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Put $p_i=\mathbb{P}(X_i=1)$ and $\lambda ={\sum }_i{p}_i$. A relative-entropy form of the Poisson approximation bound for dependent Bernoulli variables isThe last two terms form the total correlation; they vanish when the Bernoulli variables are independent.

Here $D_e$ and $H_e$ use natural logarithms.

To prove the bound, let $Q_i$ be the Poisson distribution with mean $p_i$ and let $Q={⨂}_i{Q}_i$. Expanding the Kullback-Leibler divergence against this product law givesThe supplied one-dimensional estimate bounds the first sum by ${\sum }_i{p}_i^2$. Under the addition map, $P_{X_1^n}$ becomes $P_{S_n}$, while the sum of independent Poisson random variables under $Q$ has the Poisson distribution with mean $\lambda$. The data processing inequality for relative entropy proves the displayed result.

If a bound directly in the paper's unhalved total-variation norm is desired, Pinsker's inequality also gives

For Bernoulli random variables $X_1,\dots ,X_n$, which need not be independent, put $p_i=\mathbb{P}(X_i=1)$, $S={\sum }_i{X}_i$, and $\lambda ={\sum }_i{p}_i$. ThenThe dependence penalty is the total correlation. The proof compares the joint Bernoulli law with the product of Poisson laws of means $p_i$, then applies the data processing inequality for relative entropy to addition.