Wolf Prize laureate Prof. George Daniel Mostow made his greatest scientific breakthrough while driving.

Haaretz tells the story of how Dan Mostow reached his breakthrough known as Mostow’s rigidity theorem.

Congratulations, Dan!

French-Isreali Meeting and Günterfest

More updates: If you are in Paris On Wednesday and Thursday this week there will be a lovely French-Isreali interacademic meeting on mathematics.  The problem is very interesting, and I will give a talk quite similar to my recent MIT talk on quantum computers.

In the  weekend  we will celebrate Günter Ziegler’s 50th birthday in Berlin. Günter started very very young so we had to wait long for this.

You run a single-item sealed bid auction where you sell an old camera. There are three bidders and the value of the camera for each of them is described by a certain (known) random variable: With probability 0.9 the value is 100+x where x is taken uniformly at random from the interval [-1,1]. With probability 0.1 the value is 300+x where x is as before. The 300 value represents the case that the item has a special nostalgic value for the buyer.

The values of the camera to the three bidders are thus i.i.d random variables. (The reason for adding this small random x is to avoid ties, and you can replace the interval [-1,1] with [-ε, ε] for a small ε, if you wish.) If you don’t want to worry about ties you can simply think about the value being 100 with probability 0.9 and 300 wit probability 0.1.

The basic question

The basic questions for you the seller is:

What is the highest expected revenue you, the seller, can guarantee and what is your optimal allocation policy.

I’d like to see the answer for this question. But let me challenge your intuition with two simpler questions.

1) Can the seller guarantee an expected revenue of 120  or more?

2) What (roughly) is the optimal allocation policy

a) Highest bidder wins.

b) Highest bidder wins if his bid passes some reserve price.

c) The item is allocated at random to the bidders with probabilities depending on their bids.

Background: Myerson’s paper and his revenue equivalence theorem

The most relevant paper to this post is a well-known paper Optimal auction design by Roger Myerson. Myerson considered the case of a single-item sealed-bid auction where the bidders’ values for the item are independent identical random variable.

Note that I did not specify the price that the winning bidder is going to pay for the camera. The reason is that according to a famous theorem by Myerson (referred to as the revenue equivalence theorem), when the bidders are strategic, the expected revenues for the seller are determined by the allocation rule and are independent from the pricing policy! (Well, you need to assume a reasonable pricing policy…)

For example, if the item is allocated to the highest bidder then the expected price will be the second highest price. If the price is determined by the second highest bid (Vickery’s auction) then each bidder has no incentive to give a bid which is different from its value. But even if the item will be allocated to the first bidder for the highest price, the expected revenues will still be the same! When you analyze an auction mechanism like ours you can assume that the payments are set in a way that the bidders have no incentive not to bid the true value the camera has. This is sometimes referred to as the revelation principle.

Myerson considered a mechanism which sometimes lead to higher revenues compared to allocating the item to the highest bidder: The seller set a reserve price and the item is allocated to the highest bidder if the bid passes this reserve price, and is not allocated at all otherwise. In the paper Myerson also considered more complicated allocation rules which are important in the analysis where the item is allocated to bidders according to some probabilities based on their bids.

Polls

This time we have two questions and two polls:

Once again this is a game-theory question. I hope it will lead to interesting discussion like the earlier one on tic-tac-toe.

A little more Background: Auctions and Vickery’s second price auction.

(From Wikipedia) An auction is a process of buying and selling goods or services by offering them up for bid, taking bids, and then selling the item to the highest bidder. In economic theory, an auction may refer to any mechanism or set of trading rules for exchange.

In our case, we consider an auction of a single item (the camera) and each bidder is giving a sealed bid.

(Again from Wikipedea) A Vickrey auction is a type of sealed-bid auction, where bidders submit written bids without knowing the bid of the other people in the auction, and in which the highest bidder wins, but the price paid is the second-highest bid. The auction was first described academically by Columbia University professor William Vickrey in 1961 though it had been used by stamp collectors since 1893.^[2] This type of auction is strategically similar to an English auction, and gives bidders an incentive to bid their true value.

A commentator named Oz proposed the following question: You have a box with n red balls and n blue balls. You take out each time a ball at random but, if the ball was red, you put it back in the box and take out a blue ball. If the ball was blue, you put it back in the box and take out a red ball.

You keep doing it until left only with balls of the same color. How many balls will be left (as a function of n)?

Peter Shor wrote in a comment “I’m fairly sure that there is not enough bias to get , but it intuitively seems far too much bias to still be . I want to say . At a wild guess, it’s either or , since those are the simplest exponents between  and .”  The comment followed by a heuristic argument of Kevin Kostelo and computer experiments by Lior Silberman that supported the answer .

