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More of a lurking variable than a connection, but: Both Cramér-Rao and the Uncertainty Principle follow from the Cauchy-Schwarz inequality.

(The sort of obvious answer from teaching statistics several times:) The sum of two independent normal random variables is again normal, i.e., the shape of the distribution is unchanged under addition …

Here's an approach to determine if $\det(C)$ is of this form with $F$ analytic and $n=2$. By Sylvester's criterion a matrix $\begin{pmatrix} a& b \\ b & c\end{pmatrix}$ is symmetric positive semidefin …

Let $A_t$ be the event that $S_1$ is a substring of $S_2$, $S_2=pS_1q$, where the length of $p$ is $t$. Then the probability of $\cup_t A_t$ can be found by inclusion-exclusion as $$\sum P(A_t)-\sum P …

It seems that Wolfram Alpha understands this integral. Its answers are given in terms of modified Bessel functions of the first kind $I_k$. It seems that the result is, for $A(k,c):=I(n,c)$, with $n …

Let $V_t$ and $W_t$ be independent standard Wiener processes ($t\ge 0$, $W_t,V_t\in\mathbb R$). Let $C$ be the event that there is a continuous function $f$ such that for all $s$, $t$, $$ W_t=W_s\iff …

Sinve you mentioned famous examples, here is a famous counterexample: power law distributions.

As you point out, the colors observed will have the same distribution with each of these models. Statistical tests involve asking "what is the probability of what is observed according to various …

No, let $\mathcal S_0=\{\{0,1\},\emptyset\}$ and let $\mathcal S_1$ be the powerset of $\{0,1\}$. Let $\mathcal F_n=\mathcal S_0^{n-1}\times\mathcal S_1 \times\mathcal S_0^{\infty}$. Let us write $X …

It is true. Let $k=\binom{N}{N/2-c\sqrt N}$ and let $K$ be a randomly* selected subset of $k$ of size $k\lambda$. Then conditionally on $|X|>N/2-c\sqrt N$, the difference $|X|-(N/2-c\sqrt N)$ is unbou …

It seems that in general this is an almost arbitrarily hard problem. Consider the simpler countably infinite case $X=\mathbb N$, $Y=\{0,1\}$. Thus $H=\{x:h(x)=1\}$ is a "random set". Fix $g:X\to\{0, …

Let $a=1$ or any other positive constant. Then $$E(M(\inf\{t:M(t)\ge a\}))=a$$ since $$P(M(\inf\{t:M(t)\ge a\})=a)=1.$$ And $a\ne M(t)$ with positive probability (actually probability 1).

Yes, your bound is optimal and here's how to attain it. Fix a linear ordering $L$ of $\Omega$. Pick a number $x \in [0,1]$ under the uniform measure and the value of our coupling shall be $(M(x),N(x) …

It just means that $$\mu\left(\{f: K_f>y\}\right)<\frac{\alpha}{y^\beta}$$ and $$\mu\left(\{f: \rho(x_0,f(x_0))>y\}\right)<\frac{\alpha}{y^\beta}$$

Yes, for integer $b$ it's a reformulation of, and exactly the same as, normality to base $b$. Wikipedia even has a section of its article about normal numbers stating exactly this, and giving credit …