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For nonnegative $a_i,b_i$, with $a={\sum }_i{a}_i$ and $b={\sum }_i{b}_i$, the log-sum inequality isIt proves both Gibbs inequality and the data processing inequality for relative entropy.

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

Let $K(y\mid x)$ be a Markov kernel, and let $P_Y=PK$, $Q_Y=QK$ be the output probability distributions. Applying the log-sum inequality for each $y$ to $a_x=P(x)K(y\mid x)$ and $b_x=Q(x)K(y\mid x)$ givesSumming over $y$ and using ${\sum }_{y}K(y\mid x)=1$ yields the data processing inequality for relative entropy

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.