<p><strong>Answer to 1.</strong></p>
<blockquote>
<p><span class="math-container">$$ x_l(\mathbf{p},w) = \frac{\partial x_l(\mathbf{p},w)}{\partial w}p_l \cdot \frac{w}{p_l} $$</span>
Let <span class="math-container">$\alpha_l = \frac{\partial x_l(\mathbf{p},w)}{\partial w}p_l$</span>.</p>
</blockquote>
<p>This is nice, now use the homogeneity properties.</p>
<p><strong>Answer to 2.</strong></p>
<p>Could do solution manual be wrong? A quick non conclusive trial:</p>
<p>For a Cobb-Douglas type function (where the cross price effects will be zero) <span class="math-container">$U(x_1,x_2) = x_1x_2$</span> the Marshallian demands are
<span class="math-container">$$
\begin{align*}
x_1(p_1,p_2,w) &amp; = \frac{1}{2} \frac{w}{p_1} \\
x_2(p_1,p_2,w) &amp; = \frac{1}{2} \frac{w}{p_2}.
\end{align*}
$$</span>
The function <span class="math-container">$x_1(p_1,p_2,w)$</span> indeed</p>
<ol>
<li>is homogenous of degree 0,</li>
<li>is homogenous of degree 1 w.r.t. <span class="math-container">$w$</span>, i.e.
<span class="math-container">$$
x_1(p_1,p_2,w) = w \cdot x_1(p_1,p_2,1),
$$</span></li>
<li>does not depend on <span class="math-container">$p_2$</span>, we could write
<span class="math-container">$$
x_1(p_1) \triangleq x_1(p_1,p_2,1).
$$</span></li>
</ol>
<p>Furthermore, it is possible to write
<span class="math-container">$$
\begin{align*}
x_1(p_1,p_2,w) &amp; = w \cdot x_1(p_1) = w \cdot \frac{1}{2} \frac{1}{p_1} \\
x_1(p_1,p_2,w) &amp; = w \cdot x_2(p_2) = w \cdot \frac{1}{2} \frac{1}{p_2},
\end{align*}
$$</span>
and then the alphas do indeed sum up to <span class="math-container">$1$</span>. Though this was only one case (and a special one at that) the solution manual could be - and I think is - correct.</p>
<hr />
<p>The root of the misunderstanding seems to be in this line:</p>
<blockquote>
<p>For example, I do not think we can write <span class="math-container">$x_l(\mathbf{p},w) =  x_l(p_l)$</span>;</p>
</blockquote>
<p>You are correct, this is not possible; but it is also not the claim. The claim</p>
<blockquote>
<p><span class="math-container">$x_l(\mathbf{p},1)$</span> is actually a function of <span class="math-container">$p_l$</span> alone.</p>
</blockquote>
<p>is not about the general range of the function <span class="math-container">$x_l$</span>. First they restrict <span class="math-container">$w$</span> to <span class="math-container">$1$</span>, thus they are only looking at the cases where <span class="math-container">$x_l(\mathbf{p},1)$</span>. 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 <span class="math-container">$x_l(\mathbf{p},1)$</span>.</p>
<p>E.g. looking at <span class="math-container">$x_1(p_1,p_2,w)$</span> again
<span class="math-container">$$
x_1(p_1,20,w) - x_1(p_1,10,w) =
\int_{p_2 = 10}^{20} \frac{\partial x_1(p_1,p_2,w)}{\partial p_2} \text{d} p_2
$$</span>
and by the no cross price effect assumption
<span class="math-container">$$
\frac{\partial x_1(p_1,p_2,w)}{\partial p_2} = 0
$$</span>
so the whole thing is zero, <span class="math-container">$x_1(p_1,p_2,w)$</span> is not affected by any change in <span class="math-container">$p_2$</span>.</p>
<p>So <span class="math-container">$x_l(\mathbf{p},w)$</span> is homogeneous of degree 1 w.r.t. <span class="math-container">$w$</span>, is not affected by the prices <span class="math-container">$p_k$</span> where <span class="math-container">$p_k \neq p_l$</span>. We also know that the general function is homogeneous of degree 0. This is only possible if changes in <span class="math-container">$p_l$</span> counter the effects of the changes in <span class="math-container">$w$</span>. It is shown via the solution manual's calculation that this in fact means that <span class="math-container">$x_l(\mathbf{p},w)$</span> is homogeneous of degree <span class="math-container">$-1$</span> w.r.t <span class="math-container">$p_l$</span>.</p>
<p>For the two good case the equation is, essentially, <span class="math-container">$\forall \alpha&gt;0$</span>:
<span class="math-container">$$
\begin{eqnarray*}
x_1(p_1,p_2,w) &amp; = &amp; x_1(\alpha \cdot p_1,\alpha \cdot p_2, \alpha \cdot w) \tag{hom. of deg. 0}
\\ \\
&amp; = &amp; \alpha \cdot x_1(\alpha \cdot p_1,\alpha \cdot p_2, w) \tag{hom. of deg. 1 w.r.t $w$}
\\ \\
&amp; = &amp; \alpha \cdot x_1(\alpha \cdot p_1, p_2, w) \tag{$p_2$ has no effect}
\\ \\
\alpha^{-1} \cdot x_1(p_1,p_2,w) &amp; = &amp; x_1(\alpha \cdot p_1, p_2, w) \tag{division by $\alpha$}
\end{eqnarray*}
$$</span>
This is the definition of <span class="math-container">$x_1$</span> being homogeneous of degree <span class="math-container">$-1$</span> w.r.t. <span class="math-container">$p_1$</span>.</p>

