<ol>
<li><p>In the original problem, for <span class="math-container">$L=2$</span>, the consumption set was <span class="math-container">$(-\infty,\infty) \times \mathbb{R}_{+}$</span>. Now, the consumer is restricted to <span class="math-container">$[0,\infty)\times \mathbb{R}_{+} = \mathbb{R}_{+}^2$</span>. Fix a price <span class="math-container">$p = (1,p_2)$</span>, and now see what happens to Walrasian demand (i.e. <span class="math-container">$x(p,w) = (x_1,x_2))$</span> as we vary <span class="math-container">$w$</span>.</p></li>
<li><p>You saw from part (a) that a consumer with these preferences consume in the following order.</p>

<p>(i) Spend all money on good 2 until we get to <span class="math-container">$x_2(p,0)$</span> (since the amount of good 2 demanded is independent of <span class="math-container">$w$</span>).</p>

<p>(ii) If you have money leftover, spend it on good 1 (i.e. <span class="math-container">$x_1(p,w) \geq 0$</span>). Otherwise, in order to balance the budget, consume negative amounts of good 1 (i.e. <span class="math-container">$x_1(p,w) &lt; 0$</span>).</p></li>
</ol>

<p>Let us now fix <span class="math-container">$p = (1,p_2)$</span>. Now, we are unable to do case (ii) when you have overspent on good 2. You can only overspend when:
<span class="math-container">$$p_2 \times x_2(p,0) &gt; w$$</span></p>

<p>Using the usual marginal rate of substitution equal price ratio, we have:
<span class="math-container">$$
\frac{\eta'(x_2)}{1} = \frac{p_2}{1}
$$</span>
Hence,
<span class="math-container">$$x_2(p,0) = \eta'^{-1}(p_2)$$</span></p>

<p>Going back to the first equation, <span class="math-container">$p_2\times \eta'^{-1}(p_2) &gt; w$</span>.</p>

<p>Since she spends all her money on good 2, the budget constraint
<span class="math-container">$$x_1(p,w) + p_2 x_2(p,0) = w$$</span>
becomes
<span class="math-container">$$p_2 x_2(p,0) = w$$</span>
or
<span class="math-container">$$x_2(p,0) = \frac{w}{p_2}$$</span></p>

<p>Hence, her Walrasian demand (up to <span class="math-container">$p_2\times \eta'^{-1}(p_2) &gt; w$</span>) is
<span class="math-container">$$x(p,w) = (0,\frac{w}{p_2})$$</span>
And once <span class="math-container">$w$</span> is large enough such that <span class="math-container">$p_2\times \eta'^{-1}(p_2) \leq w$</span>,
<span class="math-container">$$x(p,w) = (w - x_2(p,0),x_2(p,0))$$</span></p>

<p>EDIT: Lets try a different way to get the answer using the calculus approach.</p>

<p>Our new maximisation problem is:
<span class="math-container">\begin{align*}
\max_{x_1,x_2} \, &amp;x_1 + \eta(x_2) \\
\text{s.t. } &amp;x_1 + p_2 x_2 = w &amp;\text{ (Budget constraint)}\\
&amp;x_1 \geq 0 &amp;\text{ (Non-negativity constraint) }
\end{align*}</span>
I put equality on the budget constraint because I assume monotonicity.
The associated Lagrangian is 
<span class="math-container">$$
\mathcal{L} = x_1 + \eta(x_2) + \lambda(w - x_1 - p_2 x_2) + \mu(x_1)
$$</span>
Taking first order conditions:
<span class="math-container">$$
1 - \lambda + \mu = 0 \\
\eta'(x_2) - p_2\lambda = 0
$$</span>
Now, suppose the non-negativity constraint <strong>does not bind</strong> (i.e. <span class="math-container">$x_1 &gt; 0$</span>). Then, by complimentary slackness, <span class="math-container">$\mu = 0$</span>, which implies <span class="math-container">$\lambda = 1$</span> and <span class="math-container">$x^*_2 = \eta'^{-1}(p_2)$</span>.
We can clearly see that the optimal choice of <span class="math-container">$x_2$</span> does not depend on <span class="math-container">$w$</span> at all when the constraint is non-binding.</p>

