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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]

For a random sample $X_i\mid \theta \sim N(\theta ,{\sigma }^2)$ with known $\sigma >0$ and prior distribution $\theta \sim N(0,{\tau }^2)$ with $\tau >0$, compare $H_0:\theta <0$ to $H_1:\theta \ge 0$. The prior odds are one by symmetry. Normal-normal conjugacy gives a normal posterior distribution withIts negative and nonnegative masses are $\Phi (-z)$ and $\Phi (z)$, whereSince the Bayes factor is the posterior odds divided by the prior odds,Here $\Phi$ is the cumulative distribution function of the standard normal distribution. The ratio decreases with the sample mean. For any threshold $b>0$, the inequality $B_{01}<b$ is equivalent to $\Phi (z)>1/(1+b)$ and hence toFor $b=1/10$, the quantile is ${\Phi }^{-1}(10/11)\approx 1.33518$, and the equivalent posterior probability threshold for $H_0$ is $1/11$. The symmetry and proper normalization of the two hypothesis-conditional priors are essential to replacing the Bayes factor by the posterior odds directly.

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]

Behavior $B$ is evidence for a disposition $D$, given a situation $S$, to the extent that its conditional likelihood ratio $P(B\mid D,S)/P(B\mid \neg D,S)$ differs from one. By Bayes' theorem this ratio multiplies the prior odds to give the posterior odds. If a constraint forces the same behavior regardless of the disposition, the ratio is one and the behavior leaves those odds unchanged. This does not exclude diagnostic information in other features, such as wording or voluntary choices. The argument identifies the assumptions needed to call a dispositional attribution biased.

For two models with finite positive prior odds, a sequence of data with identical conditional likelihoods under both models has Bayes factor one. The posterior odds therefore equal the prior odds. This requires equality of the probability model for all the data considered, not merely that both models fit a selected data set.

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.

Scientific underdetermination limits what evidence alone establishes, but its practical consequences depend on the strength of the equivalence being claimed. Rival theories fitting present data need not agree on future observations. Rival formulations agreeing on every observable in a specified domain present a more persistent limitation within that domain.

With finitely many data points $(x_i,y_i)$, suppose an interpolation polynomial $p$ fits all $n$ measurements. Thenalso fits them for every $\lambda$. This proves nonuniqueness for a deliberately broad model class. It does not prove that all these functions are equally plausible physical theories. Constraints from independently tested mechanisms, bounds on derivatives, symmetries and the measurement process can sharply restrict the class. A new observation at a different point can also distinguish the displayed alternatives.

The Duhem problem concerns an additional kind of latitude: a prediction tests a theory together with auxiliary hypotheses. A mismatch can be assigned to different components. This is a reason to vary apparatus, calibrate independently and use multiple kinds of evidence. It is not a demonstration that any preferred theory can be made equally well supported by arbitrary repairs.^[ref]

Exact empirical equivalence requires more. If two accounts supply identical probability distributions for all observations allowed by a stated setup, observations of that kind cannot favour one through their likelihoods. Bayesian inference makes the limitation explicit: posterior odds equal prior odds when the likelihood ratio is one. Declaring a prior preference does not transform it into new experimental discrimination.

Theories can nevertheless differ in explanatory organization, computational usefulness or how they connect with a larger domain. Those are legitimate questions, provided their role is stated. If a proposed new intervention distinguishes them, the equivalence did not cover that enlarged domain. If the difference is instead only a change of redundant description, treating it as two competing physical worlds may create a spurious dispute; gauge symmetry provides an important case requiring care.

For scientific practice, the appropriate response is model comparison with explicit assumptions, searches for discriminating tests and honest limits on what has been established. For scientific realism, enduring empirical equivalence challenges any claim that the evidence uniquely fixes a full ontology. It does not make precise predictions or experimental decisions impossible. Underdetermination is consequently a constraint on inference whose seriousness must be demonstrated case by case, rather than a general proof that all scientific reasoning is arbitrary.

Dispositional attribution is most biased when a person's conduct is easy to encode as a trait but the constraints on that conduct are poorly represented or insufficiently used. The bias is conditional on attention, cognitive load, expectations and learned explanatory practices. Calling an inference biased requires a defensible benchmark: behavior sometimes really is informative about a person's enduring preferences. Correspondence bias concerns excessive correspondence between observed conduct and inferred disposition after relevant situational causes are taken into account, rather than the mere existence of a trait judgment.

The classic demonstration separates the direction of a political essay from whether its author chose that position. In the assigned-position condition, readers nevertheless inferred attitudes corresponding to the essay. Readers also distinguished choice from assignment, so the result supports insufficient discounting rather than complete blindness to constraint. The experiment by Edward E. Jones and colleagues makes the situation explicit and therefore challenges the explanation that perceivers simply had no situational information. It does not show that all information in an assigned essay is necessarily nondiagnostic: enthusiasm, wording or willingness to comply could still be informative under a different task model.^[ref]

Conditional diagnosticity in attribution provides a useful normative benchmark. Let $D$ denote a disposition, $B$ the observed behavior and $S$ the situation. By Bayes' theorem, posterior odds equal prior odds times a conditional likelihood ratio:If assignment makes the same behavior almost equally likely under either disposition, its likelihood ratio is near one. Treating this behavior as if it had been freely chosen exaggerates its diagnosticity. Conversely, repeated voluntary actions across varied situations can justify a strong dispositional attribution. The benchmark must therefore distinguish an inference from a single constrained act from an inference supported by a history of choices.

