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Old evidence challenges a simple account of confirmation as actual conditional updating, but does not refute every probabilistic treatment of the raven paradox. The strongest response distinguishes support from newly acquired information and makes the sampling procedure explicit.

Let $H$ say that every raven is black. Its contrapositive says that every nonblack thing is a nonraven. Since these statements are logically equivalent, evidence should support them equally. Combining this equivalence requirement with the unrestricted rule that every positive instance confirms a universal generalization appears to make discovering a white shoe support $H$. The tension is between that result and the judgement that such a discovery is irrelevant.^[ref]

Bayesian confirmation replaces an automatic instance rule with a comparison of likelihoods. For background $K$, positive prior probability for both $H$ and its alternative, and evidence $E$ of positive probability,The equivalence follows directly from Bayes' theorem. A statement's grammatical appearance as a positive instance does not determine either likelihood.

Consider an intentionally simple finite model. There are $r$ ravens and $M$ nonblack objects, with these counts fixed under both hypotheses. Under $H$ none of the nonblack objects is a raven; under the sole alternative $H_1$, exactly one is. Selecting uniformly among the nonblack objects and discovering a nonraven has likelihoods $1$ and $1-1/M$. Selecting uniformly among the ravens and discovering a black one has likelihoods $1$ and $1-1/r$. With equal prior probabilities, the respective posterior probabilities of $H$ areIf $M\gg r$, the first discovery supplies far less support. For $M=10^6$ and $r=100$, the increases over $1/2$ are approximately $2.5\times 10^{-7}$ and $0.00251$. This illustrates a possible resolution of the intuitive disparity, not a theorem that every nonblack nonraven always confirms $H$. More complex alternatives can reverse the comparison, and examining something already known to be a shoe need supply no relevant new information.^[ref]

Now suppose $E$ is already in $K$. Then $P(E\mid K)=1$, and ordinary conditioning givesThere is no further increase from learning the same fact again. Yet an already established observation can become an important reason to accept a subsequently developed theory. Clark Glymour's problem of old evidence exposes this mismatch between actual updating and scientific assessments of evidential support.^[ref]

One response evaluates the observation relative to an earlier or counterfactual background that does not contain it. That can represent how strongly a theory accounts for the observation compared with alternatives. It must specify that background rather than pretend to remove knowledge uniquely from a person's present beliefs. Another response concerns learning that a theory successfully accounts for the observation. That discovery can be new even when the observation is old. Modelling it requires attention to limited reasoning and changing hypotheses; an ideal agent already knowing every logical consequence cannot simply learn an entailment it previously overlooked.

Neither manoeuvre licenses counting the same evidence twice, or ignoring that parameters were selected to fit it. Neither establishes a unique complete theory of confirmation. Nevertheless, the old-evidence objection concerns the interpretation and completeness of a probabilistic framework; it does not invalidate the likelihood calculation for genuinely new sampling information.

The preferable solution combines a background-sensitive account of support with rejection of unrestricted instance confirmation. It preserves logical equivalence while allowing a white shoe to provide tiny support, no support or contrary support under different justified models. This is more explanatory than assigning confirmation solely from logical form, although assessing old evidence requires additional work beyond routine conditioning.

Scientific confirmation is evidential support rather than deductive proof. Two distinct problems concern, first, what counts as relevant positive evidence and, second, which part of a tested theoretical package that evidence supports. A defensible response makes background information and the production of evidence explicit. The following two problems are independent: resolving the logical puzzle about positive instances does not eliminate ambiguity about auxiliary hypotheses.

Problem one: the raven paradox. Write the hypothesis aswhere $R$ means “is a raven” and $B$ “is black”. Its contrapositive isThe hypotheses are logically equivalent. A plausible equivalence principle says that identical information must support them equally. A plausible instance principle says that an observed $F$ that is $G$ confirms “every $F$ is $G$”. Combining the principles makes a nonblack nonraven, such as a white shoe, support the raven hypothesis. That conflicts with the intuition that inspecting shoes tells us nothing about ravens. Hempel develops the puzzle in his analysis of instance-based scientific confirmation.^[ref]

My preferred response retains logical equivalence while rejecting an unconditional instance rule. By Bayes' theorem, for evidence $E$, background $K$ and positive relevant probabilities,An instance's bearing depends on how likely it was under the competing possibilities, including how the object was selected. No rule depending only on the grammatical form of an instance can settle that comparison. Branden Fitelson and James Hawthorne analyse the role of background assumptions and Bayesian confirmation in this puzzle.^[ref]