<p>Now, suppose the non-negativity constraint <strong>does bind</strong> (i.e. <span class="math-container">$x_1 = 0$</span>). Using the budget constraint, <span class="math-container">$0 + p_2x_2 = w$</span>, thus <span class="math-container">$x_2 = \frac{w}{p_2}$</span>.</p>

<p>Finally, when does the non-negativity constraint matter? It matters only when we would like to set <span class="math-container">$x_1 &lt; 0$</span>. Again, using the budget constraint, 
<span class="math-container">$$
0 &gt; x_1 = w - p_2x_2^* = w - p_2 \eta'^{-1}(p_2) \Leftrightarrow w &lt; p_2 \eta'^{-1}(p_2)
$$</span></p>

<p>Hence, when <span class="math-container">$w &lt; p_2 \eta'^{-1}(p_2)$</span>, the constraint <strong>binds</strong> thus <span class="math-container">$x_1 = 0$</span> and <span class="math-container">$x_2 = \frac{w}{p_2}$</span>.</p>

<p>When <span class="math-container">$w \geq p_2 \eta'^{-1}(p_2)$</span>, the constraint <strong>does not bind</strong> thus <span class="math-container">$x_2 = \eta'^{-1}(p_2)$</span> and <span class="math-container">$x_1 = w - \eta'^{-1}(p_2) \geq 0$</span>.</p>

CC BY-SA 4.0
2019-10-28T01:03:39.670Z
Thanks a lot! I have two questions about your answer: 1. what is x2(p,0)? Is it the optimal choice of the utility maximization problem when there's no nonnegative constraint on x1? 2. why x2(p,0) is positive? Is it possible that x1(p,0)=x2(p,0)=0?
HXW
CC BY-SA 4.0
2019-10-28T23:40:47.987Z
@Huaixin 1. That is indeed correct. 2. I am implicitly making the assumption that preferences are monotone.
Walrasian Auctioneer
CC BY-SA 4.0
2019-10-29T04:13:17.277Z
Thanks, I see it. But I am still confused: 1. why the consumer will spend all her money on good 2 when there's nonnegative constraint on good 1?(since when w is small and x1 is nonnegative, she cannot afford x2(p,0))  2. why the budget constraint becomes p2x2(p,0)=w? I think with nonnegative constraint, we have to solve the utility maximization problem again and it seems there's a gap when moving to the nonnegative situation. Your explanation is very helpful and intuitive, but I don't understand it in calculus...
HXW
CC BY-SA 4.0
2019-10-30T16:10:02.230Z
@Huaixin I've added some edits that will hopefully help.
Walrasian Auctioneer
CC BY-SA 4.0
2019-10-31T06:50:36.403Z
Got it, thanks for your patience and very clear answer!
HXW
CC BY-SA 4.0
2019-10-26T20:49:58.170Z
2019-10-30T16:09:36.393Z
2019-10-30T16:09:36.393Z
Walrasian Auctioneer
<p>I'm doing exercises of Chapter3 of MWG, there's a problem that I don't understand (I didn't figure out the solution manual either...).</p>

<p>It is about exercise 3.D.4, the full statement of the exercise is as follows:</p>

<p>Let <span class="math-container">$(-\infty,\infty)\times R_+^{L-1}$</span> denote the consumption set, and assume that preferences are strictly convex and quasilinear. Normalize <span class="math-container">$p_1=1$</span>.</p>

<p>(a) Show that the Walrasian demand functions for goods <span class="math-container">$2,...,L$</span> are independent of wealth. What does this imply about the wealth effect of demand for good 1?</p>

<p>(b) Argue that the indirect utility function can be written in the form <span class="math-container">$v(p,w)=w+\Phi(p)$</span> for some function <span class="math-container">$\Phi(·)$</span>.</p>

<p>(c) Suppose, for simplicity, that <span class="math-container">$L=2$</span>, and write the consumer's utility function as <span class="math-container">$u(x_1,x_2)=x_1+\eta(x_2)$</span>. Now, however, let the consumption set be <span class="math-container">$R_+^2$</span>, so that there is a nonnegative constraint on consumption of the numeraire <span class="math-container">$x_1$</span>. Fix prices <span class="math-container">$p$</span>, and examine how the consumer's Walrasian demand changes as wealth <span class="math-container">$w$</span> varies. When is the nonnegativity constraint on the numeraire irrelevant?</p>

