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A Bayes factor measures how data change the relative support for two specified statistical models. Write $m_j(D)=p(D\mid H_j)$ for their Bayesian model evidence and define the direction explicitly:For positive finite $m_0(D),m_1(D)$, a value above one favors $H_1$ over $H_0$; a value below one favors $H_0$ over $H_1$. Reversing the comparison gives $B_{01}={\mathrm{BF}}_{01}=1/{\mathrm{BF}}_{10}$, the opposite direction used in some formulas below.

For mutually exclusive, exhaustive $H_0,H_1$ with positive prior probabilities, divide their two Bayes' theorem formulas. Their common data normalizer cancels, leavingThe Bayes factor therefore multiplies prior odds into posterior odds. It is evidence conditional on the specified statistical models and their within-model prior distributions; it is not $\mathbb{P}(H_1\mid D)$. Model prior probabilities supply a separate input.^[ref]

For an original stipulated example, consider ten independent and identically distributed Bernoulli trials, with heads as success. Compare $H_0:p=0.4$ with $H_1:p=0.7$, both fixed before seeing the data. Take $D$ to be one particular ordered sequence containing eight heads and two tails. The likelihood function under either point model is $p^8(1-p{)}^2$, soIf $\mathbb{P}(H_1)=0.1$, the prior odds for $H_1$ against $H_0$ are $1/9$. Hence the posterior odds are $21.9909706/9\approx 2.4434412$, andEqual model prior probabilities would instead give ${\mathrm{BF}}_{10}/(1+{\mathrm{BF}}_{10})\approx 0.956505$. The same Bayes factor can thus accompany different posterior probabilities. These statements condition on the two-model comparison; they do not establish that either statistical model describes a real coin.

Using the heads count $K=8$ instead of the ordered sequence changes the sample space. The binomial distribution assigns it mass $\left(\genfrac{}{}{0ex}{}{10}{8}\right)p^8(1-p{)}^2$. The same binomial coefficient multiplies both models' Bayesian model evidence and cancels in their Bayes factor. Mixing an ordered-sequence numerator with a count denominator would introduce an erroneous factor of $45$.

Fix the candidate statistical models, their parameter prior distributions, and model prior probabilities before using the current data for this update. The Bayes factor is useful because it separates evidence from model prior odds, but it still depends on the parameter prior distributions. An improper prior has an arbitrary multiplicative normalization: multiplying its kernel by $c$ multiplies its formal marginal likelihood by $c$, even if the within-model posterior distribution is normalizable. Without a separately justified convention, the ordinary Bayes factor is then undefined up to arbitrary constants.^[ref]

The Bayesian model evidence is the prior predictive distribution evaluated at the data $D$. For a statistical model $H_j$ with a normalized prior density ${\pi }_j$, its marginal likelihood isThis is an expected value of the likelihood function under the prior distribution; for discrete statistical parameters, replace the integral by a sum. For continuous observations, $m_j(D)$ is a probability density function value with respect to a specified data reference measure, rather than a probability of one exact observation. Comparisons must use the same data and reference measure. The same $m_j(D)$ normalizes the within-model Bayesian posterior and supplies the between-model comparison in a Bayes factor.^[ref]

A normalized prior distribution makes the Bayesian model evidence an average, so it cannot exceed the largest likelihood function value on its support. A broad prior distribution can assign little mass to values that predict the data well, even when one value gives an excellent fit. The worked uniform-alternative binomial Bayes factor below makes the difference between averaging and maximum likelihood estimation explicit.

Let $\pi$ be a proper prior density, let $L>0$ for prior-almost every parameter, and let $Z=\int L\pi \in (0,\infty )$. For posterior draws ${\theta }_i$, the harmonic mean estimator isThe identity ${\mathbb{E}}_{\pi (\cdot \mid y)}[1/L]=1/Z$ follows by cancelling $L$ in the posterior density. The variance of $1/L$ is often infinite, making the estimate unstable even when an ergodic average is consistent.^[ref]

The support qualification matters. If $a={\int }_{\{L>0\}}\pi <1$, cancellation instead gives $a/Z$. For a uniform prior on $[0,1]$ and $L(\theta )=1_{[0,1/2]}(\theta )$, the evidence is $1/2$, but every posterior likelihood value is one and the displayed estimator is always one. An improper prior also leaves the absolute evidence normalization undefined.

Suppose $X_i\sim N(\mu ,1)$ independently, and compare $H_0:\mu =0$ with $H_1:\mu \sim N(0,{\tau }^2)$ for fixed $\tau >0$, using equal prior model probabilities. Put $z=\sqrt{n}\overline{X}$. The posterior probability of a Gaussian point null isHolding $z$ fixed while $n\to \infty$ leaves the two-sided p-value $2\Phi (-|z|)$ unchanged but makes the null posterior probability tend to one. For $n=100$, $\tau =1$ and $z=2$, the p-value is about $0.0455$, while the null posterior is about $0.581$.

The two procedures answer different model-based questions: a tail probability under the point null and a comparison of that null with a specified alternative prior distribution. The alternative's fixed prior scale spreads mass over parameter values much wider than the increasingly precise likelihood. This is the Jeffreys-Lindley phenomenon, not a conversion formula from p-values to posterior probabilities.^[ref]

Compare an unrestricted model $H_1$ with parameters $(\psi ,\lambda )$ to $H_0:\psi ={\psi }_0$. Suppose the likelihoods agree under that restriction, the priors are proper, and the nuisance prior under $H_0$ equals the conditional prior ${\pi }_1(\lambda \mid {\psi }_0)$ under $H_1$. Using compatible density versions with $0<{\pi }_1({\psi }_0)<\infty$, Bayes' theorem givesIndeed, the numerator of the marginal posterior density at ${\psi }_0$ is ${\pi }_1({\psi }_0)\int L(y\mid {\psi }_0,\lambda ){\pi }_1(\lambda \mid {\psi }_0)d\lambda ={\pi }_1({\psi }_0)m_0(y)$.

This ratio concerns densities in the unrestricted model, not a posterior atom at the null. A separately chosen nuisance prior under the null can invalidate the simple ratio; matching marginal nuisance priors is insufficient when $\psi$ and $\lambda$ are prior-dependent. Density values on a measure-zero restriction also require an agreed continuous or otherwise specified version.^[ref]