# MWG Exercise 2.E.5

## Exercise

Suppose that $$x(\mathbf{p},w)$$ is a demand function which is homogeneous of degree one with respect to $$w$$ and satisfies Walras' law and homogeneity of degree zero. Suppose also that all the cross-price effects are zero, that is $$\frac{\partial x_l(\mathbf{p},w)}{\partial p_k} = 0$$ whenever $$k \neq l$$. Show that this implies that for every $$l$$, $$x_l(\mathbf{p},w) = \frac{\alpha_lw}{p_l}$$, where $$\alpha_l > 0$$ is a constant independent of $$(\mathbf{p},w)$$.

## My Attempt

This is how I proceed. (I got stuck in the end and couldn't manage to show that $$\alpha_l$$ is a constant independent of $$(\mathbf{p},w)$$.)

By Proposition 2.E.2, since all the cross-price effects are zero, we have that \begin{align*} p_l\frac{\partial x_l(\mathbf{p},w)}{\partial p_l} + x_l(\mathbf{p},w) = 0,\ \text{for}\ l = 1,\dots,L. \end{align*} By Proposition 2.E.1, again since all the cross-price effects are zero, we have that \begin{align*} \frac{\partial x_l(\mathbf{p},w)}{\partial p_l}p_l + \frac{\partial x_l(\mathbf{p},w)}{\partial w}w = 0,\ \text{for}\ l = 1,\dots,L. \end{align*} Then, \begin{align*} x_l(\mathbf{p},w) = \frac{\partial x_l(\mathbf{p},w)}{\partial w}w, \end{align*} so that \begin{align*} x_l(\mathbf{p},w) = \frac{\partial x_l(\mathbf{p},w)}{\partial w}p_l \cdot \frac{w}{p_l} \end{align*} Let $$\alpha_l = \frac{\partial x_l(\mathbf{p},w)}{\partial w}p_l$$

From here, I don't know how to proceed next. Could someone please help me out? If we were to follow this method, how to show that the $$\alpha_l$$ is a constant independent of $$(\mathbf{p},w)$$?

## Answer from the Solution Manual

I looked up and find the following answer from the Solution Manual. However, I am confused by some of its steps.

Since $$x(\mathbf{p},w)$$ is homogeneous of degree one with respect to $$w$$, $$x(\mathbf{p},\alpha w) = \alpha x(\mathbf{p},w)$$ for every $$\alpha > 0$$. Thus, $$x_l(\mathbf{p},w) = wx_l(\mathbf{p},1)$$. Since $$\frac{\partial x_l(\mathbf{p},1)}{\partial p_k} = \frac{\partial\varphi_l(p)}{\partial p_k} = 0$$ whenever $$k \neq l$$, $$x_l(\mathbf{p},1)$$ is actually a function of $$p_l$$ alone. So we can write $$x_l(\mathbf{p},w) = x_l(p_l)$$. Since $$x(\mathbf{p},w)$$ is homogeneous of degree zero, $$x_l(p_l)$$ must be homogeneous of degree $$-1$$ (in $$p_l$$). Hence, there exists $$\alpha_l > 0$$ such that $$x_l(p_l) = \frac{\alpha_l}{p_l}$$. By Walras' law, $$\sum_{l}p_l\left(\frac{\alpha_l}{p_l}\right)w = w\sum_{l}\alpha_l = w$$. We must thus have $$\sum_l\alpha_l = 1$$.

I start to feel confused after it said that "$$x_l(\mathbf{p},1)$$ is actually a function of $$p_l$$ alone.". For example, I do not think we can write $$x_l(\mathbf{p},w) = x_l(p_l)$$; I cannot see how $$x_l(p_l)$$ must be homogeneous of degree $$-1$$ (in $$p_l$$); and the last sentence about applying Walras' law is definitely not right. Here is my understanding of this answer from the Solution Manual. I would like to know if it is correct.

