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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 $R>\alpha /(1-\beta )$.

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 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 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.

The raven paradox combines a plausible equivalence principle with an overly unrestricted account of positive-instance scientific confirmation. Let $R(x)$ and $B(x)$ mean that $x$ is a raven and black respectively. ThenA white shoe is a positive instance of the right-hand generalization. If every positive instance confirms its universal generalization, and logically equivalent statements receive the same evidential support, inspecting the shoe seems to confirm the original raven claim. There is no contradiction in the equivalence itself. The tension concerns the evidential rule and the intuition that an arbitrary shoe teaches us nothing about birds.^[ref]

The right response is to make background knowledge and sampling explicit. Under Bayesian confirmation, evidence increases support for $H$ precisely when its likelihood is greater under $H$ than under the alternative, assuming nondegenerate prior probabilities. A fact can therefore be compatible with $H$ yet provide no new support. The mere existence of a white shoe does not determine the needed likelihood ratio.^[ref]

Consider a deliberately simple finite model. There are $N$ objects, exactly $r$ ravens, and $M=N-r$ nonravens, with $M>r>1$. All nonravens are white. Under $H_0$ all ravens are black; under $H_1$ exactly one raven is white. Draw uniformly from the entire population. The event of drawing a white nonraven has probability $M/N$ under either hypothesis. Its Bayes factor is one and it leaves the prior odds unchanged. A black raven instead has likelihoods $r/N$ and $(r-1)/N$, so gives Bayes factor $r/(r-1)$ for $H_0$.

Now change the protocol: draw uniformly from the white objects, then discover that the selected object is a nonraven. This result has likelihoods $1$ and $M/(M+1)$. It gives the small positive Bayes factorFor $N=1000$ and $r=10$, this is $991/990$, whereas sampling a black raven from the full population gives $10/9$. The white nonraven can now help because the search could have revealed a white raven. The two experiments have different selection information, so their differing evidential values do not violate logical equivalence.

This model establishes possibilities, not a universal numerical law about shoes. Different alternatives, population information or selection procedures change the result. If the shoe's being a nonraven was already known, discovering that it is white need not add anything. Conversely, ruling out a potential counterexample can supply a small amount of information. Retain logical equivalence, but reject the unconditional instance-confirmation rule. Once the complete evidential situation is specified, neither zero support nor weak positive support is paradoxical, and neither amounts to proof of a universal law.

Empirical equivalence must be specified relative to a domain and an account of what counts as observable. Two theories may fit all evidence collected so far while predicting different future results; that is ordinary underdetermination by a finite record. A stronger equivalence requires the same observable consequences for every permitted experiment in the stated domain. For stochastic predictions, matching a set of possible outcomes is not enough: the theories must assign the same outcome probability distributions under corresponding preparations and testing assumptions.

A clear physical example at the level of descriptions occurs in electromagnetism. Write the electric field and magnetic field using a scalar potential $\varphi$ and magnetic vector potential $\mathbf{A}$:For a smooth, single-valued function $\chi$, the gauge transformationleaves both fields unchanged: the curl of a gradient vanishes, and the mixed derivatives cancel in the electric-field expression. With the same charges, boundary conditions and force law, these descriptions give the same classical measurements. Agreement is structural, not an accidental fit to a few readings. However, many physicists regard such descriptions as representing one physical situation, so this alone is not an example of two genuinely incompatible physical theories. The smooth global transformation assumption also matters; equality of local fields is not an unrestricted license to identify every globally different potential configuration.

A second example exhibits possible surplus structure. For particles interacting by Newtonian forces depending only on relative positions, a common Galilean transformationleaves relative positions and accelerations unchanged. Accordingly, adding a privileged but completely uncoupled state of uniform rest does not alter any internal mechanical prediction. Rival choices of that unobservable rest state cannot be distinguished by these experiments. This is a deliberately specified completion of the mechanical model: it does not show that rotation or acceleration is undetectable, or that Newtonian mechanics and special relativity are empirically equivalent. The latter give different high-speed predictions. Whether undetectable absolute rest makes a genuinely different theory or a redundant representation is precisely one philosophical issue raised by the example.

Thus there are concrete cases of empirical redundancy, while the existence of equally well-motivated, fully distinct rivals with identical consequences across all physics is a much stronger claim. Historical theories often agree only in limited regimes or after additional assumptions. One must not convert that restricted agreement into an equivalence theorem for every possible test.

The philosophical relevance does not depend on finding a complete historical pair. A constructed rival can expose the logical limits of evidence, although an artificial extension need not deserve the same credibility as an economical account. If two specified models have equal likelihoods for every admissible data set $D$, thenfor finite positive prior odds. The Bayes factor is one; the data do not discriminate between them. Choices about simplicity or interpretation would be additional considerations. Constructive empiricism treats empirical adequacy as the relevant belief commitment in acceptance, while a scientific realist must explain what further warrants belief in the unobservable structure. The equivalence argument identifies a limit of evidential discrimination; it does not establish that every theoretical claim is equally good or that every unobservable claim is false.^[ref]

Bayes' theorem solves a precise part of the scientific confirmation problem: it gives a coherent relation between prior commitments, evidential likelihoods and updated probabilities. It does not select those commitments from no premises, and it should not be identified with every meaning of confirmation.

