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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!

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  • $\begingroup$ 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 $p_l$. $\endgroup$
    – Bertrand
    Jan 31 at 20:46

1 Answer 1

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Answer to 1.

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

This is nice, now use the homogeneity properties.

Answer to 2.

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. $p_1$.

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  • $\begingroup$ I found a dozen typos before going to bed, I feel confident that there are more; feel free to edit. $\endgroup$
    – Giskard
    Jan 31 at 0:16
  • $\begingroup$ 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? $\endgroup$
    – Shenron
    Jan 31 at 2:00
  • $\begingroup$ 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? $\endgroup$
    – Shenron
    Jan 31 at 3:08
  • $\begingroup$ Can you please give the definition of $x_1(p_1)$ as you understand it and do you understand what I wrote otherwise? $\endgroup$
    – Giskard
    Jan 31 at 6:29
  • $\begingroup$ "$\alpha_l$ 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 $w$? $\endgroup$
    – Giskard
    Jan 31 at 10:00

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