Search Results

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 …

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 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$[z_i(w)]>0$.

$Y_i\sim^\text{iid} \text{Exp}(1)$ random variables, $f(Y)=\min_i Y_i$ has Exponential distribution with parameter $n$ with $f$ 1-Lipschitz. Then $E[f(Y)]=1/n$ but $$P(f(Y)-E[f(Y)]>t)=\exp(-n(t+1/n))$ …

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 …

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 …

First $\pi_nP_n^t = \pi_n$ for all $n,t$. Since for the limiting matrix $P$ the distance to stationarity $d(t) = \sup_\mu \|\mu P^t - \pi\|_{TV}$ converges to 0 as $t\to+\infty$, there exists $t_0$ su …

This expands comments by AnthonyQuas. Start with the observation from Anthony Quas that the event of interest is $\|Au\|>t$ for an orthogonal projection matrix $A$ whose image is distributed according …

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 …

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

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 …

Assume iid $N(0,1)$ entries, assume $C$ diagonal, and focus first on the non-diagonal terms: $G=\sum_j \sum_{l\ne j} w_j^Tw_l w_j^TCw_l = \sum_{j\ne l, ik} w_{ji}w_{li} c_i w_{jk} w_{lk}$. Write this …

Write the SVD of $\Gamma$, say $\Gamma = \sum_i q_i s_i v_i^T$. with $s_1,...,s_n>0$ the singular values and $q_i, v_i$ are the left and right singular vectors. If $Q=[q_1|...|q_n]$, $B=diag(s_1,...,s …

The distribution of $v^TB^{-1}AB^{-1}v$ is the same for every vector $v$ in the unit sphere either deterministic or independent of $W$. Once this is established, you are allowed to take $v=z/\|z\|$ in …