This is correct!

Good intuition, Peter!

In our student probability day a couple of weeks ago, Yuval Peres told me the origin of this problem. The way it was originally posed by D. Williams and P. McIlroy used roughly the following story (I modified it a little): There are two groups of n gunmen that shoot at each other. Once a gunman is hit he stops shooting, and leaves the place happily and peacefully. How many gunmen will be left after all gunmen in one team had left. The problem was solved by Kingman and Volkov in this paper.

J. F. C. Kingman, and S. E. Volkov,  Solution to the OK Corral model via decoupling of Friedman’s urn, J. Theoret. Probab. 16 (2003), no. 1, 267–276. A nice presentation of the result entitled: Internal erosion and the exponent 3/4 was given by Lionel Levine and Yuval Peres.

           

Yeshu Kolodni and Lord Kelvin

The question

In 1862, the physicist William Thomson (who later became Lord Kelvin) of Glasgow published calculations that fixed the age of Earth at between 20 million and 400 million years. Later in the 1890s Kelvin calculated the age of Earth by using thermal gradients, and arrived at an estimate of 100 million years old which he later reduced to 20 million years. (For much more interesting details see this Wikipedia article.)

The question was: what was the main reason for Lord Kelvin’s wrong estimation

a) Radioactivity – Heat produced by radioactive decay; this was a process unknown to science for decades to come.

b) Convection – The transfer of heat not through radiation or heat-conduction but through the movement of hot parts to the surface; this is a process familiar in home cooking.

Here is the answer and some discussion mainly based on what Yeshu Kolodny have told me.

The short answer: Convection

The short answer is that the main mistake in Lord Kelvin’s estimates was convection. 

This was also the most popular answer in our poll with 48%, radioactivity got 44%, and 6% voted for both these elements have similar effect.

The longer answer and Kelvin’s computation of the age of the sun

There is more to be said, of course. And let me from now on simply quote what Yeshu told me directly:

“Kelvin (previously Thomson) was certainly one of the greatest physicists and engineers of the 19th century. He directed the laying of the first transatlantic telegraph cable, formulated the second law of thermodynamics and wrote hundreds of influential papers. At very young age he was one of the first to study and understand Fourier’s heat-flow treatment, and it is this that led him  to calculating first the age of the sun, then that of the earth. In the age of the sun a major error was obviously neglecting heat production by nuclear reactions; in the age of he earth – convection.

Kelvin assumed that the source of the Sun’s energy was gravitational – a collapse (accumulation) of many meteors, and thus conversion of mechanical energy into heat (Joule’s heat equivalent was already known). This way he calculated the Sun’s age first to be around 32 millions years, later put an upper limit of 300 million years and a “most probable age of 100 million years). Kelvin (correctly) assumed that the earth cannot be older than the sun, and is most likely of the same age. The prevalent hypothesis for the formation of Sun and planet formation was that of Kant-Laplace, which involved  the above-mentioned “nebular  hypothesis” (an improved version of it is still accepted today). So  when he got the “age of the earth” to be within error of what he assumed to be the “age of the sun” both arrived by different approaches, he was very happy.

The geologists victory and where was Lord Kelvin right

It is rather impressive that this genius of physics was opposed by a bunch of geologists (high boots, shorts, field hammers, limited perspective) and that the latter turned out right, but it is also true that Kelvin showed two important points:

The first is that the age of the earth is NOT INFINITE as many thought in the 19th century, and the second is that the age of the earth is calculable from physical principles. At that time, several geologists claimed that “physics cannot be applied to geology,” and in this they were wrong! We owe him much for teaching us these two points.”

What an exciting, spirit-lifting story it is! And there are more facets to it.  Of course, radioactivity played a major role in modern estimates for the age of the earth.

The age of the earth

(Thanks to Yeshu Kolodny) We now know that the age of the earth is 4.54±1% Billion years.

From Wikipedea: In 1862, the physicist William Thomson (who later became Lord Kelvin) of Glasgow published calculations that fixed the age of Earth at between 20 million and 400 million years. He assumed that Earth had formed as a completely molten object, and determined the amount of time it would take for the near-surface to cool to its present temperature. His calculations did not account for heat produced via radioactive decay (a process then unknown to science) or convection inside the Earth, which allows more heat to escape from the interior to warm rocks near the surface.

