Springer’s copyright agreement is, according to Springer, compatible with posting your article on the arXiv under the CC-BY license

It is far from clear to me that Springer’s copyright agreement (quoted below) is compatible with posting your paper on the arXiv under the CC-BY license. Happily, a Springer representative has just confirmed for me that this is allowed:

Let me first say that the cc-by-0 license is no problem at all as it allows for other publications without restrictions. Second, our copyright statement of course only talks about the version published in one of our journals, with our copyright line (or the copyright line of a partner society if applicable, or the author’s copyright if Open Access is chosen) on it.

At least if you are publishing in a Springer journal, and more generally, I would strongly encourage you to post your papers to the arXiv under the more permissive CC-BY-0 license, rather than the minimal license the arXiv requires.

As a question to any legally-minded readers: does copyright law genuinely distinguish between “the version published in one of our journals, with our copyright line”, and the “author-formatted post-peer-review” version which is substantially identical, barring the journals formatting and copyright line?

 

Pious penal probability puzzle

Here is a fun problem, with a great story and a surprising answer.

According to the Talmud, in order for the Sanhedrin to sentence a man to death, the majority of them must agree to it. However

R. Kahana said: If the Sanhedrin unanimously find [the accused] guilty, he is acquitted. (Babylonian Talmud, Tractate Sanhedrin, Folio 17a)

Scott Alexander has a devious mind and considers how he would respond to this rule as a criminal:

[F]irst I’d invite a bunch of trustworthy people over as eyewitnesses, then I’d cover all available surfaces of the crime scene with fingerprints and bodily fluids, and finally I’d make sure to videotape myself doing the deed and publish the video on YouTube.

So, suppose you were on a panel of N judges, all of whom had seen overwhelming evidence of the accused’s guilt, and wanted to make sure that a majority of you would vote to convict, but not all of you. And suppose you cannot communicate. With what probability p would you vote to convict?

Test your intuition by guessing an answer now, then click below:

My gut instincts were that (1) we should choose p really close to 1, probably approaching 1 as N \to \infty and (2) there is no way this question would have a precise round answer. As you will see, I was quite wrong.

Tumblr user lambdaphagy is smarter than I was and wrote a program. Here are his or her results:

As you can see, it appears that p is not approaching 1, or even coming close to it, but is somewhere near 0.8. Can we explain this?

Heuristic solution

We want to avoid two events: unanimity, and a majority vote to acquit. The probability of unanimity is p^N.

The probability of a majority vote to acquit is \sum_{k=0}^{\lfloor N/2 \rfloor} \binom{N}{k} p^k(1-p)^{N-k}. Assuming that p>0.5, and it certainly should be, almost all of the contribution to that sum will come from terms where k \approx N/2. In that case, \binom{N}{k} \approx 2^N/\sqrt{N}. And we’ll roughly care about \sqrt{N} such terms. So the odds of acquittal are roughly \sqrt{N} \tfrac{2^N}{\sqrt{N}} p^{N/2} (1-p)^{N/2} = \sqrt{4p(1-p)}^N.

So we roughly want p^N+\sqrt{4p(1-p)}^N to be as small as possible. For N large, one of the two terms will be much larger than the other, so it is the same to ask that \max(p, \sqrt{4p(1-p)})^N be as small as possible.

Here is a plot of \max(p, \sqrt{4 p(1-p)}):

maxPlot
Ignore the part with p below 1/2; that’s clearly wrong and our approximation that \sum_{k=0}^{\lfloor N/2 \rfloor} \binom{N}{k} p^k(1-p)^{N-k} is dominated by k \approx N/2 won’t be good there. Over the range p>1/2, the minimum is where p=\sqrt{4p (1-p)}.

Let’s do some algebra: p = \sqrt{4p(1-p)}, p^2= 4 p(1-p), p = 4(1-p) (since p=0 is clearly wrong), p = 4/5. Holy cow, 4/5 is actually right!

Lessons learned

First of all, actually do some computations.

Secondly, I was wrongly thinking that failing by acquittal would be much more important than failing by unanimity. I think I was mislead because one of them occurs for N/2 values of k and the other only occurs for one value. I should have realized two things (1) the bell curve is tightly peaked, so it is really only the k very close to N/2 which matter and (2) exponentials are far more powerful than the ratio between N or \sqrt{N} and 1 anyway.

Rigorous computation

Finally, for the skeptics, here is an actual proof. Assuming p>1/2, we have
\displaystyle{ \sum_{k=0}^{\lfloor N/2 \rfloor} \binom{N}{k} p^k(1-p)^{N-k} \leq  \sum_{k=0}^{\lfloor N/2 \rfloor} \binom{N}{k} p^{N/2} (1-p)^{N/2} \leq  2^N p^{N/2} (1-p)^{N/2}}.
The main step is to replace each p^k(1-p)^{N-k} by the largest it can be.

But also,
\displaystyle{ \sum_{k=0}^{\lfloor N/2 \rfloor} \binom{N}{k} p^k(1-p)^{N-k} \geq \binom{N}{\lfloor N/2 \rfloor} p^{\lfloor N/2 \rfloor} (1-p)^{\lceil N/2 \rceil} \geq \frac{1}{N+1} \sqrt{\frac{1-p}{p}} \ 2^N p^{N/2} (1-p)^{N/2}} .
Here we have lower bounded the sum by one of its terms, and then used the easy bound \binom{N}{\lfloor N/2 \rfloor} \geq \frac{1}{N+1} 2^N since it is the largest of the N+1 entries in a row of Pascal’s triangle which sums to 2^N.

