As I reviewed producer theory, I found a couple of things that I should have asked you a month ago:

1 On page 60 of the latest slides, I found the first bullet used to be “Closure ensures that \pi(p) is actually achieved.” Is it because the unboundedness or the asymptotic case might still prevent the non-emptiness of the correspondence even though Y is closed?

Yes.

2 On page 50, the proof for homogeneity of degree one for the profit function is assumes the existence of maximum. I wonder why we would do this here.

The “max” should be a “sup.” Otherwise, the proof goes through precisely. Good catch.

3 On page 51, in the proof for homogeneity of y, can I argue that

\argmax_{y \in Y} \lambda p \cdot y = \lambda \argmax_{y \in Y} p \cdot y = \lambda y(p) ?

I will have to assume argmax exists, though.

No… and the conclusion is untrue! (Note your proof is that y(.) is homogeneous of degree one, when it’s actually homogeneous of degree zero.)

When you “take the lambda out of the argmax,” it doesn’t appear in front, it just disappears. Think of argmax (-2 x^2) = argmax (-x^2). (In this manner, you can cancel any strictly increasing monotone transformation from an argmax or argmin.)

FollowupSorry, some questions are not well formulated.

For 3, can I write

\argmax_{y \in Y} \lambda p \cdot y = \argmax_{y \in Y} p \cdot y ?

Because the objective function is just a strictly monotone transformation.

Yes

4. On page 49, can I use the same proof to demonstrate that if f(\cdot) is homogeneous of degree k, then its gradient is homogeneous of degree k-1?

Yes.

5. On page 77, the substitution matrix. I know the diagonal elements must be non-negative. However, I put down in my notes that the derivatives of the output with respect to input prices must be negative. I wonder if that’s true and if so, why.

We would generally expect this to be the case (and I may have said as much in lecture), but it need not be true. Consider the following (non-differentiable) example:

Y = {(0,0,0) , (2,-1,0) , (3,0,-3) }

Then y(3,2,2) = {(2,-1,0)} and pi(3,1,2) = 4

while y(3,4,2) = {(3,0,-3)} and pi(3,4,2) = 3

That is, an increase in an input price lead to an increase in output.

6. On pages 120 and 121, why do we say X*(t) is non-decreasing in t in “stronger set order” rather than “strong set order”? Is it because of multidimensionality of X?

Yes. Strong set order means that a in A and b in B implies min{a,b} in A and max{a,b} in B. Stronger set order means that a in A and b in B implies meet{a,b} in A and join{a,b} in B. “min” and “max” are not well-defined operations on partially ordered sets such as R^n for n > 1.

If X is lattice and partially ordered, can we say that each variable x_i is non-decreasing in t? I think so because of the positive feedback loop among all variables.

I don’t understand the premise of the question.

In my notes I put down something “F is supermodular in (x_1, x_2, …, x_n, t_1, t_2, …, t_m”, rather than the stated conditions in the theorem. Does it matter?

This is the stated condition in the first version I stated of multivariate Topkis’ Theorem (“F is supermodular”). As I state on the next slide, this is a stronger than is required… we can dispense with ID between t_i and t_j.

Also, it seems the notes only provided the Multivariate Topkis Thm when t is fully ordered, i.e., only the version on page120. In that case, do we still need to check if F has increasing differences in x_i and t ? Intuitively I definitely should, but I wonder why it’s not in the theorem.

I am confused. You always need to consider ID in x_i and t_j, as well as in x_i and x_j. Not sure what you mean that “it’s not in the theorem.”

FollowupFor 6, I meant to say, for example, if X has 3 variables x_1, x_2, x_3, and we have two parameters t_1 and t_2. If F is supermodular in all x_i, and has ID in all (x_i , t_j), then by Multivariate Topkis X*(t) is non-decreasing in t in the stronger set order.

So can we conclude x_1 increases in strong set order, x_2 increases in strong set order, and x_3 increases in strong set order, in t= (t_1 and t_2) ?

Intuitively I think we can, at least weakly so. Like in the beer, wine, wealth example. Otherwise many previous analysis wouldn’t go through. But I don’t know how to think about proving it even heuristically. If X increases in t in Ser SO, how can I conclude each element of X, x_i, is increasing in t in SSO?

I think we can. Suppose t’ > t; then by Topkis x*(t’) >= x*(t) in the stronger SO. That is, for any (x1 x2) in x*(t) and (x1′ x2′) in x*(t’), we have (max{x1, x1′} max{x2, x2′}) in x*(t’) and (min{x1, x1′} min{x2, x2′}) in x*(t’). Thus max{x1 x1′} in x1*(t’) and min{x1 x1′} in x1*(t). Thus x1*() is increasing in t in the strong SO.

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