Since $$x(\mathbf{p},w)$$ is homogeneous of degree one with respect to $$w$$, $$x(\mathbf{p},\alpha w) = \alpha x(\mathbf{p},w)$$ for every $$\alpha > 0$$. Thus, $$x_l(\mathbf{p},w) = wx_l(\mathbf{p},1)$$. Since $$\frac{\partial x_l(\mathbf{p},1)}{\partial p_k} = \frac{\partial\varphi_l(p)}{\partial p_k} = 0$$ whenever $$k \neq l$$, $$x_l(\mathbf{p},1)$$ is actually a function of $$p_l$$ alone. Thus, we can write $$x_l(\mathbf{p},w) = x_l(p_l,w)$$. Since $$x(\mathbf{p},w)$$ is homogeneous of degree zero and is homogeneous of degree one with respect to $$w$$, we have \begin{align*} x_l(\alpha p_l, w) &= x_l(p_1,\dots,\alpha p_l,\dots,p_L,w)\\ &= x_l\left(\alpha \cdot \left(\frac{1}{\alpha}p_1,\dots,p_l,\dots,\frac{1}{\alpha}p_L,\frac{1}{\alpha}w\right)\right)\\ &= x_l\left(\frac{1}{\alpha}p_1,\dots,p_l,\dots,\frac{1}{\alpha}p_L,\frac{1}{\alpha}w\right)\\ &= x_l\left(p_l,\frac{1}{\alpha}w\right)\\ &= \frac{1}{\alpha}x_l(p_l,w). \end{align*} Therefore, $$x_l(p_l,w)$$ is homogeneous of degree $$-1$$ with respect to $$p_l$$. Hence, \begin{align*} x_l(\mathbf{p},w) &= x_l(p_1,\dots,p_l,\dots,p_L,w)\\ &= \frac{1}{p_l}x_l(p_1,\dots,1,\dots,p_L,w)\\ &= \frac{w}{p_l}x_l(p_1,\dots,1,\dots,p_L,1). \end{align*} Let $$\alpha_l = x_l(p_1,\dots,1,\dots,p_L,1)$$. Since $$x_l$$ is independent of $$p_k$$ for all $$k \neq l$$, we have that $$\alpha_l$$ is a constant independent of $$(\mathbf{p},w)$$.

## My Questions

So, as I have mentioned earlier, I have two questions:

1. If we were to follow my method (in the My Attempt section), is it possible for us to proceed to show that $$\frac{\partial x_l(\mathbf{p},w)}{\partial w}p_l$$ is a constant independent of $$(\mathbf{p},w)$$?
2. I feel that the answer in the Solution Manual is not correct, and I provided my understanding of his method. I would like to know whether my understanding of the answer from the Solution Manual is correct or not?

Thank you very much in advance!

• The proof given in the Solution Manual is correct, except the typo in the equation at the end of line 3, which should read $x_l(\mathbf{p},w) = x_l(p_l)w$. Then, because $x_l(\mathbf{p},w)$ is homogeneous of degree zero in $(\mathbf{p},w)$, it implies that $x_l(p_l)w$ is homogeneous of degree zero in $(p_l,w)$, which in turn implies that $x_l(p_l)$ must be homogeneous of degree $-1$ in $p_l$. Jan 31 at 20:46

$$x_l(\mathbf{p},w) = \frac{\partial x_l(\mathbf{p},w)}{\partial w}p_l \cdot \frac{w}{p_l}$$ Let $$\alpha_l = \frac{\partial x_l(\mathbf{p},w)}{\partial w}p_l$$.

This is nice, now use the homogeneity properties.

Could do solution manual be wrong? A quick non conclusive trial:

For a Cobb-Douglas type function (where the cross price effects will be zero) $$U(x_1,x_2) = x_1x_2$$ the Marshallian demands are \begin{align*} x_1(p_1,p_2,w) & = \frac{1}{2} \frac{w}{p_1} \\ x_2(p_1,p_2,w) & = \frac{1}{2} \frac{w}{p_2}. \end{align*} The function $$x_1(p_1,p_2,w)$$ indeed

1. is homogenous of degree 0,
2. is homogenous of degree 1 w.r.t. $$w$$, i.e. $$x_1(p_1,p_2,w) = w \cdot x_1(p_1,p_2,1),$$
3. does not depend on $$p_2$$, we could write $$x_1(p_1) \triangleq x_1(p_1,p_2,1).$$

Furthermore, it is possible to write \begin{align*} x_1(p_1,p_2,w) & = w \cdot x_1(p_1) = w \cdot \frac{1}{2} \frac{1}{p_1} \\ x_1(p_1,p_2,w) & = w \cdot x_2(p_2) = w \cdot \frac{1}{2} \frac{1}{p_2}, \end{align*} and then the alphas do indeed sum up to $$1$$. Though this was only one case (and a special one at that) the solution manual could be - and I think is - correct.