CC BY-SA 4.0
2024-01-31T00:16:10.253Z
I found a dozen typos before going to bed, I feel confident that there are more; feel free to edit.
Giskard
CC BY-SA 4.0
2024-01-31T02:00:11.243Z
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?
Shenron
CC BY-SA 4.0
2024-01-31T03:08:46.280Z
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?
Shenron
CC BY-SA 4.0
2024-01-31T06:29:40.203Z
Can you please give the definition of $x_1(p_1)$ as you understand it and do you understand what I wrote otherwise?
Giskard
CC BY-SA 4.0
2024-01-31T10:00:31.117Z
"*$\alpha_l$ 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 $w$?
Giskard
CC BY-SA 4.0
2024-01-31T16:32:24.177Z
As for "Answer to 1", do you mean the following: We have had $\alpha_l=\frac{\partial x_l(\mathbf{p},w)}{\partial w}p_l=x_l(\mathbf{p},1)p_l=x_l(p_l,1)p_l$, where the last equality comes from the fact that there is no cross-price effect. Since $x_l(\mathbf{p},w)=x_l(p_l,w)$ is homogeneous of degree 1 in $w$, we have $\alpha=x_l(p_l,1)p_l=x_l(p_l,p_l)=x_l(1,1)$, where the last equality holds because $x_l(\mathbf{p},w)=x_l(p_l,w)$ is homogeneous of degree 0. Therefore, $\alpha$ is a constant independent of $(\mathbf{p},w)$.
Shenron
CC BY-SA 4.0
2024-01-31T16:56:16.030Z
As for deriving $x_l(p_l)$ is homogeneous of degree $-1$ in your "Answer to 2", do you mean the following: The $x_l(p_l)$ in the Solution Manual should have been written as $x_l(p_l,1)$. Let $k\in\mathbb{R}_+$. Since, by the homogeneity of degree 0 and the homogeneity of degree 1 in $w$, $x_l(kp_l,1)=x_l(p_l,\frac{1}{k})=\frac{1}{k}x_l(p_l,1)$, we obtain that $x_l(p_l,1)$ is homogeneous of degree $-1$ in $p_l$. Basically, the effect of a $k$-percentage change in $p_l$ should offset the effect of a same $k$-percentage change in $w$, in order to make the demand function homogeneous of degree 0.
Shenron
CC BY-SA 4.0
2024-01-31T18:03:02.527Z
This is all correct except for "*in the Solution Manual should have been written*" because their current notation is also fine. They explicitly explain that they are introducing new notation. You don't have to like their notation, other people might, but the notation is defined, and the properties given for it are correct.
Giskard
CC BY-SA 4.0
2024-01-30T23:51:11.957Z
2024-01-31T00:15:10.383Z
2024-01-31T00:15:10.383Z
Giskard
<h2><strong>Exercise</strong></h2>
<blockquote>
<p>Suppose that <span class="math-container">$x(\mathbf{p},w)$</span> is a demand function which is homogeneous of degree one with respect to <span class="math-container">$w$</span> and satisfies Walras' law and homogeneity of degree zero. Suppose also that all the cross-price effects are zero, that is <span class="math-container">$\frac{\partial x_l(\mathbf{p},w)}{\partial p_k} = 0$</span> whenever <span class="math-container">$k \neq l$</span>. Show that this implies that for every <span class="math-container">$l$</span>, <span class="math-container">$x_l(\mathbf{p},w) = \frac{\alpha_lw}{p_l}$</span>, where <span class="math-container">$\alpha_l &gt; 0$</span> is a constant independent of <span class="math-container">$(\mathbf{p},w)$</span>.</p>
</blockquote>
<h2><strong>My Attempt</strong></h2>
<p>This is how I proceed. (I got stuck in the end and couldn't manage to show that <span class="math-container">$\alpha_l$</span> is a constant independent of <span class="math-container">$(\mathbf{p},w)$</span>.)</p>
<blockquote>
<p>By Proposition 2.E.2, since all the cross-price effects are zero, we have that
<span class="math-container">\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*}</span>
By Proposition 2.E.1, again since all the cross-price effects are zero, we have that
<span class="math-container">\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*}</span>
Then,
<span class="math-container">\begin{align*}
x_l(\mathbf{p},w) = \frac{\partial x_l(\mathbf{p},w)}{\partial w}w,
\end{align*}</span>
so that
<span class="math-container">\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*}</span>
Let <span class="math-container">$\alpha_l = \frac{\partial x_l(\mathbf{p},w)}{\partial w}p_l$</span></p>
</blockquote>
<p>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 <span class="math-container">$\alpha_l$</span> is a constant independent of <span class="math-container">$(\mathbf{p},w)$</span>?</p>
<h2><strong>Answer from the Solution Manual</strong></h2>
<p>I looked up and find the following answer from the Solution Manual. However, I am confused by some of its steps.</p>
<blockquote>
