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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....

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1 Answer 1

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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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