<p>You are right, in this case, we use <em>Roy's Identity</em>:</p>
<p><span class="math-container">$$
x_{l}(p, w)=-\frac{\frac{\partial v(p, w)}{\partial p_{l}}}{\frac{\partial v(p, w)}{\partial w}}
$$</span></p>
<p>Since we want to show this for <span class="math-container">$w=1$</span>, we need to could rewrite it for this particular case as follows:</p>
<p><span class="math-container">$$
x_{l}(p, 1)=\frac{\frac{\partial v(p, 1)}{\partial p_{l}}}{-\frac{\partial v(p, 1)}{\partial w}}
$$</span></p>
<p>To sum up, it is enough to show that:</p>
<p><span class="math-container">$$
-\frac{\partial v(p, 1)}{\partial w}=\sum_{k=1}^{L} p_{k} \frac{\partial v(p, 1)}{\partial p_{k}}
$$</span></p>
<p>In order to do so, we use <em>Walras' Law</em>:</p>
<p><span class="math-container">$$
\begin{aligned}
\sum_{k=1}^{L} p_{k} x_{k}(p, w) &amp; =w \\
\sum_{k=1}^{L} p_{k}\left(-\frac{\frac{\partial v(p, w)}{\partial p_{k}}}{\frac{\partial v(p, w)}{\partial w}}\right) &amp; =w \\
\frac{-1}{\frac{\partial v(p, w)}{\partial w}}\left(\sum_{k=1}^{L} p_{k} \frac{\partial v(p, w)}{\partial p_{k}}\right) &amp; =w \\
\sum_{k=1}^{L} p_{k} \frac{\partial v(p, w)}{\partial p_{k}} &amp; =-w \frac{\partial v(p, w)}{\partial w}
\end{aligned}
$$</span></p>
<p>Setting <span class="math-container">$w=1$</span> we get exactly what we wanted to show.</p>

CC BY-SA 4.0
2024-06-19T11:40:46.963Z
2024-06-19T11:40:46.963Z
Lorenzo Castagno
<p>I need some hint in here:</p>
<p>I am trying to show that an individual that has preferences that satisfy the usual assumptions: that is, use the assumptions, this is satisfied:</p>
<p><span class="math-container">$$
x_{l}(p, 1)=\frac{\frac{\partial v(p, 1)}{\partial p_{l}}}{\sum_{k=1}^{L} p_{k} \frac{\partial v(p, 1)}{\partial p_{k}}} \quad \forall l=1, \ldots, L
$$</span></p>
<p>I have been thinking to use Roy's identity, but I am missing something....</p>

<microeconomics><game-theory>
Using Roy's Lemma and Walras' Law to show that
DanIsabel
# Using Roy's Lemma and Walras' Law to show that

(Asked by: username_0 on 2024-06-19)

I need some hint in here:

I am trying to show that an individual that has preferences that satisfy the usual assumptions: that is, use the assumptions, this is satisfied:

$$
x\_{l}(p, 1)=\frac{\frac{\partial v(p, 1)}{\partial p\_{l}}}{\sum\_{k=1}^{L} p\_{k} \frac{\partial v(p, 1)}{\partial p\_{k}}} \quad \forall l=1, \ldots, L
$$

I have been thinking to use Roy's identity, but I am missing something....

**Tags:** microeconomics, game-theory

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## 1 answers

### Answer by username_1 on 2024-06-19. Score: 6 | Accepted answer

You are right, in this case, we use *Roy's Identity*:

$$
x\_{l}(p, w)=-\frac{\frac{\partial v(p, w)}{\partial p\_{l}}}{\frac{\partial v(p, w)}{\partial w}}
$$

Since we want to show this for $w=1$, we need to could rewrite it for this particular case as follows:

$$
x\_{l}(p, 1)=\frac{\frac{\partial v(p, 1)}{\partial p\_{l}}}{-\frac{\partial v(p, 1)}{\partial w}}
$$

To sum up, it is enough to show that:

$$
-\frac{\partial v(p, 1)}{\partial w}=\sum\_{k=1}^{L} p\_{k} \frac{\partial v(p, 1)}{\partial p\_{k}}
$$

In order to do so, we use *Walras' Law*:

$$
\begin{aligned}
\sum\_{k=1}^{L} p\_{k} x\_{k}(p, w) & =w \\
\sum\_{k=1}^{L} p\_{k}\left(-\frac{\frac{\partial v(p, w)}{\partial p\_{k}}}{\frac{\partial v(p, w)}{\partial w}}\right) & =w \\
\frac{-1}{\frac{\partial v(p, w)}{\partial w}}\left(\sum\_{k=1}^{L} p\_{k} \frac{\partial v(p, w)}{\partial p\_{k}}\right) & =w \\
\sum\_{k=1}^{L} p\_{k} \frac{\partial v(p, w)}{\partial p\_{k}} & =-w \frac{\partial v(p, w)}{\partial w}
\end{aligned}
$$

Setting $w=1$ we get exactly what we wanted to show.


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