For background information $K$, a hypothesis $H$ and newly considered evidence $E$, let $p=P(H\mid K)$, $a=P(E\mid H,K)$ and $b=P(E\mid \neg H,K)$. With a positive denominator,The first formula requires a positive denominator; the odds formula additionally assumes $0<p<1$ and $b>0$. The second equation separates prior odds from the Bayes factor. For $0<p<1$, the evidence raises the probability of $H$ exactly when $a>b$. It is not enough that $E$ is likely under $H$; it must be more likely than under the relevant alternative mixture. This guards against an inverse-probability error in evaluating scientific tests.

For example, let the prior be 0.01 and the two likelihoods be 0.8 and 0.08. The Bayes factor is ten, but the posterior probability is only $10/109\simeq 0.0917$. The evidence strongly favours $H$ relative to its previous odds without making $H$ probable in an absolute sense. Evidence can therefore confirm a hypothesis while leaving a rational investigator doubtful. The comparison also depends on which alternatives are included in $\neg H$ and how their probabilities are distributed.

The raven paradox shows another benefit. Logically equivalent claims receive identical probabilities in the same probabilistic model, but this does not force every investigation of a white object to provide the same evidence about ravens. If an object is selected because it is already known to be a shoe, observing that it is not a raven may have equal likelihood under the rival hypotheses and provide a Bayes factor of one. Random selection from a specified population can give different likelihoods. The sampling procedure and background information must be represented, rather than letting a grammatical positive-instance rule determine evidential support.^[ref][ref]

The problem of old evidence is a genuine limit of the simplest updating account. If $E$ already belongs to $K$, conditioning on it again produces no increase. Yet an old measurement can be important support for a newly formulated theory. One may assess support relative to a specified earlier or counterfactual background, or model learning a previously unrecognized connection between theory and evidence. These are substantive additions. Simply applying the same conditioning rule to an already certain fact does not represent the new intellectual achievement.^[ref]

Other limitations concern model construction and priors. A hypothesis assigned zero prior probability remains at zero after ordinary conditioning on evidence of positive total probability. A badly specified likelihood can generate confident but unreliable results. Different reasonable priors need not disappear under finite or non-discriminating evidence. Nor does the theorem supply a noncircular solution to the problem of induction: the model's connection to future physical events remains an empirical commitment.

Thus the Bayesian framework gives a rigorous account of probability change, comparison of alternatives and some apparent paradoxes. These are real achievements. It is incomplete as a general recipe for scientific judgment unless the hypotheses, backgrounds, likelihoods and forms of learning are also defended. A mathematical rule of coherent updating can organize evidential reasoning without establishing that every coherent input represents nature correctly.

Scientific underdetermination occurs when the relevant evidence does not select a unique theoretical account. Its force depends on the scope of “the evidence”. Several models can agree with a finite data set while disagreeing about new measurements; a suitable experiment can then discriminate them. The stronger claim of empirical equivalence says that rivals agree on every possible observation within a stated domain, so further evidence of that same kind cannot separate them.

The Duhem problem gives a related but distinct source of ambiguity. If a target theory $T$ and auxiliary hypotheses $A$ jointly imply an observation $E$, then observing $\neg E$ shows $\neg (T\wedge A)$, not uniquely $\neg T$. Independent evidence about the apparatus or background assumptions can redistribute support. It does not make the original conjunction logically true; it provides additional information relevant to which revision is warranted.

A finite-data example makes the limits clear. If $p_0$ interpolates observations at distinct $x_1,\dots ,x_n$, then every polynomial $p_c(x)=p_0(x)+c{\prod }_{i=1}^n(x-x_i)$ has the same sampled values. Thus polynomial interpolation does not determine a unique prediction outside the sampled points when the degree is unrestricted. A degree bound of $n-1$ would instead give the familiar uniqueness result, so the allowed model class matters. That does not show that every interpolant is equally credible for extrapolation. Smoothness assumptions, independently motivated mechanisms and further measurements can matter. But those extra commitments must be stated rather than smuggled into the claim that the original data uniquely determined a model.

In a probabilistic comparison, Bayes' theorem giveswhen the quantities are defined. Equal likelihoods leave the prior odds unchanged by this evidence. Neither equality nor unequal priors follows merely from both theories being logically compatible with $E$. Thus compatibility, equal fit in a particular measure and full empirical equivalence should not be conflated.

The philosophical significance is a constraint on inference, not a universal proof of irrationality. An empiricist may use it to motivate restraint about hidden structure; a scientific realist may defend further explanatory or intervention-based reasons for belief. A convincing sceptical argument must identify genuine rivals and the relevant equivalence, while a convincing realist reply must explain why its additional virtues track the claims at issue. Scientific knowledge can be selective and fallible even when evidence does not deductively choose a unique complete theory.

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.

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