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Chernoff information measures how quickly repeated independent evidence can distinguish two fixed probability distributions. Suppose a statistical hypothesis test observes $n$ independent and identically distributed random variables, with common law $P$ under one simple hypothesis and $Q$ under the other. Give the two hypotheses fixed positive prior probabilities, and let $p_n^{\ast }$ be the smallest achievable average error probability. For distinct positive probability mass functions on a common finite alphabet,Here $\ln$ is the natural logarithm, so the error exponent is measured in nats per observation. This optimal rate is unchanged by choosing different fixed positive prior probabilities; the best decision rule and finite-sample error can still change.^[ref]

The interpretation is $p_n^{\ast }=\exp (-nC(P,Q)+o(n))$: a larger Chernoff information means fewer observations are needed at the leading exponential scale. It does not assert the exact finite-sample identity $p_n^{\ast }=e^{-nC(P,Q)}$. The finite-sample Chernoff error bound and the explicit Chernoff information for complementary Bernoulli distributions make this distinction concrete. This is useful when designing repeated measurements or comparing sensors whose output probability distributions differ under two candidate conditions.

For probability mass functions with zero entries, define the affinity for $0<s<1$ and use $C(P,Q)=-\ln {\inf }_{0<s<1}{\sum }_{x}P(x{)}^{s}Q(x{)}^{1-s}$, taking endpoint limits when needed. This avoids an ambiguous $0^0$ convention. Identical laws give $C=0$; disjoint probability supports give $C=\infty$, because one observation separates them perfectly. The displayed finite-rate theorem concerns fixed laws with common positive support, not laws that change with $n$ or arbitrarily dependent repeated observations.

Let $P$ be the Bernoulli distribution with success probability $0.2$ and $Q$ the Bernoulli distribution with success probability $0.8$. For $0\le s\le 1$,The two summands have product $0.16$. Their arithmetic mean is at least their geometric mean, with equality at $s=1/2$. ThusThe same argument gives $C=-\ln (2\sqrt{p(1-p)})$ for complementary parameters $0<p<1$. At $p=1/2$ the laws agree and $C=0$.

With equal prior probabilities, count the successes $K$ among $n$ observations. The likelihood ratio is $Q^{\otimes n}/P^{\otimes n}=4^{2K-n}$, so choose $Q$ when $K>n/2$ and $P$ when $K<n/2$. If $n$ is even, either decision or a fair tie randomization gives the same average error at $K=n/2$. Under $P$, $K$ has the binomial distribution with parameters $n,0.2$. Symmetry therefore gives the exact optimal average errorThe last term applies only for even $n$; it represents half the tie probability. Direct finite sums give $p_1^{\ast }=0.2$, $p_5^{\ast }=0.05792$ and $p_{20}^{\ast }\simeq 0.00157912$. Twenty observations reduce the optimal average error to about 0.158%. The finite-sample Chernoff error bound at $n=20$ is $\frac{1}{2}(0.8{)}^{20}\simeq 0.00576461$, a guarantee rather than the exact error.

Reducing the exact error from $p_{20}^{\ast }$ to $p_{50}^{\ast }\simeq 1.29363\times 10^{-6}$ illustrates the value of repeated evidence. The example assumes a single fixed coin law throughout the experiment. If repeated sensor readings share an unmodelled common error, increasing their number need not supply this much independent information.