Elliot Lipnowski
banner
elliotlip.bsky.social
Elliot Lipnowski
@elliotlip.bsky.social
To be specific, in the Dawes, Tomes, Mousely, Grubbs Fidelity Fiduciary Bank.

http://elliotlipnowski.com/
I’m sorry to learn your baby isn’t a theorist
September 30, 2025 at 9:28 AM
That is good too! But I meant noticing you can replace an argument with a bunch of derivative-taking and epsilon-counting to one where an application of Tarski or Krein-Milman does most of the work.
September 28, 2025 at 11:48 AM
The same mathematical result, not the same result as the sex. Perverts.
September 17, 2025 at 8:55 PM
I think there’s still an opportunity for price discrimination since the thing you’re screening on is how much time people are happy to spend on consuming the same content. And extra steps take time—so it’s similar to coupon clipping as price discrimination.
September 13, 2025 at 3:42 PM
My reasoning: the marginal cost is negligible relative to revenue, so it's a demand question. I think charging more to listen double-speed is effective price discrimination, since I would guess the people that want the fast version are less price-elastic.
September 13, 2025 at 11:11 AM
This could be an argument for fairness of the "=rp" answer.
September 12, 2025 at 11:43 PM
Your way of phrasing is way better: if I listen to a book at double speed, should Spotify want to charge me more (>rp), less (<rp), or the same (=rp) as if I listen at regular speed?
September 12, 2025 at 11:14 PM
I think I disagree, but I’m curious about your reasoning.
September 12, 2025 at 10:50 PM
Thank you!
September 12, 2025 at 9:32 PM
Fun micro #econsky problem set question: If p is the price Spotify charges for unit-speed listening (replacing a budget with a price for simplicity) and r>1, should they price r-speed listening =rp, >rp, or
September 12, 2025 at 9:23 PM
I was under the impression these were called "siblings"
August 28, 2025 at 7:37 PM
I didn’t know this!
August 14, 2025 at 2:00 AM
I should check!
August 14, 2025 at 1:59 AM
A kind of mechanical approach modifies Hart and Schmeidler's program to check, for every product set B of action profiles, whether some strict CE has marginal supports B_i. Then A^S is the largest B that passes the test (or empty if nothing does).

But there must be something more interpretable!
August 13, 2025 at 1:44 PM
1) The set CE^S is convex.
2) There's a product set A^S of action profiles such that (i) CE^S is contained in Δ(A^S), and (ii) if CE^S is nonempty, then something in CE^S has support A^S.
3) The closure of CE^S is just the intersection of CE and Δ(A^S).

So it all comes down to what A^S is.
August 13, 2025 at 1:44 PM
Thank you! This is all super helpful.
August 2, 2025 at 10:12 AM