Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate discrete distributions ▷ Discrete uniform distribution ▷ Kullback-Leibler divergence

Theorem: Let $X$ be a random variable. Assume two discrete uniform distributions $P$ and $Q$ specifying the probability distribution of $X$ as

$\label{eq:dunis} \begin{split} P: \; X &\sim \mathcal{U}(a_1, b_1) \\ Q: \; X &\sim \mathcal{U}(a_2, b_2) \; . \end{split}$

Then, the Kullback-Leibler divergence of $P$ from $Q$ is given by

$\label{eq:duni-KL} \mathrm{KL}[P\,||\,Q] = \ln \frac{b_2-a_2+1}{b_1-a_1+1} \; .$

Proof: The KL divergence for a discrete random variable is given by

$\label{eq:KL-disc} \mathrm{KL}[P\,||\,Q] = \sum_{x \in \mathcal{X}} p(x) \, \ln \frac{p(x)}{q(x)} \; .$

This means that the KL divergence of $P$ from $Q$ is only defined, if for all $x \in \mathcal{X}$, $q(x) = 0$ implies $p(x) = 0$. Thus, $\mathrm{KL}[P\,\vert\vert\,Q]$ only exists, if $a_2 \leq a_1$ and $b_1 \leq b_2$, i.e. if $P$ only places non-zero probability where $Q$ also places non-zero probability, such that $q(x)$ is not zero for any $x \in \mathcal{X}$ where $p(x)$ is positive.

If this requirement is fulfilled, we can write

$\label{eq:duni-KL-s1} \mathrm{KL}[P\,||\,Q] = \sum_{x=-\infty}^{a_1} p(x) \, \ln \frac{p(x)}{q(x)} + \sum_{x=a_1}^{b_1} p(x) \, \ln \frac{p(x)}{q(x)} + \sum_{x=b_1}^{+\infty} p(x) \, \ln \frac{p(x)}{q(x)}$

and because $p(x) = 0$ for any $x < a_1$ and any $x > b_1$, we have

$\label{eq:duni-KL-s2} \mathrm{KL}[P\,||\,Q] = \sum_{x=-\infty}^{a_1} 0 \cdot \ln \frac{0}{q(x)} + \sum_{x=a_1}^{b_1} p(x) \, \ln \frac{p(x)}{q(x)} + \sum_{x=b_1}^{+\infty} 0 \cdot \ln \frac{0}{q(x)} \; .$

Now, $(0 \cdot \ln 0)$ is taken to be $0$ by convention, such that

$\label{eq:duni-KL-s3} \mathrm{KL}[P\,||\,Q] = \sum_{x=a_1}^{b_1} p(x) \, \ln \frac{p(x)}{q(x)}$

and we can use the probability mass function of the discrete uniform distribution to evaluate:

$\label{eq:duni-KL-s4} \begin{split} \mathrm{KL}[P\,||\,Q] &= \sum_{x=a_1}^{b_1} \frac{1}{b_1-a_1+1} \cdot \ln \frac{\frac{1}{b_1-a_1+1}}{\frac{1}{b_2-a_2+1}} \\ &= \frac{1}{b_1-a_1+1} \cdot \ln \frac{b_2-a_2+1}{b_1-a_1+1} \sum_{x=a_1}^{b_1} 1 \\ &= \frac{1}{b_1-a_1+1} \cdot \ln \frac{b_2-a_2+1}{b_1-a_1+1} \cdot (b_1-a_1+1) \\ &= \ln \frac{b_2-a_2+1}{b_1-a_1+1} \; . \end{split}$
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Metadata: ID: P425 | shortcut: duni-kl | author: JoramSoch | date: 2023-11-17, 15:19.