The remaining proofs would surely have been easier to write.
06.12.2025 18:14 β π 2 π 0 π¬ 0 π 0@elliotlip.bsky.social
To be specific, in the Dawes, Tomes, Mousely, Grubbs Fidelity Fiduciary Bank. http://elliotlipnowski.com/
The remaining proofs would surely have been easier to write.
06.12.2025 18:14 β π 2 π 0 π¬ 0 π 0Aren't we all
05.12.2025 16:43 β π 1 π 0 π¬ 0 π 0It's equilibrium existence, so fitting that it's about as popular at the party as a theorist is.
05.12.2025 16:43 β π 2 π 0 π¬ 1 π 0Did you work more on this after first posting?
I think I tried it at the time (working on this paper was what prompted my question), and the flow chart it gave was way sparser than this one (which is pretty great though not complete).
Incredibly good??
05.12.2025 15:42 β π 1 π 0 π¬ 3 π 0New paper!
elliotlipnowski.com/wp-content/u...
Please join us on 20/Nov to hear Marina Halac (Yale) present "Agreeing to Implement", a joint work with Elliot Lipnowski and Doron Ravid. Panellists: Matteo Camboni and Sam Kapon.
@elliotlip.bsky.social @doronravid.bsky.social @samkapon.bsky.social
Iβm sorry to learn your baby isnβt a theorist
30.09.2025 09:28 β π 2 π 0 π¬ 0 π 0That 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.
28.09.2025 11:48 β π 1 π 0 π¬ 0 π 0#mathsky notation question. We feel totally comfortable with a decoration being a function (e.g., for a number x, let \bar{x}:=x^2). But what about a case/font/format change as a function (e.g., for a number x, let \mathbf{x} denote the constant function on [0,1] taking value x)?
21.09.2025 11:22 β π 5 π 1 π¬ 6 π 0Jimmy Kimmel has learned his lesson.
by Susan Collins
The same mathematical result, not the same result as the sex. Perverts.
17.09.2025 20:55 β π 2 π 0 π¬ 0 π 0Sex is great but have you tried replacing annoying quantitative bookkeeping arguments in a proof with qualitative arguments to get the same result??
17.09.2025 20:54 β π 2 π 0 π¬ 3 π 0I think thereβs still an opportunity for price discrimination since the thing youβre screening on is how much time people are happy to spend on consuming the same content. And extra steps take timeβso itβs similar to coupon clipping as price discrimination.
13.09.2025 15:42 β π 1 π 0 π¬ 0 π 0My 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.
13.09.2025 11:11 β π 0 π 0 π¬ 1 π 0This could be an argument for fairness of the "=rp" answer.
12.09.2025 23:43 β π 1 π 0 π¬ 0 π 0Your 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?
12.09.2025 23:14 β π 0 π 0 π¬ 1 π 0I think I disagree, but Iβm curious about your reasoning.
12.09.2025 22:50 β π 0 π 0 π¬ 1 π 0Thank you!
12.09.2025 21:32 β π 1 π 0 π¬ 0 π 0Fun 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 <rp?
12.09.2025 21:23 β π 2 π 0 π¬ 1 π 0Spotify account gives a budget of audiobook hours before needing to pay for more hours, but if you listen to a 10-hour book at double-speed (so 5 listening hours), it counts as 10 hours of your budget. So they price r-speed listening at exactly r times that of unit-speed listening.
12.09.2025 21:23 β π 2 π 0 π¬ 2 π 0Wednesday Addams is fashionable af
06.09.2025 22:50 β π 2 π 0 π¬ 0 π 0Boat question.
15.08.2025 15:55 β π 408 π 58 π¬ 4 π 5Schumer, Jeffries propose strongly-worded letter to counter RFK Jr.'s purge at the CDC.
29.08.2025 03:14 β π 901 π 79 π¬ 46 π 12I was under the impression these were called "siblings"
28.08.2025 19:37 β π 2 π 0 π¬ 0 π 0I didnβt know this!
14.08.2025 02:00 β π 0 π 0 π¬ 0 π 0I should check!
14.08.2025 01:59 β π 1 π 0 π¬ 0 π 0A 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!
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 Ξ(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 Ξ(A^S).
So it all comes down to what A^S is.
Has anybody developed the theory of STRICT correlated equilibria of a finite game?
#econsky #mathsky #ECsky #TCSsky
I was playing around with it a bit, and here are some things that are easy to show about the set CE^S of strict correlated equilibrium distributions.