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For a finite measurable partition $(B_j)$ with positive probabilities, and an event $A$ of positive probability,Its beauty is that a model predicting observations from a hidden state can also describe uncertainty about the state after an observation. Each numerator is a joint probability; the denominator, supplied by the law of total probability, makes the answers sum to one. This is a direct consequence of conditional probability, rather than an extra axiom.

Select the coin from law of total probability once, then toss that same coin repeatedly, with conditionally independent tosses given its identity. Let $B$ denote the $3/4$-head coin, initially of prior probability $1/5$. After one head $H$, its posterior probability isThe likelihood of a head is larger under $B$, but $B$ remains less likely than the other coin because its prior probability was smaller. A large likelihood for an observation does not itself imply a large posterior probability for the hypothesis.

After an ordered sequence containing $h$ heads and $t$ tails, its likelihood ratio between the two coins is $(3/2{)}^h(1/2{)}^t$. Bayes' theorem therefore gives posterior oddsEach head multiplies the odds by $3/2$ and each tail by $1/2$. Three heads give posterior probability $27/59$, still below one half; four give $81/145$, above one half. The finite-model update and the distinction between prior probability, likelihood and posterior probability are set out in MIT 18.05, Reading 11: Bayesian Updating with Discrete Priors.

The hidden selection also creates dependence. Each toss has marginal head probability $11/20$, but two heads have joint probability $(4/5)(1/2{)}^2+(1/5)(3/4{)}^2=5/16$, greater than $(11/20{)}^2=121/400$. Their indicator covariance is $1/100$. Seeing a head makes the head-favoring coin more likely, so it also raises the chance of a future head. This is covariance induced by a shared latent variable, despite conditional independence of the tosses given the coin.

Historically, Thomas Bayes's essay on an unknown success probability appeared in 1763 after his death, communicated by Richard Price. Its opening problem asks how observed successes and failures constrain that unknown probability; Price's introductory letter explains the connection to reasoning from experience. That is a more substantial inference problem than simply rearranging today's elementary identity. ^[ref]

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 positive result's posterior odds of true to false areThus being more likely true requires $\boxed{R>\alpha /(1-\beta )}$. The question's weak inequality is necessary; equality gives a tie rather than a strict preference.

For these two simple hypotheses the Bayes factor is the ratio of their normal probability densities:Therefore $\boxed{B_{10}(y)=e^{2.49(y-1.245)}}$, and the posterior odds are $R{B}_{10}(y)$.

For $0<\alpha ,\beta <1$, both event probabilities tend to one, so the posterior odds tend to $\boxed{R}$. If $1-\beta >\alpha$, the event's Bayes factor decreases toward one. To see monotonicity, put $a=1-\alpha >\beta$ and writeThe last quotient is a weighted average of the decreasing values $(\beta /a{)}^k$, with weights $a^k$. Adding the next, smaller value decreases the average.

Thus the selection rule becomes less informative as more teams try: eventually a positive study is almost certain even under the null. This explains the hotter-field interpretation in Ioannidis's model. It describes evidence retained by this particular selection event, rather than proving that every individual study becomes less reliable or that the full accumulated evidence gets weaker.

Consistent independent replications should strengthen confidence; contradictory results can weaken it. The change depends on the actual evidence. If $k$ of $n$ studies are positive and all labels are available, the posterior odds areFor the normal model, retaining the full independent statistics instead givesEach study multiplies the existing odds by its own Bayes factor. Reading a complete synthesis is therefore different from learning only that somebody obtained a positive result. Publication bias, dependence between studies and changes in study design must be accounted for before applying the independent-product formula.

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.

Integrating the positive component's normal distribution prior gives its prior predictive distribution:The negative component gives $Y_i\mid -\sim N(0,1)$. Let $f_1(y)={\varphi }_{1+V}(y)$ and $f_0(y)={\varphi }_1(y)$ denote those centred normal densities. The point-null mixture prior assigns component odds $q/(1-q)$, so the posterior odds of the positive component areIts posterior probability is $O_i/(1+O_i)$. Evidence depends on $y_i^2$, so large effects of either sign favour the positive component. This is the marginal distribution before observing $y_i$, not the posterior predictive law for a second observation conditional on it.

Write $H_0,H_1$ for the two simple hypotheses, with prior probabilities ${\pi }_0,{\pi }_1$ and densities $f_0,f_1$. Under zero-one loss, the posterior expected losses of choosing $H_0$ and $H_1$ are respectively the posterior probabilities of $H_1$ and $H_0$. A Bayes decision rule therefore chooses $H_1$ ifchooses $H_0$ for the reverse inequality, and may randomize at equality. With positive priors this is a likelihood-ratio test with cutoff $k={\pi }_0/{\pi }_1$.

