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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 $p_{\mathrm{old}}$ and $p_{\mathrm{new}}$ be the underlying full-recovery probabilities. The two-sided statistical hypothesis test isUse the Pearson chi-squared test of homogeneity for the two independent groups. Under $H_0$, the pooled recovery estimate is $\hat{p}=56/100=0.56$, so each row has expected counts 28 recovered and 22 not recovered. All expected counts comfortably exceed five. The Pearson chi-squared statistic for contingency tables isThe asymptotic reference distribution is a chi-squared distribution with $(2-1)(2-1)=1$ degree of freedom. Since $1.461<3.84$, do not reject equal recovery probabilities at the 5% level. The observed recovery proportions, 0.50 and 0.62, do not provide statistically significant evidence of a difference in this trial. This conclusion does not prove that the procedures have identical recovery probabilities.