<p><strong>My question</strong> is about the (c) part: <strong>1.</strong> what does it mean? <strong>2.</strong> Can anyone explain the solution to it?</p>

<p>The solution to (c) (from the solution manual) is:</p>

<p>The non-negativity constraint is binding if and only if <span class="math-container">$p_2x_2(p,0)&gt;w$</span>. Note that <span class="math-container">$x_2(p,0)=(\eta')^{-1}(p_2)$</span>, because <span class="math-container">$p_1=1$</span>. Hence the constraint is binding if and only if <span class="math-container">$p_2(\eta')^{-1}(p_2)&gt;w$</span>. If so, the Walrasian demand is given by <span class="math-container">$x(p,w)=(0,w/p_2)$</span>. Thus, as <span class="math-container">$w$</span> changes, the consumption level of the first good is unchanged and the consumption of the second good changes at rate <span class="math-container">$1/p_2$</span> with <span class="math-container">$w$</span> until the non-negativity constraint no longer binds.</p>

<microeconomics><utility><preferences>
A question about MWG Exercise 3.D.4
HXW
# A question about MWG Exercise 3.D.4

(Asked by: username_0 on 2019-10-26)

I'm doing exercises of Chapter3 of MWG, there's a problem that I don't understand (I didn't figure out the solution manual either...).

It is about exercise 3.D.4, the full statement of the exercise is as follows:

Let $(-\infty,\infty)\times R\_+^{L-1}$ denote the consumption set, and assume that preferences are strictly convex and quasilinear. Normalize $p\_1=1$.

(a) Show that the Walrasian demand functions for goods $2,...,L$ are independent of wealth. What does this imply about the wealth effect of demand for good 1?

(b) Argue that the indirect utility function can be written in the form $v(p,w)=w+\Phi(p)$ for some function $\Phi(·)$.

(c) Suppose, for simplicity, that $L=2$, and write the consumer's utility function as $u(x\_1,x\_2)=x\_1+\eta(x\_2)$. Now, however, let the consumption set be $R\_+^2$, so that there is a nonnegative constraint on consumption of the numeraire $x\_1$. Fix prices $p$, and examine how the consumer's Walrasian demand changes as wealth $w$ varies. When is the nonnegativity constraint on the numeraire irrelevant?

**My question** is about the (c) part: **1.** what does it mean? **2.** Can anyone explain the solution to it?

The solution to (c) (from the solution manual) is:

The non-negativity constraint is binding if and only if $p\_2x\_2(p,0)>w$. Note that $x\_2(p,0)=(\eta')^{-1}(p\_2)$, because $p\_1=1$. Hence the constraint is binding if and only if $p\_2(\eta')^{-1}(p\_2)>w$. If so, the Walrasian demand is given by $x(p,w)=(0,w/p\_2)$. Thus, as $w$ changes, the consumption level of the first good is unchanged and the consumption of the second good changes at rate $1/p\_2$ with $w$ until the non-negativity constraint no longer binds.

**Tags:** microeconomics, utility, preferences


---

## 1 answers

### Answer by username_1 on 2019-10-26. Score: 3 | Accepted answer

1. In the original problem, for $L=2$, the consumption set was $(-\infty,\infty) \times \mathbb{R}\_{+}$. Now, the consumer is restricted to $[0,\infty)\times \mathbb{R}\_{+} = \mathbb{R}\_{+}^2$. Fix a price $p = (1,p\_2)$, and now see what happens to Walrasian demand (i.e. $x(p,w) = (x\_1,x\_2))$ as we vary $w$.
2. You saw from part (a) that a consumer with these preferences consume in the following order.

   (i) Spend all money on good 2 until we get to $x\_2(p,0)$ (since the amount of good 2 demanded is independent of $w$).

   (ii) If you have money leftover, spend it on good 1 (i.e. $x\_1(p,w) \geq 0$). Otherwise, in order to balance the budget, consume negative amounts of good 1 (i.e. $x\_1(p,w) < 0$).