Here is an explicit illustrative model. There are 100 labelled objects: ten known ravens and ninety known nonravens. Under $H$ all ten ravens are black; under the sole alternative $A$, exactly one of the ten is nonblack, with its identity uniformly distributed. Nonraven colours are already known and identical under both possibilities, and the prior probabilities of $H,A$ are equal. Selecting a known white shoe according to a rule independent of $H,A$ has the same likelihood under each, so the posterior probability of $H$ stays $1/2$. A uniformly selected raven, however, is black with probability $1$ under $H$ and $9/10$ under $A$. After observing it to be black,If sampling is instead from known nonblack objects whose membership is uncertain, discovering a nonraven can rule out a possible counterexample and provide support. There is no context-independent prohibition on support from a nonraven. The response explains both the familiar irrelevance intuition and why it is not universally correct. Its limit is that the prior probabilities and sampling assumptions need independent defence; the calculation is a transparent model, not a theorem that all shoes are irrelevant.

Problem two: the Duhem problem of testing a package. Suppose the target hypothesis $T$ and auxiliary hypotheses $A$ jointly imply an outcome $E$:A failure gives $\neg (T\wedge A)$, equivalently $\neg T\vee \neg A$, rather than $\neg T$ alone. Success also does not establish either premise: that would affirm the consequent. Pierre Duhem's point is especially pressing for physical experiments whose predictions require theories of the apparatus as well as the target system.^[ref]

My preferred response combines independent calibration, deliberately varied tests and explicit comparison of packages. For a simple model, suppose a detector has independently calibrated false-positive probability $f$ and detection efficiency $d$. A target predicts a signal on each trial with probability $s$. Then the positive-output probability isUnder a no-signal rival it is $p_0=f$. If $s=1/2$, $d=9/10$, and $f=1/10$, then $p_T=1/2$. One positive reading supplies a Bayes factor of $5$ for the specified target package against the specified no-signal package. But an uncalibrated faulty detector with false-positive probability $1/2$ fits the same positive-reading rate. Target data alone do not distinguish those packages.

Signal-absent calibration trials test the false-positive assumption directly. If ten independent such trials are all negative, their likelihood ratio in favour of $f=1/10$ against $f=1/2$ isThis does not magically isolate $T$: the calibration relies on its own assumptions, and correlated trials would invalidate the power calculation. It shows concretely how additional evidence can discriminate explanations that the initial data leave unresolved. Repeating the target measurement with apparatus based on different physical processes further reduces dependence on one shared failure mode.

Thus the Duhem problem defeats automatic attribution of support or failure, not rational experimental comparison. Where rivals remain empirically equivalent, suspension of judgement about their difference is appropriate. Together, the two responses defend scientific confirmation as a structured, fallible comparison using background information, rather than either pure instance counting or a mechanical verdict on an isolated hypothesis.

Bayes' theorem updates a parameter distribution using measured data:The likelihood describes the measurement process, the prior distribution encodes information before the present data, and the denominator normalizes the posterior distribution. Include instrumental noise and known selection effects in the likelihood. nuisance parameters, such as an uncertain background, should be jointly inferred and marginalized, rather than silently fixed.

For $n$ independent readings with a Gaussian distribution with known common variance ${\sigma }^2$ and prior distribution $\theta \sim N({\mu }_0,{\tau }_0^2)$, completing the square givesThe posterior distribution is $N({\mu }_n,{\tau }_n^2)$: prior distribution and data contribute in proportion to precision. A broad prior distribution recovers the usual estimate $\bar{x}$ with standard deviation $\sigma /\sqrt{n}$. Report a posterior distribution estimate and credible interval, specify the prior distribution, and examine sensitivity when the data are weak. A credible interval is a posterior distribution probability statement about the parameter conditional on the model; it is not defined by the repeated-sampling coverage of a confidence interval.

The likelihood is $L(\theta ;d)=p(d\mid \theta )$ considered as a function of parameters at the observed data. It is not itself a probability distribution over $\theta$. Bayes' theorem givesThe prior represents parameter information before these data; the denominator normalizes the posterior and can compare fully specified models with compatible prior assumptions. Detector response, backgrounds, dependence and selection must be included in the data model.

For independent measurements $x_i\sim N(\theta ,{\sigma }_i^2)$ with known errors, $\log L=\mathrm{constant}-\frac{1}{2}{\sum }_i(x_i-\theta {)}^2/{\sigma }_i^2$. Maximum likelihood minimizes this chi-squared and gives $\hat{\theta }={\sum }_i{w}_i{x}_i/{\sum }_i{w}_i$, $w_i={\sigma }_i^{-2}$. A Gaussian prior $N({\mu }_0,s_0^2)$ instead yieldsThis explicitly combines data and prior precision, approaching the weighted mean when prior information is weak. Unknown calibration parameters should be jointly inferred and marginalized, not silently fixed. posterior intervals are conditional on the assumed model and prior; posterior predictive checks and sensitivity to plausible priors help expose poor fits and weakly identified parameters. The maximum-likelihood estimate and a Bayesian posterior answer different inferential questions even when their numerical centres agree.

