impossiblewizardry

Rick Durrett:

The results for E[η_{n,m}] are useful for population genetics, but are not really relevant to cancer modeling. To investigate genetic diversity in the exponentially growing population of humans, you would sequence the DNA of a sample of individuals from the population. However, in the study of cancer each patient has their own exponentially growing cell population, so it is more interesting to have the information provided by Theorem 1 about the fraction of cells in the population with a given mutation.

The results he seems to think are so useless still seem to be the only results in the paper that has been used in data analysis (in this paper).

What Rick Durrett doesn’t seem to have realized is that, two years prior to the publication of this paper, people started doing DNA sequencing of individual cells from patient’s tumors. So, the study of a single patient became a population genetics problem, and results from a population genetics perspective were exactly what was needed.

Rick Durrett can be forgiven for not noticing that. He’s a mathematician, and can’t be expected to keep up with the latest advances in genomics. But as a consequence he should worry less about whether his results are applicable to the few problems he’s familiar with from his collaborators.

I prefer the spirit he demonstrated in this paper with Foo and Leder, where they provide extensive information about the growth of cancer in a space of 3 or more dimensions.

Zeilberger:

Ewens and Wilf are very right when they claim that P(r, n,m) and Q(r, n,m) are very far apart around the “tail” of the distribution, but who cares about the tail? Definitely not a scientist and even not an applied mathematician. It turns out, empirically (and we did extensive numerical testing, see Procedure HowGoodPA1(R0,N0,Incr,M0,m,eps) in BallsInBoxes), that whenever P(r, n,m) is not extremely small, it is very well approximated by Q(r, n,m), and using the latter (it is so much faster!) gives very good approximations, and enables one to construct the “center” of the probability distribution (i.e. ignoring the tails) very accurately.

Statistical geneticists care about the tails. Imagine a gene association study, where you have some disease, and you want to know, for each of the 20,000 or so genes in the human genome, whether rare variants in those genes increase risk of the disease. You use the Bonferroni correction: instead of using 0.05 as a p-value cutoff, you use 0.05/20,000. So your question is about whether a probability is above or below a number which is around 1 in a million.

The approximation Zeilberger is considering does seem to be good even to probabilities around 10^-7 in the example he looks at, so it seems it can handle another order of magnitude more tests, but this could certainly occur in a study with a larger scope. For example, if instead of one disease, we consider 10, or 100.

Ewens knows this, because he works in statistical genetics. Zeilberger has no reason to think he knows more than Ewens about what matters in scientific practice. Zeilberger knows very little about what matters in practice, and doesn’t need to. He just needs to keep making his solutions more general, so that whatever does come up in practice will turn out to be a special case of something.

Which is what he does in fact do, so I have no problem with how Zelberger manages his research, it’s just ridiculous for him to pepper his papers with baseless value judgments.

I think of rotations as, in a way, not real transformations. Instead of imagining the object itself rotating, I just imagine the axes rotating. I’m just choosing to describe the same object, in a different way.

Recently I’ve had to get used to thinking of reflections in the same way too. Imagine a map of a city, and you describe points on the map as how far east of the center they are, and how far north of the center they are. Now you decide to describe them instead in terms of how far east and how far south. What you’ve done to the coordinates is a reflection. But you have not changed the map; it’s not a reflected inverted map; instead it’s only one of the basis vectors which has been reflected.

This perspective has helped me understand the terminology people use in principal component analysis. In PCA, you diagonalize the correlation matrix:

C = P D P’

The matrix P is described as the rotation matrix. But rotations all have determinant 1; might this have determinant -1 instead? At first I tried to prove somehow that the determinant must be 1 instead of -1, to explain why they call it a rotation.

But I realized, they probably make the determinant 1 by convention. You can always flip the sign of the determinant by flipping the sign of one of the columns of P. And you can always do this because P must be a basis made of eigenvectors of C, and that doesn’t change if you flip the sign of one of the basis vectors.

