I'm self-studying Vershynin's High-Dimensional Probability. The aim of exercise 5.4.14 is to provide a sharpness result related to the matrix Bernstein inequality. It reads:
Let $X$ be an $n\times n$ random matrix that takes values $e_ke_k^T$, $k=1,\dots,n$, with probability $1/n$ each. ($e_k$ is the canonical basis for $\mathbb{R}^n$). Let $X_1,\dots,X_N$ be independent copies of $X$. Consider the sum $S=\sum_{i=1}^N X_i$, which is a diagonal matrix. (b) Relating this to the classical coupon collector's problem, show that if $N\asymp n$, then $$E||S||\asymp \frac{\log n}{\log\log n}$$
Here's what I tried: Part (a) asked to show that $S_{ii}$ is the number of balls in the $i$-th bin, when $N$ balls are thrown into $n$ bins independently, which is: $$P[S_{ii}=k]={{N}\choose{k}}\left(\frac{1}{n}\right)^k\left(\frac{n-1}{n}\right)^{N-k}$$ Since $N\asymp n$, I tried to work with the binomial distribution (with parameters $N\asymp n$ and $1/n$). If so, $ES_{ii}=N/n\asymp 1$. Since $S$ is a diagonal matrix, $||S||=\max_{1\leq i \leq n}S_{ii}$, so we have a maximum over dependent binomials. By these elementary techniques, I get the crude bound: $$E||S||\leq \sum_{i=1}^nES_{ii}\asymp n$$ Can you help me to prove the stronger bound given in the exercise, and to understand the connection to the coupon collector's problem?