Let $D$ denote detection. Weight each incident fraction by its detection efficiency, not by its loss probability. Bayes' theorem givesThus about 2.83% of detected particles are in the high-energy group. Energy-dependent detector dead time changes the composition of the observed sample.

Bayes' theorem updates a hypothesis using observed data. For mutually exclusive exhaustive hypotheses $H_j$ and data $D$ of nonzero probability,The prior $P(H_j)$ describes information before these data; the likelihood $P(D\mid H_j)$ describes the sampling model; the denominator is the evidence; and the posterior is the updated probability. likelihood and posterior are different conditional quantities. For a continuous parameter, sums become integrals and the posterior is a probability density.

Consider a counter registering independent events at unknown constant rate $\lambda$, observed for known exposure $T$ with negligible background and dead time. The observed count has a Poisson distribution:Take a proper gamma prior in shape-rate convention,Multiplying by the likelihood adds the count to the shape and the exposure to the rate. Normalization therefore givesHere $\beta$ has exposure units. This is Poisson-gamma conjugacy. Successive independent counting runs update the same posterior by adding their counts and exposures, so the result agrees with analysing all the data together.

A credible interval contains a specified posterior probability for $\lambda$, conditional on the chosen prior and model. For low counts, the posterior can be asymmetric and sensitive to the prior, and a symmetric Gaussian error bar can be misleading. For large $n$ and an exposure dominating prior information, the mean approaches $n/T$ and its standard deviation approaches $\sqrt{n}/T$. Background, missed events or time-dependent rates require a richer likelihood; Bayes' theorem cannot compensate for an incorrect physical counting model. The prior must be specified independently of the same data being used in the likelihood, unless its estimation is explicitly included in a larger model.

Bayes' theorem updates prior knowledge using the likelihood of new observations. For alternative hypotheses $H_j$,The prior $P(H_j)$ describes information before the data $D$; $P(D\mid H_j)$ is the sampling likelihood, and the posterior $P(H_j\mid D)$ describes the updated probabilities. Posterior odds equal prior odds times the likelihood ratio. A likelihood is a probability model for data at specified parameters, not automatically a normalized probability distribution over those parameters.

For a continuous parameter $\theta$, the same rule uses densities: $\pi (\theta \mid D)=L(D\mid \theta )\pi (\theta )/\int L(D\mid u)\pi (u)du$. Priors can encode positivity, physical limits, previous independent measurements or calibration knowledge. Their parameterization matters: a density uniform in a rate is not uniform in its logarithm. Proper priors make model evidences well defined; an arbitrary normalization in an improper prior is especially problematic for comparing models.

A counting example uses Poisson-gamma conjugacy. If a detector registers $K$ events in exposure $t$ with constant rate $\lambda$, $L\propto {\lambda }^K{e}^{-t\lambda }$. A gamma prior with shape $\alpha$ and rate parameter $\beta$, $\pi (\lambda )\propto {\lambda }^{\alpha -1}{e}^{-\beta \lambda }$, givesThe posterior mean combines the prior rate $\alpha /\beta$ with the observed rate $K/t$, weighted by the prior and actual exposures. The parameters must use a consistent shape-rate convention. Small counts can leave substantial prior sensitivity; abundant informative data usually reduce it for regular identifiable models.

Independent old data can become the prior for a new experiment, giving the same result as joint updating under a consistent model. The new data must not also have been used to construct that prior, or the evidence is counted twice. A credible interval has a specified posterior probability under the chosen model and prior; this interpretation differs from repeated-sampling coverage of a confidence interval. Predictions for a future observation average its sampling distribution over the posterior, incorporating parameter uncertainty. Prior sensitivity checks and a defensible likelihood are therefore part of the inference, rather than Bayes' theorem itself supplying missing physical knowledge.

Let $M$ denote an incident muon and $S$ a detector signal. The selected sample is conditioned on $S$. The law of total probability givesBy Bayes' theorem, the contamination, namely the conditional probability that a selected particle is not a muon, isFor example, per $10000$ incident particles the expected selected counts are $8550$ true muons and $100$ other particles. The denominator is the $8650$ selected particles; the rejected muons are outside this sample.

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

Let $s={\sum }_i{x}_i$. The Bernoulli distribution likelihood function is ${\theta }^s(1-\theta {)}^{n-s}$. Multiplication by the beta distribution prior density and Bayes' theorem giveRecognizing the normalized beta distribution,This is Beta-binomial conjugacy; the same update applies whether the data are recorded as the individual Bernoulli trials or their success count.

