# A question about MWG Exercise 3.D.4

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.

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 $$x_1 = w - \eta'^{-1}(p_2) \geq 0$$.

• 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? – Huaixin Oct 28 '19 at 1:03
• @Huaixin 1. That is indeed correct. 2. I am implicitly making the assumption that preferences are monotone. – Walrasian Auctioneer Oct 28 '19 at 23:40
• 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... – Huaixin Oct 29 '19 at 4:13
• @Huaixin I've added some edits that will hopefully help. – Walrasian Auctioneer Oct 30 '19 at 16:10
• Got it, thanks for your patience and very clear answer! – Huaixin Oct 31 '19 at 6:50