Proof: Kullback-Leibler divergence for the binomial distribution
Index:
The Book of Statistical Proofs ▷
Probability Distributions ▷
Univariate discrete distributions ▷
Binomial distribution ▷
Kullback-Leibler divergence
Metadata: ID: P420 | shortcut: bin-kl | author: JoramSoch | date: 2023-10-20, 12:49.
Theorem: Let $X$ be a random variable. Assume two binomial distributions $P$ and $Q$ specifying the probability distribution of $X$ as
\[\label{eq:bins} \begin{split} P: \; X &\sim \mathrm{Bin}(n, p_1) \\ Q: \; X &\sim \mathrm{Bin}(n, p_2) \; . \end{split}\]Then, the Kullback-Leibler divergence of $P$ from $Q$ is given by
\[\label{eq:bin-KL} \mathrm{KL}[P\,||\,Q] = n p_1 \cdot \ln \frac{p_1}{p_2} + n (1-p_1) \cdot \ln \frac{1-p_1}{1-p_2} \; .\]Proof: The KL divergence for a discrete random variable is given by
\[\label{eq:KL-disc} \mathrm{KL}[P\,||\,Q] = \sum_{x \in \mathcal{X}} p(x) \, \ln \frac{p(x)}{q(x)}\]which, applied to the binomial distributions in \eqref{eq:bins}, yields
\[\label{eq:bin-KL-s1} \begin{split} \mathrm{KL}[P\,||\,Q] &= \sum_{x=0}^{n} p(x) \, \ln \frac{p(x)}{q(x)} \\ &= p(X=0) \cdot \ln \frac{p(X=0)}{q(X=0)} + \ldots + p(X=n) \cdot \ln \frac{p(X=n)}{q(X=n)} \; . \end{split}\]Using the probability mass function of the binomial distribution, this becomes:
\[\label{eq:bin-KL-s2} \begin{split} \mathrm{KL}[P\,||\,Q] &= \sum_{x=0}^{n} {n \choose x} \, p_1^x \, (1-p_1)^{n-x} \cdot \ln \frac{ {n \choose x} \, p_1^x \, (1-p_1)^{n-x} }{ {n \choose x} \, p_2^x \, (1-p_2)^{n-x} } \\ &= \sum_{x=0}^{n} {n \choose x} \, p_1^x \, (1-p_1)^{n-x} \cdot \left[ x \cdot \ln \frac{p_1}{p_2} + (n-x) \cdot \ln \frac{1-p_1}{1-p_2} \right] \\ &= \ln \frac{p_1}{p_2} \cdot \sum_{x=0}^{n} {n \choose x} \, p_1^x \, (1-p_1)^{n-x} x + \ln \frac{1-p_1}{1-p_2} \cdot \sum_{x=0}^{n} {n \choose x} \, p_1^x \, (1-p_1)^{n-x} (n-x) \; . \end{split}\]We can now see that some terms in this sum are expected values with respect to binomial distributions:
\[\label{eq:bin-means-s1} \begin{split} \sum_{x=0}^{n} {n \choose x} \, p_1^x \, (1-p_1)^{n-x} x &= \mathrm{E}\left[ x \right]_{\mathrm{Bin}(n,p_1)} \\ \sum_{x=0}^{n} {n \choose x} \, p_1^x \, (1-p_1)^{n-x} (n-x) &= \mathrm{E}\left[ n-x \right]_{\mathrm{Bin}(n,p_1)} \; . \end{split}\]Using the expected value of the binomial distribution, these can be simplified to
\[\label{eq:bin-means-s2} \begin{split} \mathrm{E}\left[ x \right]_{\mathrm{Bin}(n,p_1)} &= n p_1 \\ \mathrm{E}\left[ n-x \right]_{\mathrm{Bin}(n,p_1)} &= n - n p_ 1 \; , \end{split}\]such that the Kullback-Leibler divergence finally becomes:
\[\label{eq:bin-KL-qed} \mathrm{KL}[P\,||\,Q] = n p_1 \cdot \ln \frac{p_1}{p_2} + n (1-p_1) \cdot \ln \frac{1-p_1}{1-p_2} \; .\]∎
Sources: - PSPACEhard (2017): "Kullback-Leibler divergence for binomial distributions P and Q"; in: StackExchange Mathematics, retrieved on 2023-10-20; URL: https://math.stackexchange.com/a/2215384/480910.
Metadata: ID: P420 | shortcut: bin-kl | author: JoramSoch | date: 2023-10-20, 12:49.