1
$\begingroup$

I am struggling a bit with the math in my first graduate microeconomics course. I'm not sure if this belongs here. If it doesn't, please direct me to a more appropriate place.

Below is one question along the with provided solution. The red part is the part I don't follow.

Solution image

I don't understand how the derivative of $v(p, e(p,\overline{u}))$ w.r.t. the price vector can evaluate to the LHS of the given equation.

Perhaps I'm confused by the vector notation, so any step explaining this equation would be very much appreciated.

$\endgroup$
3
$\begingroup$

a) The convention, when differentiating a real valued function $v$ wrt to a column vector $p$ of dimension $(G \times 1)$, is that: $$ \nabla _{p}v\left( p,w\right) \equiv \frac{\partial v}{\partial p}\left( p,w\right) =\left( \begin{array}{c} \frac{\partial v}{\partial p_{1}}\left( p,w\right) \\ \vdots \\ \frac{\partial v}{\partial p_{G}}\left( p,w\right) \end{array}% \right). $$

b) When expenditures $w$ are not constant, the column vector $p$ with respect to which we take the derivative occurs twice in $v(p, e(p,\overline{u}))$, and the chain rule has to be applied in order to obtain the total effect of the price change on the indirect utility: the first partial effect on utility is due to the price changes, and the second effect is through the adjustment in expenditures implied by the price change. Hence the expression: \begin{eqnarray*} &&\frac{\partial v}{\partial p}\left( p,e\left( p,u\right) \right) +\frac{% \partial v}{\partial w}\left( p,e\left( p,u\right) \right) \frac{\partial e}{% \partial p}\left( p,u\right) \\ &=&\left( \begin{array}{c} \frac{\partial v}{\partial p_{1}}\left( p,e(p,u)\right) \\ \vdots \\ \frac{\partial v}{\partial p_{G}}\left( p,e(p,u)\right) \end{array}% \right) +\frac{\partial v}{\partial w}\left( p,e(p,u)\right) \left( \begin{array}{c} \frac{\partial e}{\partial p_{1}}\left( p,u\right) \\ \vdots \\ \frac{\partial e}{\partial p_{G}}\left( p,u\right) \end{array}% \right). \end{eqnarray*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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