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I learned from Gian-Carlo Rota (Combinatorial snapshots) the following probabilistic motivation for looking at the Riemann zeta function: "subject to technical assumptions," the only probability measu …

No. Let's restrict our attention to $\mathbb{N}$. The hypotheses imply that if $q$ is a prime, then the probability that a random positive integer is not divisible by $q$ is $1 - \frac{1}{q}$. They al …

I've been told that in machine learning it's common to compute the singular value decomposition of matrices in order to throw out all information in the matrix except that corresponding to, say, the $ …

Yes. Here's what should be a proof: the set of pairs of elements satisfying any particular relation is Zariski closed, hence has measure zero (to show that it is not $SO(3) \times SO(3)$ it suffices …

Rota and Shen's On the Combinatorics of Cumulants ends with a conjecture which I'll restate as follows: Let $p \in \mathbb{R}[x_1, x_2, ...]$ be a polynomial such that, for any sequence $X_1, X_2 …

You can make sense of the uniform probability distribution on lots of infinite sets, notably any compact topological group $G$, where "uniform probability distribution" should mean "normalized Haar me …

Here is a result that gives the flavor of the kind of thing along these lines I hope to see in the future. Recall Tarski's undefinability of truth: under suitable assumptions, a formal system can't be …

In his notes on free probability, Terence Tao describes a general approach to non-commutative probability which prioritizes the algebra of random variables above the sample space; I find this conceptu …

One answer I've heard is that you want the notion of standard deviation to 1) apply to points in Euclidean space, and 2) to be invariant under rotation. You don't get the second property unless you s …

Equivalently, we are considering a random function $f : [n] \to [n]$ where $[n] = \{ 1, 2, \dots n \}$ is a finite set of size $n$, which assigns to each prank cigarette a pack. The second question is …

$2^{\mathbb{R}}$, being a product of compact Hausdorff groups, is a compact Hausdorff group, so it has a normalized Haar measure ("flipping uncountably many coins").

This isn't an answer to your question, but without any restrictions, the expected number of $r$-cycles in a permutation of at least $r$ elements is $\frac{1}{r}$, so the expected number of cycles in a …

it's possible to extend the analogy to the factorization of polynomials over finite fields $\mathbb{F}_q$ (see this blog post for details; one needs to take $q \to \infty$ for the statistics to match, …

Here are examples showing that unlike in the previous problem, here it does not suffice to simply use the fact that the harmonic series / the sum of the reciprocals of the primes diverges. In fact for …

Suppose I toss $n$ coins. It's natural to model this probabilistically in terms of a sample space $\{ H, T \}^n$ constructed as the product of $n$ copies of the sample space of possible outcomes of a …