Test your intuition/knowledge

What was the main reason for Lord Kelvin’s wrong estimation

a) Radioactivity – Heat produced by radioactive decay; this was a process unknown to science for decades to come.

b) Convection – The transfer of heat not through radiation or heat-conduction but through the movement of hot parts to the surface; this is a process familiar in home cooking.

Here is a link to a nice physics “test your intuition” over The Quantum Pontiff

We gave it a tentative name, “Shmulik.” Since Shmulik belongs to  a protected species, a person from the authority for wild animals came and took it to be freed in nature in the Cedar Valley near Jerusalem.

Margulis’ paper

Ramanujan graphs were constructed independently by Margulis and by Lubotzky, Philips and Sarnak (who also coined the name). The picture above shows Margulis’ paper where the graphs are defined and their girth is studied. (I will come back to the question about girth at the end of the post.) In a subsequent paper Margulis used the girth property in order to construct efficient error-correcting codes. (Later Sipser and Spielman realized how to use the expansion property for this purpose.)

The purpose of this post is to briefly tell you about new Ramanujan graphs exhibited by Adam Marcus, Daniel Spielman, and Nikhil Srivastava. Here is the paper. This construction is remarkable for several reasons: First, it is the first elementary proof for the existence of Ramanujan graphs which also shows, for the first time, that there are k-regular Ramanujan graphs (with many vertices)  when k is not q+1, and q is a prime power. Second, the construction uses a novel “greedy”-method (with further promised fruits) based on identifying classes of polynomials with interlacing real roots, that does not lead (so far) to an algorithm (neither deterministic nor randomized). Third, the construction relies on Nati Linial’s idea of random graph liftings and verify (a special case of) a beautiful conjecture of Yonatan Bilu and Linial. 

The Ramanujan property

Let us consider the adjacency matrix of a d-regular graph G with n vertices. The trivial eigenvalues of the graph are d and -d; d is always an eigenvalue and -d is an eigenvalue if and only if G is bi-partite. A graph is called a Ramanujan graph if all its nontrivial eigenvalues belong to the interval . A theorem of Alon and Boppana asserts that  for every there are only finitely many d-regular graphs that all their non-trivial eigenvalues have absolute value less than . Complete graphs and complete bipartite graphs are Ramanujan and the challenge is to find Ramanujan graphs where the degree d is fixed and the number of vertices n is large. The Ramanujan property proofed by Margulis and LPS for their graphs  relies on deep number theory.

Ramanujan graphs are superb expanders. Expander graphs can be defined as d-regular graphs for which all nontrivial eigenvalues belong to the interval where is fixed. Expander graphs were first defined (using a combinatorial expansion property) by Pinsker who also gave a probabilistic construction. The first explicit construction was achieved by Margulis. The original Ramanujan graphs (both bipartite and not) by Margulis and LPS had degrees which were of the form p+1, where p is a prime. A construction where the degree is q+1 for every prime power q was achieved by Moshe Morgenstern.

2-Lifts and the Bilu-Linial conjecture

Starting from a graph G, a 2-lift of G is obtained by replacing every vertes of G by a pair of vertices, and replacing every edge of G by two disjoint edges among the corresponding new vertices. There are two ways that this can be done. Linial and Bilu showd that the eigenvalues of the new graph are obtained by adding to the eigenvalues of the old graph certain eigenvalues of a matrix which encode the lifting. They conjectured that every d-regular graph has a 2-lifting where all the new eigenvalues are in the interval . Markus, Spielman, and Srivastava proved the conjecture for bipartite graphs.

A few words on the proof:

The proof is not difficult and it gives certainly an elementary proof that Ramanujan d-regular graphs exist, and, for the first time, also for every value of d.  The proof relies in a crucial way on the matching polynomial of a graph which was defined by Heilmann and Lieb. The method involves studying interlacing classes of polynomials with real roots. The very basic skeleton of the proof is this: you show that the sum (or average) of all adjacency polynomials for 2-lifts lies in a restricted interval. Then you use the interlacing property to deduce that there is  a single adjacency matrix with the required property. The interval is where the eigenvalues of the infinite d-regular tree lies. (This gives a clear motivation for the concept of Ramanujan graphs and various generalizations, including to Ramanujan (high-dimensional) complexes.”) MSS notice a pleasant feature of their proof that the existence of Ramanujan graphs is directly connected to the bounds on the spectral radius of the infinite d-regular tree.

