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For probability mass functions $P,Q$, their Chernoff affinity at $0<s<1$ is the sum $Z(s)={\sum }_{x}P(x{)}^{s}Q(x{)}^{1-s}$. It weights pointwise agreement between the two laws. The weighted arithmetic-geometric mean inequality gives $P(x{)}^{s}Q(x{)}^{1-s}\le sP(x)+(1-s)Q(x)$, hence $Z(s)\le 1$. With positive common support, $Z(0)=Z(1)=1$. Smaller affinity means a stronger separation at that weighting.

At $s=1/2$, $Z(1/2)={\sum }_x\sqrt{P(x)Q(x)}$ is the Bhattacharyya coefficient. The quantity $-\ln Z(1/2)$ is the Bhattacharyya distance. Allowing all weights $s$ gives Chernoff information at least this large. Symmetry under exchanging $P,Q$ sends $s$ to $1-s$, so Chernoff information is symmetric even though an optimizing weight need not be $1/2$.

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.