0
$\begingroup$

I am not sure how to word this question. I understand the envelop theorem pretty well and as a result Shepard's Lemma. The proof is relatively straightforward. I am getting stuck on the intuition behind integrating the Hicksian demand function and getting the (total) expenditure function as a result. It seems like this would be some sort of reversal of the envelope theorem.

Suppose we have a two good world. A typical Hicksian demand function can be written as a function of p1, p2, and utility. Using the fundamental theorem of calculus and the envelop theorem, I know that integrating this function over an interval with respect to p1 or p2 will give us back the total expenditure function. So if we integrate over a range with respect to p1, for example, it gives us the change in total expenditure. What's not clear to me is how expenditures on good 2 are being included (intuitively).

In a single-variable integral, it is easy to understand this intuitively. If we have x and we integrate a function f(x) over an interval it is easy to understand this as adding up the infinitely small changes times the rate of change of f'(x) to get the "total" change in f(x). In the Hicksian case, we seem to be adding up the infinitely small changes in price of good 1 times the rate of change (Hicksian demand for good 1) and assuming expenditures on good 2 are constant when they almost certainly wouldn't be. Is there a better way to understand this? Are expenditures on good 2 implicitly included?

$\endgroup$
2
$\begingroup$

Expenditures on good $2$ are, indeed, included and can vary.

We know from Shephard's lemma that whenever the marginal change in expenditure for good $1$ with respect to its price varies with the price of good $1$, the Hicksian demand for good $1$ must vary too. But since utility is fixed, changes in the Hicksian demand for good $1$ require changes in the Hicksian demand for good $2$, a move along the indifference curve. And since we do not change the price of good $2$, the expenditure on good $2$ must vary too.

$\endgroup$

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.