One mechanism is asymmetric attention. The actor is a discrete, changing object; the constraints can be dispersed across instructions, other people and background circumstances. A perceiver may accurately remember that a constraint existed without representing its causal force when interpreting the action. Explaining a nervous performance as anxiety is also economical: one familiar trait compresses many movements, whereas reconstructing an unfamiliar interview situation requires additional assumptions. This reasoning predicts more correspondence bias when the actor dominates attention, contextual information is ambiguous or perceivers lack relevant situational knowledge. Social salience alone, however, does not identify which trait will be inferred; expectations and category knowledge also shape interpretation.

The cognitive-load account receives a more direct test in Daniel T. Gilbert and colleagues' experiments. Participants watched silent videotaped clips of an apparently anxious target, with subtitles describing her discussion topics. Those required to rehearse the topics in order used this situational information less to adjust their trait judgments. The load therefore involved processing the context itself, rather than simply removing that information. Related experiments used constrained speech. The proposed sequence distinguishes identifying the behavior, deriving a characterization and correcting for context; the last step was especially vulnerable to cognitive load. This explains why merely telling observers about a constraint need not remove correspondence bias.^[ref]

The experimental manipulation supports a contribution from limited cognitive resources, but it does not prove that every social attribution follows a rigid three-stage sequence. A concurrent task may also alter attention to instructions or memory for diagnostic details. Context could influence the earliest interpretation when it is obvious and familiar. Thus the strongest claim is that effective situational use can require additional processing in these tasks, rather than that dispositional inference is universally automatic and situational inference universally controlled. A discriminating experiment would measure comprehension of the constraint as well as the final social attribution and vary processing time independently of contextual clarity.

Culture also changes the starting point of explanation. Michael W. Morris and colleagues compared American and Chinese explanations of social and physical events, using animated displays and accounts of killings in newspapers. The social explanations differed in their emphasis on individual characteristics and contextual causes; the comparison with physical events constrained an account based on a general inability to reason about causes. These findings suggest that learned explanatory frameworks influence which causes come readily to mind.^[ref]

These comparisons do not warrant treating national populations as uniform psychological types. Nationality was not randomly assigned, newspaper writing is not a transparent measure of private judgment, and language or knowledge of the events can affect the explanations elicited. Their theoretical importance is that a supposedly universal default must accommodate systematic variation. An observer who routinely represents relationships and constraints may need less later correction, rather than merely possessing more cognitive resources. Cross-cultural psychology therefore complements the resource account by asking what information is represented before correction becomes necessary.

Motivation matters through the same conditional logic. A consequential decision can encourage searching for alternatives, but a desire for a quick, blame-consistent explanation can instead favor a dispositional attribution. Neither effect follows from motivation alone; the relevant question is which information the motive makes useful. Clear constraints, sufficiently understood instructions, time to use them and observations across multiple settings should improve calibration. These are mechanistic predictions, not guarantees that deliberation removes every cognitive bias.

The evidence is best explained by an interaction between behavioral diagnosticity and the perceiver's representation of context. Correspondence bias is strongest when salient conduct is readily characterized but situational alternatives receive too little weight; it weakens when constraints are understood and effectively incorporated. This account preserves the genuine usefulness of trait inference while explaining why its accuracy varies across tasks, resources and cultural experience.

Let $p_0=\mathrm{Pr}(\theta \in {\Theta }_0)$ and $p_1=1-p_0$, assuming both are positive. The prior odds are $\alpha =p_0/p_1$. Given data $x$, the posterior odds areWith conditional joint likelihood $L(\theta ;x)={\prod }_{i}f(x_i\mid \theta )$, define the hypothesis-conditional marginal densitiesThe Bayes factor for $H_0$ relative to $H_1$ is $B=m_0(x)/m_1(x)$. The normalizations $p_0,p_1$ matter: the prior distribution within each hypothesis must have total probability one. Bayes' theorem now givesConverting posterior odds to posterior probability therefore gives

The symmetric continuous prior distribution gives $\mathrm{Pr}(\theta <0)=\mathrm{Pr}(\theta \ge 0)=1/2$, so the prior odds are one. Therefore the Bayes factor equals the posterior odds. Standardizing the normal posterior distribution, putThen $\mathrm{Pr}(H_0\mid x)=\Phi (-z)$ and $\mathrm{Pr}(H_1\mid x)=\Phi (z)$. The endpoint $\theta =0$ has zero posterior probability, so the strict/non-strict split causes no additional term. HenceIn particular, positive values of the sample mean make this Bayes factor smaller, favouring the positive-mean hypothesis.