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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.

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

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

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