For a test $\varphi$, its prior risk is ${\pi }_0{E}_0\varphi +{\pi }_1(1-E_1\varphi )$. Choose $k$ and boundary randomization so that the likelihood-ratio test ${\varphi }_k$ has size $\alpha$, and choose priors with ratio $k$. Equivalence of normal and extensive analyses makes the pointwise posterior minimizer minimize this prior risk over all randomized tests. Thus for any $E_0{\varphi }^{\prime }\le \alpha$,This proves the Neyman-Pearson lemma: a likelihood-ratio test of size $\alpha$ has greatest power among tests of size at most $\alpha$. Such a cutoff and randomization come from a quantile of the likelihood ratio under $H_0$; always reject where $f_0=0<f_1$. Endpoint cutoffs are handled directly or by their limiting rules.

For the remaining assertions use the usual common-support formulation with positive densities. A family has a monotone likelihood ratio in $T$ if for every ${\theta }_1>{\theta }_0$, $f_{{\theta }_1}(x)/f_{{\theta }_0}(x)$ is a nondecreasing function $r(T(x))$. A monotone test here is an upper-cutoff test $\varphi =1_{T>c}+\eta 1_{T=c}$, $0\le \eta \le 1$, including constant tests. It does not mean an arbitrary smoothly increasing randomized rejection function.

Put $D(\theta )=\gamma (\theta )-{\gamma }^{\prime }(\theta )$. For a finite cutoff, choose $k>0$ between the values of $r$ below and above $c$, with $k=r(c)$ when the boundary has positive probability. On $T>c$, $\varphi -{\varphi }^{\prime }\ge 0$ and $r-k\ge 0$; on $T<c$, both are nonpositive. At a boundary with positive mass $r-k=0$. ThereforeWhen there is probability on both sides, finite positive monotone ratios below the cutoff bound those above it away from zero, so $k$ can be positive and finite. A boundary with positive mass instead supplies the positive finite value $r(c)$. If the cutoff yields a constant test on the support, positivity on the common support proves strict propagation directly: a test identically zero cannot have greater power, and for a test identically one any positive-probability nonrejection by the competitor remains positive-probability at the larger parameter. Hence $D({\theta }_0)>0$ implies $D({\theta }_1)>0$.

Let ${\theta }^{\ast }=\inf \{\theta :D(\theta )>0\}$, taking ${\theta }^{\ast }=+\infty$ when this set is empty. Below ${\theta }^{\ast }$ there is no positive difference. Above a finite ${\theta }^{\ast }$ there is a smaller parameter with positive difference, which propagates upward; the same reasoning applies when ${\theta }^{\ast }=-\infty$. ThusNo assertion at ${\theta }^{\ast }$ itself is needed.

For the final request fix any cutoff $a$ and prior density $\pi$. Set $A(x)={\int }_{\theta >a}\pi (\theta )f_{\theta }(x)d\theta$ and $B(x)={\int }_{\theta \le a}\pi (\theta )f_{\theta }(x)d\theta$. The posterior odds are $A(x)/B(x)$. If $T(x_2)\ge T(x_1)$, thenEvery bracket is nonnegative by monotone likelihood ratio. Equal statistics make the bracket zero, so the ratio depends only on $T$. Dividing where the posterior is defined proves the posterior odds are nondecreasing in $T$, with extended values allowed when one posterior mass is zero. This proves that posterior odds are monotone under a monotone likelihood ratio without restricting the prior's shape.

The common positive-support condition is material to the printed strict inequality. Under the broader definition permitting zero densities, take a two-parameter family on $T\in \{0,1,2\}$ with $f_0=(1/3,1/3,1/3)$ and $f_1=(0,0,1)$. Its likelihood ratio $(0,0,3)$ is nondecreasing. The cutoff test rejecting $T\ge 1$ has greater power than the test rejecting only $T=2$ at parameter zero, but both have power one at parameter one. Thus the strict assertion needs the usual positivity convention; it is false without it.

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.

The posterior probability $q_i=\mathbb{P}({\lambda }_i>1\mid y_i)$ answers a natural directional question: how likely is this area's risk to exceed the reference risk? It accounts for uncertainty rather than declaring an increase from the point estimate alone. However, it does not distinguish a negligible increase from a clinically important one, and it combines prior beliefs with evidence from the data. With an informative prior distribution, a large probability may already have been present before observation, or a genuinely surprising count may be strongly moderated by the prior.

To measure the evidence supplied by the observations, compare posterior and prior odds. If $q_0=\mathbb{P}({\lambda }_i>1)$ under the chosen proper prior distribution, the Bayes factor for increased versus nonincreased risk isThis follows from the posterior odds identity when the two models use the original prior conditioned respectively on ${\lambda }_i>1$ and ${\lambda }_i\le 1$. For the specified prior, $q_0$ is the upper tail at one of $\mathrm{Gamma}(16,16)$; centering its mean at one does not make this tail probability exactly $1/2$. To address seriousness itself, choose a substantive excess $c>0$ and report the posterior probability of ${\lambda }_i>1+c$, together with an interval for the size of the excess. An evidence ratio and a meaningful risk threshold answer different, complementary questions.

For two exhaustive mutually exclusive models of positive prior probability, prior odds compare their probabilities before observing the data. They must be distinguished from the parameter prior distributions within those models. A Bayes factor multiplies these prior odds to produce posterior odds.