So the odds of failure are bounded between
\tfrac{1}{N+1} \sqrt{\frac{1-p}{p}} \sqrt{4p(1-p)}^N+p^N and \sqrt{4p(1-p)}^N+p^N. We further use the convenient trick of replacing a + with a \max, up to bounded error to get that the odds of failure are bounded between \max \left( \tfrac{1}{N+1} \sqrt{\frac{1-p}{p}} \sqrt{4p(1-p)}^N, p^N \right) and 2 \max \left( \sqrt{4p(1-p)}^N, p^N \right).

Now, let p be a probability greater than 1/2 other than 4/5. We claim that choosing conviction probability 4/5 will be better than p for N large. Indeed, the p-strategy will fail with odds at least \max \left( \tfrac{1}{N+1} \sqrt{\frac{1-p}{p}} \sqrt{4p(1-p)}^N, p^N \right), and the 4/5 strategy will fail with odds at most 2 \cdot (4/5)^N. Since \max(p, \sqrt{4p(1-p)}) > 4/5, one of the two exponentials in the first case is larger than 4/5, and the p-strategy is more likely to fail, as claimed.

Of course, for a Sanhedrin of 23 members, 2 (4/5)^{23} \approx 0.011, so our upper bound predicts only a one percent probability of failure. More accurately computations give 0.005. So the whole conversation deals with the overly detailed analysis of an unlikely consequence of a bizarre hypothetical event. Fortunately, this is not a problem in the study of Talmud!

Random three dimensional partitions

Back in graduate school, I read a beautiful paper of Kenyon, Okounkov and Sheffield. It started with the following physical story.

Electron microscope image of a salt crystal

This is the corner of a crystal of salt, as seen under an electron microscope. (I took the image from here, unfortunately I couldn’t find better information about the sourcing.) As you can see, the corner is a bit rounded, where some of the molecules have rubbed away. They ask the question: “What is the shape of that rounded corner?”

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3 continuing positions at the ANU, in statistics, probability, stochastic analysis, mathematical finance, biomathematics.

We’ve just posted an ad for up to 3 continuing positions at the Mathematical Sciences Institute, at the Australian National University, in Canberra. (Where I work!)

It’s up on mathjobs, but applicants will need to apply through the university website. Here’s the pitch:

The Mathematical Sciences Institute at the Australian National University is seeking to invigorate its research and teaching profile in the areas of statistics, probability, stochastic analysis, mathematical finance and/or biomathematics/biostatistics. We wish to fill several continuing positions at the Academic Level B and/or Level C (which equates to the position of Associate Professor within the United States of America). Up to 3 full time positions may be awarded.

You will be joining an internationally recognised leading team of academics with a focus on achieving excellence in research and teaching. The Institute comprises of approximately fifty academics, within seven mathematical research programs. Applicants are expected to have an outstanding record in research, teaching and administration. All positions will involve some teaching, in the specialised areas advertised and/or standard mathematics undergraduate courses, but this may be at a reduced level for several years.

It’s a great place to work, excellent opportunities for research grant funding, and a really nice place to live. Feel free to contact me if you have any questions about the job or living in Canberra.

Please pass this on to friends with relevant interests!

(Oh, and don’t forget those two postdocs I’m hiring in quantum algebra, higher category theory, subfactors, representation theory, etc.)

The shape of a random partition

In this post we will give a heuristic derivation of a result of Vershik, describing the shape of a random partition of a large integer N. (Vershik’s Russian original is available here; English translation is pay-walled.)

By a partition of N, we mean positive integers \lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_r > 0 with \sum \lambda_i = N. We draw a partition as a collection of boxes: For example, this is 4+2+1+1+1:

smallPart

Suppose we let N \to \infty, select partitions of N uniformly at random and rescale the size of the boxes by 1/\sqrt{N}, so that the diagram of the partition always has area 1. What is the shape of the most likely diagram?

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Postdoc position at ANU

Update — there are now not one, but two, positions available! The application has been extended to the end of November.

We’ve just put up an ad for a new 2 year postdoctoral position at the ANU, to work with myself and Tony Licata. We’re looking for someone who’s interested in operator algebras, quantum topology, and/or representation theory, to collaborate with us on Australian Research Council funded projects.

The ad hasn’t yet been crossposted to MathJobs, but hopefully it will eventually appear there! In any case, applications need to be made through the ANU website. You need to submit a CV, 3 references, and a document addressing the selection criteria. Let me know if you have any questions about the application process, the job, or Canberra!

Michigan Math In Action

Those of you who are interested in college math instruction may be interested in a no-longer-so-new blog “Michigan Math In Action”, which a number of our faculty started last year. (I was involved in the sense of telling people “blogs are fun!”, but haven’t written anything for them yet.) It mostly features thoughtful pieces on teaching calculus and similar courses.

Recently, Gavin Larose put up a lengthy footnoted post on the effort that goes into running our “Gateway testing” center, and the benefits we get from it. This is a room designed for proctoring computerized tests of basic skills, and we use it for things like routine differentiation or putting matrices into reduced row echelon form, which we want every student to know but which are a waste of class time. Check it out!

Register soon for ALGECOM Fall 2015!

Just a quick reminder that, if you are looking for graduate support to attend ALGECOM at the University of Michigan on Saturday October 24, or to register for the poster session, you should please send an e-mail to speyer@umich.edu by Tuesday Sept 15. (Yes, after sunset but before midnight is fine, I won’t be online during Rosh Hoshanah either.)
Even if you are not requesting support, I’d appreciate knowing that you are coming.

IMG_5474

Our speakers are Jonah Blasiak (Drexel), Laura Escobar (UIUC), Joel Kamnitzer (Toronto) and Tri Lai (IMA and Minnesota). Please see our website for more information.