A symbolic way to see this is

P₂ = P₁ F

where F flips the sign of a column. And what is F? An identity matrix, but with one of the diagonal elements changed from 1 to -1. And

det P₂ = det P₁ det F

And det F is -1. So flipping the sign of a column flips the sign of the determinant.

And to see that it still works as a diagonalization:

P₂ D P₂‘ = P₁ F D F P₁ = P₁ D P₁

C wants to become a rich doctor, her boyfriend wants to live on the moon.

Unlike C, I want a life off the beaten path, rather than something where you know pretty much exactly how it’s going to go and a lot of people have done it before, so you have confidence in your chances of success because you’ve seen other people do the same thing. But unlike C’s boyfriend, I acknowledge that the consequence of this is that I have no idea where I’ll be in 20 years, the moon or whatever.

on a sunny day, you’ll have a shadow. Until you enter the shadow of a building, where you’re illuminated not directly by the sun, but by reverberating light reflecting from surfaces all around you, which has no single source, and thus will not cast a shadow.

Except in Houston, with all of these reflective glass buildings, your shadow will actually change direction as you walk around. When you’re in the shadow of a building, and can’t see the sun, you may still be illuminated from the other direction by the reflection of the sun from a glass building, which is bright enough and coherent enough to cast its own shadow.

my grandma was telling me about a guy who would have maybe been fired for sexual harassment today, a salesman who sold to the store she worked at, who was always trying to touch her, and making comments. Her solution was, she stopped going to work the days he was going to be there. It seems like the slimes are punished more efficiently now, but how much has the average person changed? My grandma was saying, the average guy wasn’t that bad, but was patronizing, like... I don’t quite get the meaning, but like you need a lot of help and he knows what’s best for you.

fucked up with google maps today. Went to the right address in the wrong town. Ended up taking a lyft which cost >$30, and the whole trip took like 3 hours when it should have taken like 30 min to 1 hour

so a transformation from Rⁿ to Rⁿ preserves Euclidean distances:

||v - w|| = ||v’ - w’||

From that you can prove it preserves Euclidean norms of vectors:

||v|| = ||v’||

And then you expand ||v’ - w’||² and see that it preserves dot products:

v’ . w’ = v . w

Geometrically, that means if you preserve all distances, you also have to preserve all angles, which makes sense. Then... OK, it has to be linear. You can find that

||(s v)’ - s v’||² = 0

just by expanding the left side and canceling. Then you show that

||v’ + w’||² = ||v + w||²

which you do just by expanding and unexpanding, and that allows you to prove

||(v + w)’ - (v’ + w’)||² = 0

again by just expanding and canceling. So that means this transformation is linear.

And then... well, it also turns out that it’s orthogonal. Because, ok let’s call the matrix of this transformation A. Then,

v^T v = ||v||² = ||v’||² = ||Av||² = v^T A^T A v

v^T (A^T A - I) v = 0

Alright? So, that last thing on the bottom, it’s a quadratic polynomial in n variables which puts some upper bound on how many zeros it can have, or like how those zeros can be shaped or something, but it’s actually zero everywhere, so it must be the zero polynomial, meaning all the coefficients are zero, and guess what’s the matrix containing the coefficients, it’s A^T A - I. So.

A^T A = I

A is orthogonal. Alright. Well, it turns out that its got to have a determinant of 1 or -1. Why? Well, because of some determinant rules that I don’t know how to prove.

1 = det I = det (A^T A) = det(A^T) det(A)  = det(A) det(A)

So, det(A)² = 1, so it’s either 1 or -1.

jadagul reblogged your post and added:

The special orthogonal group.

Wikipedia:

In mathematics, the orthogonal group in dimension n, denoted O(n), is the group of distance-preserving transformations of a Euclidean space of dimension n that preserve a fixed point, where the group operation is given by composing transformations. ...

An important subgroup of O(n) is the special orthogonal group, denoted SO(n), of the orthogonal matrices of determinant 1. This group is also called the rotation group, because, in dimensions 2 and 3, its elements are the usual rotations around a point (in dimension 2) or a line (in dimension 3).

So THAT’S the rotations!!