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For any vertex $v$, you can take $\mu$ to be $7/8 \pi + 1/8 v$, that is, the with probability $1/8$ pick $v$ and otherwise pick a random vertex according to $\pi$. Now the distance to stationarity at …

Consider for a moment the case $m=cn$ for some constant $c$. This also yields a maximal load of $(1+o(1))\log(n)/\log(\log(n))$. Geometric distribution with mean $n$ will give you $cn$ balls with high …

What is your question? In the situation you describe, both distributions are approximately Poisson with the same parameter, so it is no wonder they also approximate each other.

Partition the unit square into small squares of area roughly $a$. Your question is equivalent to asking for which $a$ do we typically see about 1 small square with points from each of $X$,$Y$ and $Z$? …

The answer is yes. To see this, take a transient graph $G$ (say $\mathbb{Z}^3$) and glue a copy of $\mathbb{N}$ to some vertex (say $v_0$). The vertices of $\mathbb{N}$ now all have $g(v)=1$. If you …

Robert's answer is correct, if you want an asymptotic answer. However, for the question as stated in the edit (that is for all $n\ge k$...) the answer is no. Take for example the case of $p=\frac12$. …

$\newcommand{\E}{\mathbb{E}}$ You practically answered your own question. Denote by $A_k$ the event "$X_0 > k$". Then $X_0=\sum_{k=0}^{m-1} \delta(A_k)$. Denote by $B_k$ the event "$X_k > k$". Then …

When tossing $n$ balls uniformly and independently into $n$ bins, the distribution of the number of balls in any specific bin is binomial with parameters $n$ and $p=\frac{1}{n}$. Thus, the probability …

This is the Balls and Bins setting.

The crucial requirement is that $\sum_{i=0}^\infty t_i^2 < \infty$. See Kolmogorov's two-series theorem and also the more general Kolmogorov's three-series theorem.

In the setting you describe, there's generally no CLT. Let me describe a counterexample showing that $\sum_{i=1}^{k-1} X_i / \sqrt{\sum_{i=1}^{k-1} \sigma_i}$ doesn't tend to normal. I'm pretty sure t …

This is tight, at least when $p=\frac12$. You simply need to approximate $\log\big({n \choose \frac{n}{2}-\varepsilon n} \frac1{2^{n}}\big)$ using Stirling's formula and you'll see that the leading co …

There is no general bound with exponential decay just assuming decay of correlations. Consider this distribution: with probability $1/(N+1)$ all $X_i=$, and with probability $N/(N+1)$ there are exactl …

Yes. For example, if we take $Y=\lfloor X \rfloor$ and $Z=X-Y$.

If you define $Z_i=Y_i−p_i(Y_1,…,Y_{i−1})$ then the sum $Z_1+…+Z_n$ is a supermartingale and the probability of it being bigger then $t$ gives you what you wanted.