Set $S={\sum }_i{x}_i$ and $N=nm$. Conditional independence of the binomial distribution observations gives the likelihood functionThe prior distribution is the beta distribution with parameters $(1,\beta )$. By Bayes' theorem, multiplying the prior density by the likelihood function gives a posterior density proportional toRecognizing and normalizing this beta distribution givesfor $0<\theta <1$. This is Beta-binomial conjugacy; the original binomial coefficients cancel in the normalization because they are independent of $\theta$.

The exponential distribution prior has rate $\lambda =1/3$ and support $\mu \ge 0$, so its prior density is $(1/3)e^{-\mu /3}{1}_{\{\mu \ge 0\}}$. By Bayes' theorem,where $C$ is a normalizing constant independent of $\mu$.

Completing the square gives a useful exact reference for the numerical sampling. PutThen this normal mean posterior with an exponential prior is a normal distribution with location $m$ and variance $s^2$, truncated below at zero. With $\varphi$ the standard normal distribution density and $\Phi$ its cumulative distribution function,The exponential rate is $1/3$, not three; the prior mean is three. The support restriction must be retained in both the formula and the sampler.

Let $H$ mean the added ball was red, and $D$ mean the first draw was red. Their relevant probabilities are $\mathbb{P}(H)=\mathbb{P}(H^c)=1/2$, $\mathbb{P}(D\mid H)=1$, and $\mathbb{P}(D\mid H^c)=1/2$. The law of total probability gives $\mathbb{P}(D)=3/4$.

Given the observed red draw, the remaining ball is red precisely when $H$ occurred. Bayes' theorem therefore givesThe observation increases the weight of the two-red-ball box, because that box is more likely to yield a red first draw.

The two definitions of conditional probability giveSubstitution proves Bayes' theorem:The positive probabilities assumed in the question make both conditional expressions defined.

Let $C_j$ identify the selected coin and let $H$ be the event of two heads. The prior coin probabilities are equal, and the conditional probabilities of $H$ are $1/9,1/4,4/9$. By Bayes' theorem,The full posterior distribution over the three biases is $(4,9,16)/29$.

Put $D=1-(1-p)(1-q)=p+q-pq$. By the law of total probability,For $0\le t<1$, termwise differentiation of the geometric series, valid inside its radius of convergence, givesTaking $t=(1-p)(1-q)$ therefore yieldsSince $D-p=(1-p)q$, this is the zero-based geometric distribution with parameterThis illustrates Bernoulli thinning of a geometric count. Here $0<p_{\ast }<1$, as $0<p,q<1$.

For the reverse conditional distribution, Bayes' theorem gives, for $n\ge k$,ThusFor $n<k$ the conditional probability is zero. The conditional distribution of $X-k$ is the failure-count form of a negative binomial distribution, with $k+1$ successes and success parameter $r$: the geometric-binomial posterior is shifted negative binomial.

Let $D$ denote the selected die. Bayes' theorem givesOnly the first roll is conditioned on here; the second outcome has not been observed.

Let $M$ count the ringed birds among the two captures, and let $D$ be the event that the detector records at least one ring. The hypergeometric distribution gives the prior probabilitiesUnder the usual additional assumption of independent per-ring detection, with no false records, the likelihoods $\mathbb{P}(D\mid M=m)$ are $0$, $9/10$, and $1-(1/10{)}^2=99/100$. Therefore $\mathbb{P}(D)=12/25+33/250=153/250$. Applying Bayes' theorem yields $\mathbb{P}(M=0\mid D)=0,\mathbb{P}(M=1\mid D)=\frac{40}{51},\mathbb{P}(M=2\mid D)=\frac{11}{51}$. Storing only the last detected number means that a stored record does not identify how many rings were present. By label symmetry, the identity of that stored ring does not further alter this distribution.

The text specifies each ring's marginal detection probability, but does not explicitly specify independence of the detection errors. To keep this modelling qualification visible, let $\ell =\mathbb{P}(D\mid M=2)$. The marginal detection rate only implies $9/10\le \ell \le 1$. Without the extra independence assumption, the full family of possible posterior distributions isIn particular $1/5\le \mathbb{P}(M=2\mid D)\le 5/23$. The displayed numerical answer corresponds to $\ell =99/100$; the general answer exposes exactly what the detector's marginal specification alone establishes.

For a binary classifier, the conditional probability of the target condition given a positive classification. If its base rate is $p$, its true-positive probability is $s$, and its false-positive probability is $f$, Bayes' theorem gives $P(E\mid +)=sp/(sp+f(1-p))$. The denominator is positive whenever a positive classification can occur.