The root of the misunderstanding seems to be in this line:

For example, I do not think we can write $$x_l(\mathbf{p},w) = x_l(p_l)$$;

You are correct, this is not possible; but it is also not the claim. The claim

$$x_l(\mathbf{p},1)$$ is actually a function of $$p_l$$ alone.

is not about the general range of the function $$x_l$$. First they restrict $$w$$ to $$1$$, thus they are only looking at the cases where $$x_l(\mathbf{p},1)$$. Then they explain that because there are no cross price effects (silent assumptions are made about integrability), changes in the other prices do not affect $$x_l(\mathbf{p},1)$$.

E.g. looking at $$x_1(p_1,p_2,w)$$ again $$x_1(p_1,20,w) - x_1(p_1,10,w) = \int_{p_2 = 10}^{20} \frac{\partial x_1(p_1,p_2,w)}{\partial p_2} \text{d} p_2$$ and by the no cross price effect assumption $$\frac{\partial x_1(p_1,p_2,w)}{\partial p_2} = 0$$ so the whole thing is zero, $$x_1(p_1,p_2,w)$$ is not affected by any change in $$p_2$$.

So $$x_l(\mathbf{p},w)$$ is homogeneous of degree 1 w.r.t. $$w$$, is not affected by the prices $$p_k$$ where $$p_k \neq p_l$$. We also know that the general function is homogeneous of degree 0. This is only possible if changes in $$p_l$$ counter the effects of the changes in $$w$$. It is shown via the solution manual's calculation that this in fact means that $$x_l(\mathbf{p},w)$$ is homogeneous of degree $$-1$$ w.r.t $$p_l$$.

For the two good case the equation is, essentially, $$\forall \alpha>0$$: $$\begin{eqnarray*} x_1(p_1,p_2,w) & = & x_1(\alpha \cdot p_1,\alpha \cdot p_2, \alpha \cdot w) \tag{hom. of deg. 0} \\ \\ & = & \alpha \cdot x_1(\alpha \cdot p_1,\alpha \cdot p_2, w) \tag{hom. of deg. 1 w.r.t w} \\ \\ & = & \alpha \cdot x_1(\alpha \cdot p_1, p_2, w) \tag{p_2 has no effect} \\ \\ \alpha^{-1} \cdot x_1(p_1,p_2,w) & = & x_1(\alpha \cdot p_1, p_2, w) \tag{division by \alpha} \end{eqnarray*}$$ This is the definition of $$x_1$$ being homogeneous of degree $$-1$$ w.r.t. $$p_1$$.

• I found a dozen typos before going to bed, I feel confident that there are more; feel free to edit. Jan 31 at 0:16
• Thank you very much for your answer. For your "Answer to 1", here is my attempt: Since $x(\mathbf{p},w)$ is homogeneous of degree 1 in $w$, we have $\alpha_l=\frac{\partial x_l(\mathbf{p},w)}{\partial w}p_l=\frac{\partial(wx_l(\mathbf{p},1))}{\partial w}p_l=x_l(\mathbf{p},1)p_l$. Well, this indeed shows that $\alpha_l$ is independent of $w$, but why is it independent of $\mathbf{p}$ as well? Jan 31 at 2:00
• For your "Answer to 2", I still cannot understand how to derive $x_l(p_l)$ is homogeneous of degree $-1$ in the context of the answer from the Solution Manual. Could you please elaborate it? Jan 31 at 3:08
• Can you please give the definition of $x_1(p_1)$ as you understand it and do you understand what I wrote otherwise? Jan 31 at 6:29
• "$\alpha_l$ is independent of $w$, but why is it independent of $\mathbb{p}$ as well?" How about using the homogeneity property that includes both variables, not just $w$? Jan 31 at 10:00