<p>Since <span class="math-container">$x(\mathbf{p},w)$</span> is homogeneous of degree one with respect to <span class="math-container">$w$</span>, <span class="math-container">$x(\mathbf{p},\alpha w) = \alpha x(\mathbf{p},w)$</span> for every <span class="math-container">$\alpha &gt; 0$</span>. Thus, <span class="math-container">$x_l(\mathbf{p},w) = wx_l(\mathbf{p},1)$</span>. Since <span class="math-container">$\frac{\partial x_l(\mathbf{p},1)}{\partial p_k} = \frac{\partial\varphi_l(p)}{\partial p_k} = 0$</span> whenever <span class="math-container">$k \neq l$</span>, <span class="math-container">$x_l(\mathbf{p},1)$</span> is actually a function of <span class="math-container">$p_l$</span> alone. So we can write <span class="math-container">$x_l(\mathbf{p},w) =  x_l(p_l)$</span>. Since <span class="math-container">$x(\mathbf{p},w)$</span> is homogeneous of degree zero, <span class="math-container">$x_l(p_l)$</span> must be homogeneous of degree <span class="math-container">$-1$</span> (in <span class="math-container">$p_l$</span>). Hence, there exists <span class="math-container">$\alpha_l &gt; 0$</span> such that <span class="math-container">$x_l(p_l) = \frac{\alpha_l}{p_l}$</span>. By Walras' law, <span class="math-container">$\sum_{l}p_l\left(\frac{\alpha_l}{p_l}\right)w = w\sum_{l}\alpha_l = w$</span>. We must thus have <span class="math-container">$\sum_l\alpha_l = 1$</span>.</p>
</blockquote>
<p>I start to feel confused after it said that &quot;<span class="math-container">$x_l(\mathbf{p},1)$</span> is actually a function of <span class="math-container">$p_l$</span> alone.&quot;. For example, I do not think we can write <span class="math-container">$x_l(\mathbf{p},w) =  x_l(p_l)$</span>; I cannot see how <span class="math-container">$x_l(p_l)$</span> must be homogeneous of degree <span class="math-container">$-1$</span> (in <span class="math-container">$p_l$</span>); 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.</p>
<blockquote>
<p>Since <span class="math-container">$x(\mathbf{p},w)$</span> is homogeneous of degree one with respect to <span class="math-container">$w$</span>, <span class="math-container">$x(\mathbf{p},\alpha w) = \alpha x(\mathbf{p},w)$</span> for every <span class="math-container">$\alpha &gt; 0$</span>. Thus, <span class="math-container">$x_l(\mathbf{p},w) = wx_l(\mathbf{p},1)$</span>. Since <span class="math-container">$\frac{\partial x_l(\mathbf{p},1)}{\partial p_k} = \frac{\partial\varphi_l(p)}{\partial p_k} = 0$</span> whenever <span class="math-container">$k \neq l$</span>, <span class="math-container">$x_l(\mathbf{p},1)$</span> is actually a function of <span class="math-container">$p_l$</span> alone. Thus, we can write <span class="math-container">$x_l(\mathbf{p},w) = x_l(p_l,w)$</span>. Since <span class="math-container">$x(\mathbf{p},w)$</span> is homogeneous of degree zero and is homogeneous of degree one with respect to <span class="math-container">$w$</span>, we have
<span class="math-container">\begin{align*}
x_l(\alpha p_l, w) &amp;= x_l(p_1,\dots,\alpha p_l,\dots,p_L,w)\\
&amp;= 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)\\
&amp;= x_l\left(\frac{1}{\alpha}p_1,\dots,p_l,\dots,\frac{1}{\alpha}p_L,\frac{1}{\alpha}w\right)\\
&amp;= x_l\left(p_l,\frac{1}{\alpha}w\right)\\
&amp;= \frac{1}{\alpha}x_l(p_l,w).
\end{align*}</span>
Therefore, <span class="math-container">$x_l(p_l,w)$</span> is homogeneous of degree <span class="math-container">$-1$</span> with respect to <span class="math-container">$p_l$</span>. Hence,
<span class="math-container">\begin{align*}
x_l(\mathbf{p},w) &amp;= x_l(p_1,\dots,p_l,\dots,p_L,w)\\
&amp;= \frac{1}{p_l}x_l(p_1,\dots,1,\dots,p_L,w)\\
&amp;= \frac{w}{p_l}x_l(p_1,\dots,1,\dots,p_L,1).
\end{align*}</span>
Let <span class="math-container">$\alpha_l = x_l(p_1,\dots,1,\dots,p_L,1)$</span>. Since <span class="math-container">$x_l$</span> is independent of <span class="math-container">$p_k$</span> for all <span class="math-container">$k \neq l$</span>, we have that <span class="math-container">$\alpha_l$</span> is a constant independent of <span class="math-container">$(\mathbf{p},w)$</span>.</p>
</blockquote>
<h2><strong>My Questions</strong></h2>
<p>So, as I have mentioned earlier, I have two questions:</p>
<ol>
<li>If we were to follow my method (in the My Attempt section), is it possible for us to proceed to show that <span class="math-container">$\frac{\partial x_l(\mathbf{p},w)}{\partial w}p_l$</span> is a constant independent of <span class="math-container">$(\mathbf{p},w)$</span>?</li>
<li>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?</li>
</ol>
<p>Thank you very much in advance!</p>

