<p>So, the indirect utility is simply (<span class="math-container">$p$</span> is a <span class="math-container">$L\times 1$</span> vector)</p>
<p><span class="math-container">$v(p,w) =  u[x(p,w)]$</span>,</p>
<p>Namely the utility function evaluated at the Marshallian demand.</p>
<p>First, notice that, if <span class="math-container">$x(p,w)$</span> (which is a <span class="math-container">$L\times 1$</span> vector as well) satisfies the Walras' Law:</p>
<p><span class="math-container">$p\cdot x(p,w) = w \ \ \ (1)$</span></p>
<p>Differentiate this with respect to <span class="math-container">$w$</span>, we get:</p>
<p><span class="math-container">$$ \sum_{l=1}^{L}\frac{\partial x_l(p,w)}{\partial w}p_l = 1 \ \ \ (2) $$</span></p>
<p>From the UMP, we know that <span class="math-container">$\partial u[x(p,w)] / \partial x_l = \lambda p_l$</span>, where <span class="math-container">$\lambda$</span> is the Lagrange multiplier.</p>
<p>Thus</p>
<p><span class="math-container">\begin{align}
\frac{\partial v(p,w)}{\partial w} &amp;= \sum_{l=1}^{L} \frac{\partial u[x(p,w)]}{\partial x_l(p,w)}\frac{\partial [x_l(p,w)]}{\partial w} \\
&amp;=  \lambda\sum_{l=1}^{L}\frac{\partial [x_l(p,w)]}{\partial w}p_l \\
&amp;= \lambda \ \ \ (3)
\end{align}</span></p>
<p>The last equality stems from <span class="math-container">$(2)$</span></p>
<p>Differentiating the Walras' Law <span class="math-container">$(1)$</span> w.r.t. <span class="math-container">$p_l$</span> (recall the product rule), we get:</p>
<p><span class="math-container">$$ x_l(p,w) + \sum_{k=1}^{L} p_k \frac{\partial x_k(p,w)}{\partial p_l} = 0 \ \ \ (4) $$</span></p>
<p>Now, differentiate <span class="math-container">$v(p,w)$</span> w.r.t. <span class="math-container">$p_l$</span>:</p>
<p><span class="math-container">\begin{align}
\frac{\partial v(p,w)}{\partial p_k} &amp;= \sum_{k=1}^{L}\frac{\partial u[x(p,w)]}{\partial x_k(p,w)} \frac{\partial x_k(p,w)}{\partial p_l} \\
&amp;= \lambda \sum_{k=1}^{L} p_k \frac{\partial x_k(p,w)}{\partial p_l} \\
&amp;= -\lambda x_l(p,w)
\end{align}</span></p>
<p>Et voilà! The Roy's Identity is proven</p>
<p><span class="math-container">$$ x_l(p,w) = -\frac{\partial v(p,w)/ \partial p_l}{\partial v(p,w) / \partial w}$$</span></p>

CC BY-SA 4.0
2022-10-17T18:22:08.873Z
2022-10-17T18:22:08.873Z
Matteo Bulgarelli
<p>i know that the roy's identity is: <span class="math-container">${\displaystyle x_{i}^{m}(p,w)=-{\frac {\frac {\partial v}{\partial p_{i}}}{\frac {\partial v}{\partial w}}}}$</span>
but i can't understand why it works.
why the fraction between partial derivatives respect to prices and income gives us the marshallian demand curve?</p>

CC BY-SA 4.0
2024-07-25T19:34:16.480Z
The intuition that I like to use is that: $w = \sum_i p_i x_i$, therefore $\frac{\partial w}{\partial p_i} = x_i$. And then you just put $\partial v$ on both the numerator and denominator to get $\frac{\partial v / \partial p_i}{\partial v / \partial w}$. This isn't exactly right and it's missing a minus sign, but at least this makes a little bit more sense in my head.
Vibius
<microeconomics><mathematical-economics><demand>
how and why Roy's equation works?
Giovanni
# how and why Roy's equation works?

(Asked by: username_0 on 2022-10-17)

i know that the roy's identity is: ${\displaystyle x\_{i}^{m}(p,w)=-{\frac {\frac {\partial v}{\partial p\_{i}}}{\frac {\partial v}{\partial w}}}}$
but i can't understand why it works.
why the fraction between partial derivatives respect to prices and income gives us the marshallian demand curve?

**Tags:** microeconomics, mathematical-economics, demand

> **Comments:**
> **username_1**: The intuition that I like to use is that: $w = \sum_i p_i x_i$, therefore $\frac{\partial w}{\partial p_i} = x_i$. And then you just put $\partial v$ on both the numerator and denominator to get $\frac{\partial v / \partial p_i}{\partial v / \partial w}$. This isn't exactly right and it's missing a minus sign, but at least this makes a little bit more sense in my head.  

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## 1 answers

### Answer by username_2 on 2022-10-17. Score: 3

So, the indirect utility is simply ($p$ is a $L\times 1$ vector)

$v(p,w) = u[x(p,w)]$,

Namely the utility function evaluated at the Marshallian demand.

First, notice that, if $x(p,w)$ (which is a $L\times 1$ vector as well) satisfies the Walras' Law:

$p\cdot x(p,w) = w \ \ \ (1)$

Differentiate this with respect to $w$, we get:

$$ \sum\_{l=1}^{L}\frac{\partial x\_l(p,w)}{\partial w}p\_l = 1 \ \ \ (2) $$

From the UMP, we know that $\partial u[x(p,w)] / \partial x\_l = \lambda p\_l$, where $\lambda$ is the Lagrange multiplier.

Thus

\begin{align}
\frac{\partial v(p,w)}{\partial w} &= \sum\_{l=1}^{L} \frac{\partial u[x(p,w)]}{\partial x\_l(p,w)}\frac{\partial [x\_l(p,w)]}{\partial w} \\
&= \lambda\sum\_{l=1}^{L}\frac{\partial [x\_l(p,w)]}{\partial w}p\_l \\
&= \lambda \ \ \ (3)
\end{align}

The last equality stems from $(2)$

Differentiating the Walras' Law $(1)$ w.r.t. $p\_l$ (recall the product rule), we get:

$$ x\_l(p,w) + \sum\_{k=1}^{L} p\_k \frac{\partial x\_k(p,w)}{\partial p\_l} = 0 \ \ \ (4) $$

Now, differentiate $v(p,w)$ w.r.t. $p\_l$:

\begin{align}
\frac{\partial v(p,w)}{\partial p\_k} &= \sum\_{k=1}^{L}\frac{\partial u[x(p,w)]}{\partial x\_k(p,w)} \frac{\partial x\_k(p,w)}{\partial p\_l} \\
&= \lambda \sum\_{k=1}^{L} p\_k \frac{\partial x\_k(p,w)}{\partial p\_l} \\
&= -\lambda x\_l(p,w)
\end{align}

Et voilà! The Roy's Identity is proven

$$ x\_l(p,w) = -\frac{\partial v(p,w)/ \partial p\_l}{\partial v(p,w) / \partial w}$$


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