## Prior derivation in Chinese restaurant process (Dirichlet Processes)

(Link to paper: https://www.seas.harvard.edu/courses/cs281/papers/rasmussen-1999a.pdf page 3-4)

For the Chinese restaurant process, as used in Dirichlet Process mixture models, we have a prior that data point i belongs to cluster j, where c is an indicator. n represents the total number of data points that are assigned to cluster j, where subscript -i refers to all observations except i.

https://i.redd.it/47dd42pa1nh31.png

As we take k to infinity, (which represents taking the number of components in the mixture model to infinity), the expression transforms to

https://i.redd.it/1v6hz4dc1nh31.png

when that cluster already exists (which I understand). For the case that the cluster doesn’t already exist, the prior transforms to

https://i.redd.it/nugw2kxl1nh31.png

This is where my problem lies – how does alpha feature in this expression if (alpha/k) tends to 0 as k tends to infinity?

** please let me know if you’d like me to explain anything more clearly – this is my first time posting here and I’m unsure how general I should make my explanations **

submitted by /u/MightyDagger

## Reference request: Maps into the universal bundle which create the same pullback bundle are homotopic.

Hello! In a course on vector bundles we proved that homotopic maps give the same pullback bundle. The professor mentioned that when considering maps into the universal bundle, this is actually a 1-1 correspondence, i.e. any two maps which give the same pullback bundle are homotopic.

I was looking for a good reference for the proof of this statement, but was only finding proofs of the forward direction which I have already seen. Does anyone know of a good reference for this? Thanks!

submitted by /u/UpsideDownRain

## How to know if a group operation is well-defined by a set of rules

The dihedral group of order n is fully defined by [;D_{2n} = <a,b: a^n = b^2 = 1, b^{-1} a b = a^{-1} > ;], and in general you can write any group by listing its elements followed by a set of algebraic relations. Usually it seems that there are shorter ways to write the algebraic relations than by listing out the multiplication for any two elements (as in the case of the dihedral group). Is there a general technique or strategy for showing that the algebraic relations specified, e.g. the relations after the colon in the above expression, are actually sufficient to define the multiplication of any two elements of the group? It’s pretty simple to do for the dihedral group, but I was wondering if there is a systematic method for doing this.

submitted by /u/MoneyMe_MoneyNow

## A problem I just thought up: what are all the possible function which curve length can be found in basic calculus?

In calculus, we have a formula to calculate length of curve described by a function f, that is ∫sqrt(1+[f'(x)]2 )dx. Technically it’s definite integral but in basic calculus finding antiderivative is the only way to calculate it, so assume you need to do that. Now it’s well-known that thee are some rather simple-looking curves in which this have no elementary antiderivative, such as the perimeter of a non-circle ellipse. One thing I noticed is that different textbooks and online resources use pretty much a small pool of a few examples. So I wondered why that is, which lead to the question: what are all the possible function f where this can work?

Problem statement: find all f such that f is an elementary function and sqrt(1+[f'(x)]2 ) has an elementary antiderivative.

My idea of a solution is as follow. Write g=f’ and h=sqrt(1+[f'(x)]2 ). Then we have [g(x)]2 +1=[h(x)]2 . So we change to an equivalent question of finding all g,h with elementary antiderivative such that [g(x)]2 +1=[h(x)]2
Then using rational parameterization of a hyperbola we set t(x)=g(x)/(h(x)-1) then g(x)=2t(x)/([t(x)]2 -1) and h(x)=([t(x)]2 +1)/([t(x)]2 -1). So we reduce the problem to finding all elementary t(x) such that both 2t(x)/([t(x)]2 -1) and ([t(x)]2 +1)/([t(x)]2 -1) have elementary antiderivative.

Now here I’m stuck. Though this does clarify the question a bit. Here are a few examples for t(x) I could think of:

• Any rational function.

• Any rational function of an exponential function, which mean this also extend to rational function in sine and cosine (with the same linear argument), or sinh and cosh.

I can’t think of any other examples, though these cases sort of make clear why there are so few examples used in textbooks. For example, if you want to write an example and pick t to a polynomial, then you will also make t linear because even for a quadratic t then f already looks very complicated.

So anyone know how to solve this problem? Is there any other examples? Or can you prove that these are all the possible ones?

EDIT: see comment below for another class of function. This one show that t(x) might not necessarily have an antiderivative.

submitted by /u/Proof_Inspector

## Local homeomorphism lifting theorem

Hi, I’d greatly appreciate it if someone could illuminate the last three lines of this proof

Here is some context to aid in reading it:

[; C(I;Y) ;], where [; I ;] denotes the unit interval, is the space of all paths in [; Y ;], endowed with the topology generated by the following basis:

Let [; cup Ui ;], [; P = { t_0=0 ,t_1,…, t_n=1 } ;], be a collection of open sets where each [; U_i ;] comes from some basis for the topology on [; Y;] and [; P ;] is a partition on the unit interval [; I ;]. Furthermore the union is over the following set of indices: [; J = {1,2,…,n } ;]. (I wasn’t able to typeset a union no matter how hard I tried.. for whatever reason). Then the set [; A(t_0,t_1,..,t_n;U_1,U_2,..,U_n) ;] is the set of all paths [; a in C(I;Y) ;] such that [; a([t{i-1},t_i]) in U_i ;] for all [; i ;]. We endow [; C(I;X) ;] with the same topology.

[; f : X rightarrow Y ;] a local homeomorphism of topological spaces has the “unique path lifting property”, if for all paths [; a in C(I;Y) ;] and [; x in X ;] such that [; f(x) = a(0) ;], there is exactly one lifted path [; tilde{a} in C(I;X) ;] such that [; tilde{a}(0) = x ;], the lift with respect to \$f\$.

I understand all of the proof besides the third to last line, since I see that [; tilde{b}([t_0,t_1]) = phi_1(b([t_0,t_1]) ;] as both lifts agree at [;t = t_0 = 0;], by construction since [;tilde{b}(0) = x’ in U_1;], so by unique path lifting they agree on [;[t_0,t_1];]. The issue I have is why this also holds for [;i=2,3,…,n;]