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....


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.