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When the prior odds of a real effect are $R$, a study's Type I error is $\alpha$, and its statistical power is $1-\beta$, a positive report updates those odds to $R(1-\beta )/\alpha$. This follows directly from Bayes' theorem. A small false-positive rate does not itself make the effect's posterior probability high: the prior odds and power also matter. The posterior exceeds one half exactly when $R>\alpha /(1-\beta )$.

For $n$ studies conditionally independent given the truth of one relationship, the event that at least one is positive has probability $1-{\beta }^n$ under the alternative and $1-(1-\alpha {)}^n$ under the null. Its Bayes factor therefore gives the displayed posterior odds. For interior error probabilities these tend back to the prior odds $R$. If power exceeds size, the Bayes factor decreases to one: the quotient of geometric sums is a weighted average of decreasing powers $[\beta /(1-\alpha ){]}^k$.

This result concerns the coarsened selection event. Full observed results instead multiply their individual Bayes factors, and concordant evidence can accumulate without this bound. It is a useful example of information discarded by publication bias.

The inequality in the printed rule is reversed. With the stated definition of $L$, a larger score favors group one. For example, in one dimension take ${\mu }_1=1$, ${\mu }_2=-1$, $\Sigma =1$, equal priors and $x=1$. The score is $2$, and $f_1(1)>f_2(1)$, whereas the printed rule assigns group two. The correct Bayes classifier follows from minimizing the error cost as follows.

Let $R$ be the region assigned to group one, let $f_i$ be its multivariate normal density, and let each misclassification cost $c>0$, with zero cost for correct classification. The expected cost isThe permitted integral-minimization result therefore selects $R=\{x:{\pi }_1{f}_1(x)>{\pi }_2{f}_2(x)\}$. Points of equality may be assigned to either group without changing the risk. Equivalently, the Bayes classifier chooses the larger posterior probability.

For positive priors and a common positive-definite covariance matrix, the normalization constants cancel in the likelihood ratio. Expanding the two quadratic forms givesThus the correct Gaussian Bayes classifier, choosing group one on ties, isThis proves the intended classification result with the necessary correction. Equal error costs remove $c$ from the decision, while prior odds determine the threshold.

Now the prior odds are ${\pi }_f/{\pi }_a=2$. The fitted log posterior odds are $q(x)+\log 2$, so the Gaussian Bayes classifier assigns fattus when $q(x)>-\log 2$. The prior-dependent Gaussian discriminant boundary becomesIt moves upwards parallel to the original line by $0.8\log 2\simeq 0.5545$, expanding the fattus region. This is the dashed line in the sketch. The fifth individual's fitted log likelihood ratio remains zero, but its posterior odds become $2:1$, giving posterior fattus probability $2/3$.

The fifth individual is now classified as fattus. For completeness, the other four classifications remain apathus, fattus, apathus and fattus: their scores lie on the same side of the new threshold as before.

Use a multivariate normal distribution within each species, a common positive-definite covariance matrix, equal prior odds and equal misclassification costs. Assume the training observations are independent and representative of the populations. Substituting the sample means and pooled sample covariance matrix gives a fitted linear discriminant analysis rule, rather than a rule with known population parameters.

The determinant of the pooled covariance is $8$, soThe midpoint of the sample means is $(4,6{)}^T$, whose scalar product with $L$ is $-11/2$. Hence the fitted log likelihood ratio isAssign fattus when $q(x)>0$ and apathus when $q(x)<0$; either assignment is optimal on $q(x)=0$ under equal priors. The separating line is $x_2=0.4x_1+4.4$, with fattus below it. As a sign check, the scores at the two sample means are $19/4$ and $-19/4$, respectively. This uses the corrected Gaussian Bayes classifier from (a).

Let the prior odds be $\mathbb{P}(T)/\mathbb{P}(F)=R$, so $\mathbb{P}(T)=R/(1+R)$. The Type I error gives $\mathbb{P}(P\mid F)=\alpha$, while the statistical power gives $\mathbb{P}(P\mid T)=1-\beta$. By Bayes' theorem,This is a posterior probability of the null, not the false-positive probability $\alpha$ conditional on the null. The distinction is the base-rate effect in a positive study.

Let $A_n$ denote the event that at least one study is positive. Conditional independence gives $\mathbb{P}(A_n\mid T)=1-{\beta }^n$ and $\mathbb{P}(A_n\mid F)=1-(1-\alpha {)}^n$. Updating the prior odds by this event's Bayes factor givesThese are the odds under at-least-one-positive selection, not the odds using every study's actual result.

Compute the Bayesian model evidence for each model by integrating its sampling density against its parameter prior:The posterior odds are the prior odds multiplied by the Bayes factor:The parameter priors used for evidence must be normalized; arbitrary normalization constants in model-specific improper priors would make this comparison undefined.

If ${\pi }_0=\mathbb{P}(H_0)$ and ${\pi }_1=\mathbb{P}(H_1)=1-{\pi }_0$, the prior odds on the null hypothesis areThese odds concern which model is true before seeing the observations. They are separate from the prior distribution of the success probability within the alternative model.

Use the orientation that favours the null hypothesis:For a composite model the denominator is its Bayesian model evidence, obtained by integrating the likelihood against the model's prior distribution. This Bayes factor exceeds one when the observed data have greater marginal probability under $H_0$. By Bayes' theorem,Thus the posterior odds equal the prior odds multiplied by the Bayes factor. The reciprocal orientation instead gives $B_{10}=1/B_{01}$.

No. A Bayes factor favouring the alternative means $B_{01}<1$, but the posterior odds are $B_{01}{O}_{01}$. They are less than one only if $B_{01}<1/O_{01}$. For example, prior odds of $1000:1$ on the null hypothesis, combined with a Bayes factor of $100:1$ in favour of the alternative, still give posterior odds of $10:1$ on the null hypothesis. An unusual claim can require substantial evidence to overcome low initial plausibility.