Solution
= Solution
Let $q_n=(1-2/n)^n$. The product of two distinct <empty-box indicators in uniform allocation> is the <indicator random variable> of both boxes being empty, so $\mathbb E(I_iI_j)=q_n$. Thus
$$
\boxed{\operatorname{cov}(I_i,I_j)=q_n-p_n^2=\left(1-\frac2n\right)^n-\left(1-\frac1n\right)^{2n}<0.}
$$
Indeed $0\le1-2/n<(1-1/n)^2$ for $n\ge2$, and taking the positive integer power $n$ preserves this strict inequality. This proves the negative <covariance> directly.