 Solution

For each $i$, the total variation distance between $\mathrm{Bernoulli}(p_i)$ and $\mathrm{Poisson}(p_i)$ isIndeed, compare their masses at zero, one, and the Poisson tail; the positive excess of Bernoulli mass at one is $p_i(1-e^{-p_i})$. A maximal coupling therefore gives a pair $(X_i,N_i)$ with mismatch probability at most $p_i^2$. Couple these pairs independently. Thenby the union bound. Since ${\sum }_i{N}_i\sim \mathrm{Poisson}({\sum }_i{p}_i)=\mathrm{Poisson}(\nu )$, part (b) yields