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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} …

To complete what Lars said: By Theorem II.13 in Davidson and Szarek (2001), $$P(\lambda_{min}(S_n)^{1/2} \notin [1-\sqrt y - t, 1+\sqrt y + t]) \le 2 e^{-nt^2}.$$ Pick, e.g., $t= n^{-1/4}$ to get an e …

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} …

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., $$ E[x \mid x+z]= (x+z)/2. $$

Assuming $X\sim N(0,I)$, writing $A^TA=\sum_i L_i u_i u_i^T$, since $sign(u_i^TX)$ and $sign(u_k^TX)$ are independent and mean-zero, $u_i^TMu_k=0$ if $i\le k$ because the denominator $\|AX\|^2=\sum_i …

The Marcenko-Pastur resut (see, e.g., https://www.sciencedirect.com/science/article/pii/S0047259X85710512) gives you the Stieljes equation of the limiting spectral distributions of matrices of the for …

The usual lower bound (e.g., Theorem II.13 in Davidson and Szarek ("Banach space theory and local operator theory") says that if $A\in R^{n\times d}$ has iid $N(0,1)$ entries then $$ P(\sigma_{\min}(A …

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 …

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 …

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 …

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^ …

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 …

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 …

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 …

You can deduce the result if it is already known that $X_1\to^d N(0,1)$. The convergence in law of $(X_1,...,X_k)$ boils down, for fixed reals $t_1,...,t_k$, to the convergence of the characteristic f …