Let us now fix $p = (1,p\_2)$. Now, we are unable to do case (ii) when you have overspent on good 2. You can only overspend when:
$$p\_2 \times x\_2(p,0) > w$$

Using the usual marginal rate of substitution equal price ratio, we have:
$$
\frac{\eta'(x\_2)}{1} = \frac{p\_2}{1}
$$
Hence,
$$x\_2(p,0) = \eta'^{-1}(p\_2)$$

Going back to the first equation, $p\_2\times \eta'^{-1}(p\_2) > w$.

Since she spends all her money on good 2, the budget constraint
$$x\_1(p,w) + p\_2 x\_2(p,0) = w$$
becomes
$$p\_2 x\_2(p,0) = w$$
or
$$x\_2(p,0) = \frac{w}{p\_2}$$

Hence, her Walrasian demand (up to $p\_2\times \eta'^{-1}(p\_2) > w$) is
$$x(p,w) = (0,\frac{w}{p\_2})$$
And once $w$ is large enough such that $p\_2\times \eta'^{-1}(p\_2) \leq w$,
$$x(p,w) = (w - x\_2(p,0),x\_2(p,0))$$

EDIT: Lets try a different way to get the answer using the calculus approach.

Our new maximisation problem is:
\begin{align\*}
\max\_{x\_1,x\_2} \, &x\_1 + \eta(x\_2) \\
\text{s.t. } &x\_1 + p\_2 x\_2 = w &\text{ (Budget constraint)}\\
&x\_1 \geq 0 &\text{ (Non-negativity constraint) }
\end{align\*}
I put equality on the budget constraint because I assume monotonicity.
The associated Lagrangian is
$$
\mathcal{L} = x\_1 + \eta(x\_2) + \lambda(w - x\_1 - p\_2 x\_2) + \mu(x\_1)
$$
Taking first order conditions:
$$
1 - \lambda + \mu = 0 \\
\eta'(x\_2) - p\_2\lambda = 0
$$
Now, suppose the non-negativity constraint **does not bind** (i.e. $x\_1 > 0$). Then, by complimentary slackness, $\mu = 0$, which implies $\lambda = 1$ and $x^\*\_2 = \eta'^{-1}(p\_2)$.
We can clearly see that the optimal choice of $x\_2$ does not depend on $w$ at all when the constraint is non-binding.

Now, suppose the non-negativity constraint **does bind** (i.e. $x\_1 = 0$). Using the budget constraint, $0 + p\_2x\_2 = w$, thus $x\_2 = \frac{w}{p\_2}$.

Finally, when does the non-negativity constraint matter? It matters only when we would like to set $x\_1 < 0$. Again, using the budget constraint,
$$
0 > x\_1 = w - p\_2x\_2^\* = w - p\_2 \eta'^{-1}(p\_2) \Leftrightarrow w < p\_2 \eta'^{-1}(p\_2)
$$

Hence, when $w < p\_2 \eta'^{-1}(p\_2)$, the constraint **binds** thus $x\_1 = 0$ and $x\_2 = \frac{w}{p\_2}$.

When $w \geq p\_2 \eta'^{-1}(p\_2)$, the constraint **does not bind** thus $x\_2 = \eta'^{-1}(p\_2)$ and $x\_1 = w - \eta'^{-1}(p\_2) \geq 0$.

> **Comments:**
> **username_0**: Thanks a lot! I have two questions about your answer: 1. what is x2(p,0)? Is it the optimal choice of the utility maximization problem when there's no nonnegative constraint on x1? 2. why x2(p,0) is positive? Is it possible that x1(p,0)=x2(p,0)=0?  
> **username_1**: @Huaixin 1. That is indeed correct. 2. I am implicitly making the assumption that preferences are monotone.  
> **username_0**: Thanks, I see it. But I am still confused: 1. why the consumer will spend all her money on good 2 when there's nonnegative constraint on good 1?(since when w is small and x1 is nonnegative, she cannot afford x2(p,0))  2. why the budget constraint becomes p2x2(p,0)=w? I think with nonnegative constraint, we have to solve the utility maximization problem again and it seems there's a gap when moving to the nonnegative situation. Your explanation is very helpful and intuitive, but I don't understand it in calculus...  
> **username_1**: @Huaixin I've added some edits that will hopefully help.  
> **username_0**: Got it, thanks for your patience and very clear answer!  

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