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Regarding Question 1, I think the notion you are looking for is that $C$ consists of random variables having mutually singular distribution measures. Regarding Question 2 (standard name for the Fact) …

How about $Y_i\sim\text{Bernoulli}(p)$ where $$ p = \frac12 \left(1 + \frac{1}{1 + \sum_{j=1}^i X_j}\right) $$ That is, every time there is a 1 in the sequence, we get closer to fair coin randomness. …

The random variable $X$ stochastically dominates the random variable $Y$, written $X\succeq Y$, if $\Pr(X\ge y)\ge\Pr(Y\ge y)$ for all $y$. This relation is transitive. Let's write $X\gtrsim Y$ if $\ …

This is just the negative binomial distribution. So you can work with the corresponding cdf, the regularized incomplete beta function.

I think yes, it's implied by the following (from Wikipedia https://en.m.wikipedia.org/wiki/Gaussian_free_field): Similarly to Brownian motion, which is the scaling limit of a wide range of discre …

An upper bound is $${\ell\choose 2} 2^{-\ell} $$ which goes to 0. Indeed, by the union bound an upper bound is $2^{-\ell}$ times the number of positions $i<j$ such that the two matching runs start at …

This question may be more appropriate for Math StackExchange. Nevertheless, some hints: For the argument from the paper, consider what would happen if $\mathbb P(A)$ and $\mathbb P(B)$ are close t …

I don't think there's a much nicer way. You could either say Let $G_X(x)=\mathbb{P}(X <x)=\lim_{(t\rightarrow x^-)}F_X(t)$; then $$ f_Y(y)=(1-G_X(0))\delta(y-1)+G_X(0)\delta(y) $$ or change th …

Let $f=0$. Then we do not know whether the $A_n=B_n$ are independent.

Let $W:[0,1]\rightarrow\mathbb R$ be standard Brownian motion with $W(0)=0$. Let $F_n$ denote the collection of all the $2^n$ many piecewise linear continuous functions $f:[0,1]\rightarrow\mathbb R$ …

This is a follow-up to an earlier MO question. Let $W:[0,1]\rightarrow\mathbb R$ be standard Brownian motion with $W(0)=0$. Let $F_n$ denote the collection of all the $2^n$ many piecewise linear con …

For a probability 1 statement, it's $\pm\sqrt{n\log\log n}$ (times a constant), see Law of the Iterated Logarithm. There are some further refinements which may give probability strictly between 0 and …

[This answer is a followup to Anthony Quas' comment and your subsequent request for an explicit map.] Let's list all the outcomes as $x_1\prec x_2\prec\dots\prec x_{2^n}$ in the following order: $x$ …

A word over the alphabet $\{0,1\}$ of length $n$ may contain squares, cubes, and generally $k$th powers, where $2\le k\le n$. Let $O_k(w)$ denote the number of $k$th power occurrences in the word $w$. …

One limitation of the von Neumann trick is that you don't know in advance how many samples will be needed. If you limit the number of samples in advance, we can get a negative answer. That is, suppos …