Moore bounds

Let me briefly talk about something else. Let be a -regular graph with $n$ vertices and girth where is an odd integer. Recall that the girth of a graph is the length of the smallest cycle. We can ask: what is the minimum number of vertices must have. Put . Let’s look at one vertex . It has neighbors, let be the set of neighbors. Every neighbor in has “new” neighbors lets be the set of all those. We can continue this way and if there are no cycles of length smaller than all these vertices are different for . So we identified $1+k+k(k-1)+\dots+k(k-1)^{m-1}$ different vertices. This gives a lower bound known as “Moore bound”.

Erdos and Sachs used the probabilistic method to construct graphs with .

Margulis’ and LPS’ papers gave a construction of graphs with .

What is the truth?

So the correct coefficient is somewhere between 4/3 and 2. What is the truth? Is it possible to construct graphs with ?

I am aware of two general arguments (both proposed by Nati Linial) suggesting opposite answers to the question if the right answer is 2 or below 2:

1) [For <2:] The girth problem is analogous to the rate of error correcting codes. Moore bounds is just the volume bound which is usually not asymptotically optimal. Better bound is expected but may require some conceptually new ideas.

2) [For 2:] Regular graphs with large girth are analogous to linear error correcting codes on the infinite regular tree. If you consider the analogue of general codes (drop linearity) then it is easy to reach the 2 upperbound. Usually the behavior of linear and general error correcting codes is similar.

So you can test your own intuition regarding this problem. Unfortunately, a definite answer does not look immanent.

“There is an interesting (and more difficult) variation I once heard but can’t recall where:

You have a box with n red balls and n blue balls. You take out each time a ball at random as before. But, if the ball was red, you put it back in the box and take out a blue ball. If the ball was blue, you put it back in the box and take out a red ball.

You keep as before until left only with balls of the same color. How many balls will be left (as a function of n)?

1) Roughly  εn for some ε>0.

2) Roughly ?

3) Roughly log n?

4) Roughly a constant?

5) Some other behavior

1) Roughly  εn for some ε>0.

2) Roughly ?

3) Roughly log n?

4) Roughly a constant?

Here is the collective intuition regarding this problem

If you have two different boxes A with n red balls, and B with n blue balls and you take out at random with equal probabilities a ball from box A or a ball from box B, then when only one color is left the number of balls left is roughly .  But in our case, when there are more red balls left it is more likely that we take out a red ball. Precisely, if we think about our problem but leaves the red and blue balls in different boxes,  when there are a balls left in box A, and b balls left in box B you choose to take out another box from box A with probability a/(a+b). This seems to push the number of balls of each color closer together but by how much?

You can look at your drawing process as a random permutation of n red balls and n blue balls. We ask what is the distribution of the number of last balls of the same color. This is the same as the distribution D of the number of first balls of the same color. The first balls come at random with probability for red/blue very close to 1/2. This gives that the expected number of balls of the same color initial for the permutation is a constant, roughly 2, and that D is concentrated on constant numbers.

There were many interesting comments to this problem. Thanks for all the commentators! When Itai and Ronen asked me this question my spontaneous guess was wrong.

We are considering the stable marriage theorem. Suppose that there are n men and n women. If the preferences are random and men are proposing, what is the likely average women’s rank of their husbands, and what is the likely average men’s rank of their wives?

Boris Pittel proved that on average a man will be matched to the woman in place log n on his list. (Place one is his most preferred woman.) A woman will be matched on average to a man ranked n/log n on her list.

We asked in the post “Test your intuition (19)”  what is the situation if there is one additional man, and men are still proposing. This question is based on a conversation with Jacob Leshno who told me about a remarkable paper Unbalanced random matching markets (The link will be added in a few days) by Itai Ashlagi, Yash Kanoria, and Jacob Leshno.

Before getting to the answer let me mention that the answers to Test your intuition (18) and (17) are coming soon.

Here is how the collective intuition looks like.

The amazing result by Ashlagi, Kanoria, and Leshno asserts that a single additional man reverse the situation. Even when men propose, which gives each of them the best match in a stable marriage, with high probability every woman will get the man ranked ~log n in her list, and every man will get the woman ranked ~n/log n in his list. Unless the number of men and women is equal the outcomes for the best algorithm and for the worst algorithm are similar for both men and women.

And here is how things look in a computer simulation:

The picture plots the men’s average rank of wives (taking only matched men), calculated by averaging many draws of the market. The number of women is held fixed at 40, and the number of men varies across the x axis from 20 to 80. You can see that the men optimal stable matching (MOSM) and the Woman optimal stable matching (WOSM) are different when there are 40 men, but very close otherwise. The men get a low average rank when there are fewer men, and they get a high rank when there are more men. (note that given the men a totally random women would give them avg rank of (40+1)/2 ).