p_2^2}{p_1}+p_2)=m$

$x_2^*=\frac{m}{p_2}.\frac{p_2}{p_1+p_2}=\frac{m}{p_1+p_2}$

We do the same thing for $x_1$ and we get

$x_1^*=\frac{m}{p_1+p_2}$

We now have the Hickdian demands, substitute these into the objective function to get the indirect utility function and by duality, we know, $V(p,E(p,u))=u$,

$V(p,m)=(\frac{m}{p_1+p_2})^\frac{1}{2}+(\frac{m}{p_1+p_2})^\frac{1}{2}=u$

Solving this to find m,

$\frac{2m^\frac{1}{2}}{p_1^\frac{1}{2}+p_2^\frac12}=u$

$m^\frac{1}{2}=\frac{u}{2}(p_1^\frac{1}{2}+p_2^\frac{1}{2})$

$m=[\frac{u}{2}(p_1^\frac{1}{2}+p_2^\frac{1}{2})]^2$

Finally we also know by duality that $E(p,V(p,m))=m$, therefore

$E=[\frac{u}{2}(p_1^\frac{1}{2}+p_2^\frac{1}{2})]^2$

To find the Hicksian demand we use Shephards Lemma (take the partial derivative of the expenditure function)

$\frac{\partial E}{\partial p_i}=2.\frac{u}{2}(p_1^\frac{1}{2}+p_2^\frac{1}{2}).\frac{1}{2}.\frac{u}{2}.p_1^\frac{-1}{2}$

Simplifying this

$h_1^*=\frac{u^2}{4}.p_1^\frac{-1}{2}(p_1^\frac{1}{2}+p_2^\frac{1}{2})=\frac{u^2}{4}(p_1^\frac{-1}{4}+p_2^\frac{1}{2})$

Is this correct? (I've tried to make each of my steps really clear so this is easy to follow)

$\endgroup$
4
  • $\begingroup$ I dont believe your last equation where $p_1$ disappears. $\endgroup$ Dec 20, 2020 at 20:54
  • $\begingroup$ @JesperHybel You're right! I've changed it now, thanks $\endgroup$ Dec 20, 2020 at 21:11
  • $\begingroup$ Your process is correct, but it's very long. You should just try to solve the expenditure minimization problem, I've solved it in an answer below as an example $\endgroup$
    – mynameparv
    Apr 25, 2023 at 15:14
  • $\begingroup$ There is an error in your last equality $\endgroup$
    – Bertrand
    Apr 25, 2023 at 16:45

2 Answers 2

2
$\begingroup$

Here is what I am thinking (and i could have made a mistake since CES algebra is a bit tedious)

First you have the MRS condition from Lagrange which boils down to

$$p_1 x_2^{\rho-1} = p_2 x_1^{\rho-1}$$

from here I always aim to reconstruct the utility function (because it is a constant equal to the max utility or in this instance you know it will satisfy constraint and be u). So to get the utility function to appear I do this

$$p_1^{\frac{\rho}{\rho-1}} x_2^{\rho} = p_2^{\frac{\rho}{\rho-1}} x_1^{\rho}$$

I then add $p_1^{\frac{\rho}{\rho-1}} x_1^{\rho}$ to both sides

$$p_1^{\frac{\rho}{\rho-1}} x_2^{\rho} + p_1^{\frac{\rho}{\rho-1}} x_1^{\rho}= p_1^{\frac{\rho}{\rho-1}} x_1^{\rho}+p_2^{\frac{\rho}{\rho-1}} x_1^{\rho}$$

and isolate

$$p_1^{\frac{\rho}{\rho-1}}( x_2^{\rho} + x_1^{\rho})= (p_1^{\frac{\rho}{\rho-1}}+p_2^{\frac{\rho}{\rho-1}} )x_1^{\rho}$$

utility function appears on LHS and I impose constraint $$p_1^{\frac{\rho}{\rho-1}}u= (p_1^{\frac{\rho}{\rho-1}}+p_2^{\frac{\rho}{\rho-1}} )x_1^{\rho}$$

finding

$$\frac{p_1^{\frac{\rho}{\rho-1}}}{(p_1^{\frac{\rho}{\rho-1}}+p_2^{\frac{\rho}{\rho-1}} )}u=x_1^{\rho}$$

which reduces to

$$\frac{p_1^{\frac{1}{\rho-1}}}{(p_1^{\frac{\rho}{\rho-1}}+p_2^{\frac{\rho}{\rho-1}} )^\frac{1}{\rho}}u^\frac{1}{\rho}=x_1$$

Inserting value of $\rho = 1/2$ I get

$$\frac{p_1^{-2}}{(p_1^{-1}+p_2^{-1} )^2}u^2=x_1$$

$\endgroup$
0
$\begingroup$

You can just solve the following problem to obtain Hicksian Demands directly: $$\begin{aligned} &&\min_{x_1,x_2\geq0} \quad & p_1x_1+p_2x_2\\ &&\textrm{s.t.} \quad & \sqrt x_1 + \sqrt x_2 \geq \mu\\\\ &&\min_{x_1\geq0} \quad &p_1x_1+p_2(\mu-\sqrt x_1)^2 \\ \end{aligned}$$

we get the second problem involving only $x_1$ by using the substitution $\sqrt x_2 =\mu -\sqrt x_1$ since the constraint binds at optimum.

Now, let $f(x_1)=p_1x_1+p_2(\mu-\sqrt x_1)^2$ defined for $x_1\geq 0$. Let us see whether $f(x_1)$ is convex or concave over its entire domain.

$$\begin{eqnarray} f'(x_1)=p_1+p_2-\frac{p_2\mu}{\sqrt x_1}\\ f''(x_1)=\frac{p_2 \mu}{2x_1^{\frac{3}{2}}}>0 & \quad \text{for }x_1\geq 0 \end{eqnarray}$$

since $f(x_1)$ is convex over its entire domain if there exists a stationary point $x_1^*\geq 0$ then it's the solution to our problem. Stationary points can be obtained by: $f(x_1)\overset{set}{\equiv}0$ yields $x_1=\left(\frac{p_2\mu}{p_1+p_2}\right)^2\geq 0$

Therefore, hicksian demand functions are: $(x_1^h,x_2^h)(p_1,p_2,\mu)=\left(\left(\frac{p_2\mu}{p_1+p_2}\right)^2,\left(\frac{p_1\mu}{p_1+p_2}\right)^2 \right)$

$\endgroup$

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.