Gibbs' inequality

From Wikipedia, the free encyclopedia

In information theory, Gibbs' inequality is a statement about the mathematical entropy of a discrete probability distribution. Several other bounds on the entropy of probability distributions are derived from Gibbs' inequality, including Fano's inequality.

It was first presented by J. Willard Gibbs in the 19th century.

[edit] Gibbs' inequality

Suppose that

$P = \{ p_1 , \ldots , p_n \}$

is a probability distribution. Then for any other probability distribution

$Q = \{ q_1 , \ldots , q_n \}$

the following inequality between positive quantities (since the p_i and q_i are positive numbers less than one) holds

$- \sum_{i=1}^n p_i \log_2 p_i \leq - \sum_{i=1}^n p_i \log_2 q_i$

with equality if and only if

$p_i = q_i \,$

for all i.

Note that the use of base-2 logarithms is optional, and allows one to refer to the quantity on each side of the inequality as an "average surprisal" measured in bits. The difference between the two quantities is the negative of the Kullback–Leibler divergence or relative entropy, so the inequality can also be written:

$D_{\mathrm{KL}}(Q\|P) \equiv - \sum_{i=1}^n p_i \ln \frac{q_i}{p_i} \geq 0$

[edit] Proof

Since

$\log_2 a = \frac{ \ln a }{ \ln 2 }$

it is sufficient to prove the statement using the natural logarithm (ln). Note that the natural logarithm satisfies

$\ln x \leq x-1$

for all x with equality if and only if x=1.

Let $I$ denote the set of all $i$ for which p_i is non-zero. Then

$\begin{align} - \sum_{i \in I} p_i \ln \frac{q_i}{p_i} & {} \geq - \sum_{i \in I} p_i \left( \frac{q_i}{p_i} - 1 \right) \\ & {} = - \sum_{i \in I} q_i + \sum_{i \in I} p_i \\ & {} = - \sum_{i \in I} q_i + 1 \\ & {} \geq 0 \end{align}$

$- \sum_{i \in I} p_i \ln q_i \geq - \sum_{i \in I} p_i \ln p_i$

and then trivially

$- \sum_{i=1}^n p_i \ln q_i \geq - \sum_{i=1}^n p_i \ln p_i$

since the right hand side does not grow, but the left hand side may grow or may stay the same.

For equality to hold, we require:

$\frac{q_i}{p_i} = 1$ for all $i \in I$ so that the approximation $\ln \frac{q_i}{p_i} = 1 - \frac{q_i}{p_i}$ is exact.
$\sum_{i \in I} q_i = 1$ so that equality continues to hold between the third and fourth lines of the proof.