CC BY-SA 4.0
2024-01-31T20:46:53.473Z
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 $p_l$.
Bertrand
<microeconomics><mathematical-economics><consumer-theory><theory><decision-theory>
MWG Exercise 2.E.5
Shenron
# MWG Exercise 2.E.5

(Asked by: username_0 on 2024-01-30)

## **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!

**Tags:** microeconomics, mathematical-economics, consumer-theory, theory, decision-theory

> **Comments:**
> **username_1**: 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 $p_l$.  

---

## 1 answers

### Answer by username_2 on 2024-01-30. Score: 1 | Accepted answer

**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. $p\_1$.

> **Comments:**
> **username_2**: I found a dozen typos before going to bed, I feel confident that there are more; feel free to edit.  
> **username_0**: 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?  
> **username_0**: 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?  
> **username_2**: Can you please give the definition of $x_1(p_1)$ as you understand it and do you understand what I wrote otherwise?  
> **username_2**: "*$\alpha_l$ 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 $w$?  
> **username_0**: As for "Answer to 1", do you mean the following: We have had $\alpha_l=\frac{\partial x_l(\mathbf{p},w)}{\partial w}p_l=x_l(\mathbf{p},1)p_l=x_l(p_l,1)p_l$, where the last equality comes from the fact that there is no cross-price effect. Since $x_l(\mathbf{p},w)=x_l(p_l,w)$ is homogeneous of degree 1 in $w$, we have $\alpha=x_l(p_l,1)p_l=x_l(p_l,p_l)=x_l(1,1)$, where the last equality holds because $x_l(\mathbf{p},w)=x_l(p_l,w)$ is homogeneous of degree 0. Therefore, $\alpha$ is a constant independent of $(\mathbf{p},w)$.  
> **username_0**: As for deriving $x_l(p_l)$ is homogeneous of degree $-1$ in your "Answer to 2", do you mean the following: The $x_l(p_l)$ in the Solution Manual should have been written as $x_l(p_l,1)$. Let $k\in\mathbb{R}_+$. Since, by the homogeneity of degree 0 and the homogeneity of degree 1 in $w$, $x_l(kp_l,1)=x_l(p_l,\frac{1}{k})=\frac{1}{k}x_l(p_l,1)$, we obtain that $x_l(p_l,1)$ is homogeneous of degree $-1$ in $p_l$. Basically, the effect of a $k$-percentage change in $p_l$ should offset the effect of a same $k$-percentage change in $w$, in order to make the demand function homogeneous of degree 0.  
> **username_2**: This is all correct except for "*in the Solution Manual should have been written*" because their current notation is also fine. They explicitly explain that they are introducing new notation. You don't have to like their notation, other people might, but the notation is defined, and the properties given for it are correct.  

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