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For two measurable sets $A,B$, let $p=\mu(A)$ and $q=\nu(B)$. Consider any coupling of a Bernoulli$(p)$ and a Bernoulli$(q)$, say, $C:\{0,1\}^2 \to [0,1]$. Then we can find a coupling $(X,Y)$ of $\mu …

No: If $x=y$ and $x,z$ are iid $N(0,1)$ (jointly normal) then $x \mid x+z$ is also normal with mean $(x+z)/2$, i.e., $$ E[x \mid x+z]= (x+z)/2. $$

Claim: if $X_1=X_2=1 \Rightarrow X_3=X_4=1$ holds then $X_3\ne X_4 \Rightarrow X_1=X_2=0$. Indeed, by pairwise independence, $$ E\Big[ X_1\Big(\frac{X_3+X_4}{2} - X_2\Big) + X_2\Big(\frac{X_3+X_4}{2} …

I believe the probability is at least $\approx0.343$. Let $\mu_n$ be a probability measure giving $q_n = P_n[ X_1 + X_2 + X_3 < 2X_4]$. Consider now $(Y_i)_{i\le 4}$ Bernoulli$(p)$. The $Y_i$'s produc …

We can invert $P=P_d(\lambda)$ easily because it is diagonal: $$ P^{-1} = \frac{1}{1-\lambda + \lambda/d} e_1e_1^T + \frac{d}{\lambda} \sum_{j\ge 2} e_j e_j^T. $$ Write $P^{-1}$ as $\frac{d}{\lambda} …

The claim that $P(Y=\tilde Y)=1$ is incorrect. But I do not see this claim in the book you linked to. Yes, the algorithm will oversample some states if one does not "reuse previous randomness". With t …

A ``down-to-earth'' observation to see what goes wrong with method of moments is this: When considering applying the method of moment to $(X_1,...,X_n)$, you may as well apply the method of moments to …

It is not possible to get $|\frac1n\sum_i z_i(w) - E[z_i(w)]|=O_P(r_n)$ with $r_n \lll n^{-1/2}$ even for a single $w$, since it would contradict the CLT whenever Var$[z_i(w)]>0$.

If $Z\sim N(0,1)$ and $X$ is subgaussian with subgaussian norm less than $1/\sqrt 2$ then $$P(|X|>t)\le 2 e^{-t^2} \le K P(|Z|>t) $$ for some numerical constant $K>1$, thanks to lower bound on $P(Z>t) …

I would start with Theorems 4.1 and 4.2 in [1]. A statement of Theorem 4.2 is as follows: if $\nu_n(B(r))$ is the number of zeros within the disk $B(r)$ of radius $r$ when the degree is $n$, then the …

If $H$ has iid $N(0,1)$ entries, write $\kappa=\langle Z^TZ, H\rangle$ using the usual Frobenius inner product $\langle A,B\rangle = trace[A^TB]$. Conditionally on $Z$, $\kappa\sim N(0,\|Z^TZ\|_F^2)$ …

Theorem 2 in Iain M. Johnstone. Zongming Ma. "Fast approach to the Tracy–Widom law at the edge of GOE and GUE." Ann. Appl. Probab. 22 (5) 1962 - 1988, October 2012. https://doi.org/10.1214/11-AAP819 …

Adding to Will's answer: The $b=[EY]=0$ condition can be dropped if one is allowed to take the normalizing $1/\sigma$ to be complex. For any candidate normalization $1/\sigma$, write it in polar coord …

Assume real entries. Let $b$ be a column of $A$ and write $\tilde A$ the matrix $A$ with that column removed so that $AA^T = \tilde A\tilde A^T + bb^T$. By the https://en.wikipedia.org/wiki/Sherman%E …

It is true under the assumption $k,d\to+\infty$ while $k/d\to 0$, in the sense that $R^T\in R^{d\times k}$ has obviously iid $N(0,1)$ entries, and satisfies $$ \|\sqrt d R^+ - R^T/\sqrt d\|_{op} \to^ …