<p>Although my teacher has yet to verify my solution for d) (which I believe is incorrect), he shared his answer and the condition <span class="math-container">$x_2&lt;k$</span> derives from the positivity of <span class="math-container">$V(p_1,p_2,b)$</span> and from the budget inequality <span class="math-container">$p_1x_1+p_2x_2≤b$</span>. The other answers are correct.</p>

CC BY-SA 4.0
2020-05-19T20:16:16.797Z
2020-05-19T20:16:16.797Z
Pedro Cunha
<p>everyone.</p>

<p>I have the following question:</p>

<p>A consumer has the following indirect utility function:</p>

<p><span class="math-container">$ V(p_1,p_2,b) = (p_2k-b)p_1^{-1} \left[ \frac{2p_2k - 2b}{p_2} \right]^{-2}, x_2 &lt; k$</span></p>

<p>a) Find the Marshalian demand for good 2.</p>

<p>b) Find the Hicksian demand for good 2.</p>

<p>c) Show that the Slutsky equation holds for good 2.</p>

<p>d) Why is it necessary to have <span class="math-container">$x_2 &lt; k$</span></p>

<p>For a) I found:</p>

<p><span class="math-container">$ \boxed{\boxed{ x_2(p_1,p_2,b) = \frac{2b-p_2k}{p_2} }} $</span></p>

<p>For b):</p>

<p><span class="math-container">$ \boxed{\boxed{ x_2^c(p_1,p_2,\bar{U}) = k - \frac{P_2}{2P_1 \cdot \bar{U} }}} $</span></p>

<p>For c) after replacing <span class="math-container">$E(p_1,p_2, \bar{U})$</span> with <span class="math-container">$p_1 \cdot x_1(p_1,p_2,b) + p_2 \cdot x_2(p_1,p_2,b)$</span> I found <span class="math-container">$\bar{U} = \frac{p_2^2}{(p_2k-b) 4p_1}$</span> and using that in</p>

<p><span class="math-container">$\frac{\partial x_2(p_1,p_2,b)}{\partial p_2} = \frac{\partial x_2^c(p_1,p_2,\bar{U})}{\partial p_2} - \frac{\partial x_2(p_1,p_2,b)}{\partial b} \cdot x_2(p_1,p_2,b)$</span> </p>

<p>I was able to show that both sides are equal.</p>

<p>For d) here's what I've tried:</p>

<p>using Roy's identity we can find</p>

<p><span class="math-container">$x_1(p_1,p_2,b) = \frac{p_2k - b}{p_1}$</span></p>

<p>assuming the demand for good 1 is positive, we must then have:</p>

<p><span class="math-container">$ \frac{p_2k - b}{p_1} &gt; 0  $</span></p>

<p>Since <span class="math-container">$p_1 &gt; 0$</span>, we can re-write the inequality as</p>

<p><span class="math-container">$ p_2k - b &gt; 0 $</span></p>

<p>but</p>

<p><span class="math-container">$b = p_1x_1 + p_2x_2$</span>, </p>

<p>so</p>

<p><span class="math-container">$ p_2k - p_1x_1 - p_2x_2 &gt; 0 \therefore p_2 \cdot (k - x_2) &gt; p_1x_1 \therefore k-x_2 &gt; \frac{p_1x_1}{p_2}$</span>, because <span class="math-container">$p_2$</span> is positive.</p>

<p>Now, since <span class="math-container">$p_1 &gt; 0$</span> and assuming <span class="math-container">$x_1 &gt; 0$</span>, we have <span class="math-container">$\frac{p_1x_1}{p_2} &gt; 0$</span>, which in turn implies that</p>

<p><span class="math-container">$k-x_2 &gt; 0 \therefore \boxed{\boxed{x_2 &lt; k, Q.E.D.}} $</span></p>

<p>Are my answers correct? In c) is there a faster way to find <span class="math-container">$\bar{U}$</span> so I can replace it in the Hickisian demand and show that the Slutsky equation holds for good 2?</p>

<p>I appreciate any inputs.</p>

<p>Have a good one,
Pedro. </p>

<consumer-theory><demand><slutsky-equation>
Marshalian and Hickisian Demands and Slutsky Equation
Pedro Cunha
# Marshalian and Hickisian Demands and Slutsky Equation

(Asked by: username_0 on 2020-05-02)

everyone.

I have the following question:

A consumer has the following indirect utility function:

$ V(p\_1,p\_2,b) = (p\_2k-b)p\_1^{-1} \left[ \frac{2p\_2k - 2b}{p\_2} \right]^{-2}, x\_2 < k$

a) Find the Marshalian demand for good 2.

b) Find the Hicksian demand for good 2.

c) Show that the Slutsky equation holds for good 2.

d) Why is it necessary to have $x\_2 < k$

For a) I found:

$ \boxed{\boxed{ x\_2(p\_1,p\_2,b) = \frac{2b-p\_2k}{p\_2} }} $

For b):

$ \boxed{\boxed{ x\_2^c(p\_1,p\_2,\bar{U}) = k - \frac{P\_2}{2P\_1 \cdot \bar{U} }}} $

For c) after replacing $E(p\_1,p\_2, \bar{U})$ with $p\_1 \cdot x\_1(p\_1,p\_2,b) + p\_2 \cdot x\_2(p\_1,p\_2,b)$ I found $\bar{U} = \frac{p\_2^2}{(p\_2k-b) 4p\_1}$ and using that in

$\frac{\partial x\_2(p\_1,p\_2,b)}{\partial p\_2} = \frac{\partial x\_2^c(p\_1,p\_2,\bar{U})}{\partial p\_2} - \frac{\partial x\_2(p\_1,p\_2,b)}{\partial b} \cdot x\_2(p\_1,p\_2,b)$

I was able to show that both sides are equal.

For d) here's what I've tried:

using Roy's identity we can find

$x\_1(p\_1,p\_2,b) = \frac{p\_2k - b}{p\_1}$

assuming the demand for good 1 is positive, we must then have:

$ \frac{p\_2k - b}{p\_1} > 0 $

Since $p\_1 > 0$, we can re-write the inequality as

$ p\_2k - b > 0 $

but

$b = p\_1x\_1 + p\_2x\_2$,

so

$ p\_2k - p\_1x\_1 - p\_2x\_2 > 0 \therefore p\_2 \cdot (k - x\_2) > p\_1x\_1 \therefore k-x\_2 > \frac{p\_1x\_1}{p\_2}$, because $p\_2$ is positive.

Now, since $p\_1 > 0$ and assuming $x\_1 > 0$, we have $\frac{p\_1x\_1}{p\_2} > 0$, which in turn implies that

$k-x\_2 > 0 \therefore \boxed{\boxed{x\_2 < k, Q.E.D.}} $

Are my answers correct? In c) is there a faster way to find $\bar{U}$ so I can replace it in the Hickisian demand and show that the Slutsky equation holds for good 2?

I appreciate any inputs.

Have a good one,
Pedro.

**Tags:** consumer-theory, demand, slutsky-equation


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## 1 answers

### Answer by username_0 on 2020-05-19. Score: 1 | Accepted answer

Although my teacher has yet to verify my solution for d) (which I believe is incorrect), he shared his answer and the condition $x\_2 derives from the positivity of $V(p\_1,p\_2,b)$ and from the budget inequality $p\_1x\_1+p\_2x\_2≤b$. The other answers are correct.


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