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The Kolmogorov axioms for a probability measure $\mathbb{P}$ are nonnegativity, normalization $\mathbb{P}(\Omega )=1$, and countable additivity on pairwise disjoint events. For $\mathbb{P}(B)>0$, conditional probability is defined byBecause the events $A\cap B_i$ are pairwise disjoint and their union is $A$, countable additivity gives the law of total probabilityFor any $j$ with $\mathbb{P}(A)>0$, the definition of conditional probability and this identity give Bayes theorem:

The prior probabilities of choosing the two urns are equal. Their white-ball probabilities areBy Bayes theorem,

Urn $i$ has $2^i$ balls, so a draw from it is white with probability $2^{-i}$. Given that the first ball is white, Bayes theorem gives the posterior probabilityAfter replacement, the second draw is conditionally independent of the first given the urn. HenceConsequently $\boxed{p(n)\to 1/3}$.

By Bayes theorem, with respect to Lebesgue measure on $(0,1{)}^2$,The omitted constant includes the binomial coefficient and the normalizing constant of the Beta distribution; the uniform prior contributes only the support indicator.

Conditional independence givesFor each satellite, Bayes theorem gives $P(d_s\mid m)\propto P(m\mid d_s)/P(m)$. ThereforeIf ${\hat{p}}_0(m)$ is a kernel density estimator for the simulated marginal masses and ${\hat{p}}_s(m)$ estimates the posterior based on satellite $s$, thenThe one-dimensional integrals can be evaluated by numerical integration on a common mass grid.

Let ${\pi }_0$ be the probability density function of the unknown and $\rho$ that of the noise, both relative to Lebesgue measure. By independence of random variables, the conditional distribution of the measurement given $u$ has density $\rho (m-Au)$. Thus its likelihood function and the joint density areThe Bayesian model evidence is the marginal data densityBayes theorem gives the posterior density solving this Bayesian inverse problem:This formula applies when $0<Z(m)<\infty$. Since $\int Z(m)dm=1$ by Tonelli theorem, these conditions hold almost everywhere under the marginal data law. Values of a conditional distribution at exceptional data are not determined by the joint law. The posterior mean or a maximum a posteriori estimate can then give a point estimate, while the full posterior density describes the remaining uncertainty.

The notation $P_0=\mathcal{N}(0,I)$ denotes generalized Gaussian white noise over $H=L^2({\mathbb{R}}^2)$, not a Gaussian measure supported on $H$: the identity is not a trace-class operator there. For example, realize the observation on a suitable space of tempered distributions. Its Cameron-Martin space of a Gaussian measure is $H$. For deterministic $h\in H$, the stochastic pairing $W_m(h)$ has normal distribution $\mathcal{N}(0,\Vert h{\Vert }_H^2)$ under $P_0$ and is the isonormal Gaussian process indexed by $H$.

Interpret the stated maps between Sobolev spaces as bounded linear operators, as usual. In particular $K:H\to H^2({\mathbb{R}}^2)\subset H$ is bounded, so its adjoint operator $K^{\ast }$ is bounded on $H$. Hence $A=K^{\ast }K$ maps $H$ into $H$, which is the property needed here. The stronger smoothing follows by Sobolev duality: boundedness of $K:H^{-4}\to H^{-2}$ gives $K^{\ast }:H^2\to H^4$, and therefore $A:H\to H^4\subset H$. As $\Pi (H)=1$, the shift $Au$ belongs to the noise Cameron-Martin space of a Gaussian measure for $\Pi$-almost every $u$.

The white-noise likelihood for a square-integrable shift, obtained from the Cameron-Martin theorem for a Gaussian measure in its white-noise form, isHere $P_u$ is the law of $Au+\eta$. In an orthonormal basis $(e_j)$ of $H$, with white-noise coordinates $m_j$, the pairing under $P_0$ isFor fixed $u$ this has mean-square convergence and converges almost surely, since ${\sum }_j|⟨Au,e_j⟩{|}^2=\Vert Au{\Vert }_H^2$. The stipulated joint measurability allows the likelihood to be used under ${\nu }_0(du,dm)=\Pi (du)P_0(dm)$. One must not replace this expression by a finite $\Vert m-Au{\Vert }_H^2$, because white noise is not $H$-valued. On the unbounded domain ${\mathbb{R}}^2$, membership of white noise in a global unweighted negative Sobolev space must not be assumed either; the stochastic pairing avoids that issue.

Let $\nu (du,dm)=\Pi (du)P_u(dm)$ be the actual joint law. The shift formula gives $\nu \ll {\nu }_0$ with Radon-Nikodym derivative $L$. DefineFor each fixed admissible $u$, the moment-generating function of a normal distribution givesBy Tonelli theorem, $\int Z(m)P_0(dm)=1$, so $Z(m)<\infty$ for $P_0$-almost every $m$. The likelihood is finite and strictly positive for ${\nu }_0$-almost every $(u,m)$; Fubini's theorem then gives $Z(m)>0$ for $P_0$-almost every $m$. The marginal observation law is $Z(m)P_0(dm)$ and is an equivalent probability measure to $P_0$.

The Bayes formula for a dominated observation model therefore definesTo verify that it is the conditional distribution, for measurable sets $B$ of unknowns and $C$ of data,Thus the posterior distribution is well defined for almost every observation under the actual data law. This is the appropriate almost-everywhere assertion supplied by Bayes theorem; the stated assumptions do not prescribe a canonical posterior at every exceptional generalized datum.

A cipher has perfect secrecy when the conditional probability of every plaintext given a ciphertext equals its prior probability: $\mathrm{Pr}(M=m\mid C=c)=\mathrm{Pr}(M=m)$ whenever $\mathrm{Pr}(C=c)>0$. Equivalently, the plaintext and ciphertext are independent random variables.

Fix a ciphertext $c$ that can occur. Perfect secrecy requires every possible plaintext $m$ to remain possible after observing $c$. For a deterministic decryption rule, two different plaintexts producing $c$ require two different keys, so the key space has cardinality at least that of the message space.

In a one-time pad, the message and key belong to the same finite additive group, the key is uniform and independent of the message, and encryption is $C=M+K$. For every $m,c$ there is exactly one key $k=c-m$, soindependent of $m$. Bayes theorem then gives $\mathrm{Pr}(M=m\mid C=c)=\mathrm{Pr}(M=m)$, proving perfect secrecy.

For $\mathbb{P}(B)>0$, the conditional probability isHenceBecause the $B_k$ form a partition, the law of total probability givesSubstitution proves Bayes theorem in the required form:

A Bayesian inverse problem consists of a prior distribution ${\mu }_0$ on the unknown $u\in X$, a reference measure on the data space $Y$, and a jointly measurable likelihood $L(u;y)$ such that $L(u;\cdot )$ is a probability density function for ${\mu }_0$-almost every $u$. For observed data $y$, Bayes theorem defines the posterior distribution byprovided $0<Z(y)<\infty$.

Every posterior given by Bayes theorem is absolutely continuous with respect to its prior ${\mu }_0$, so it belongs to $\mathrm{Prob}(X,{\mathcal{B}}_X,{\mu }_0)$. If $y_n\to y$, total-variation well-posedness givesPart c, now using the common dominating measure ${\mu }_0$, yieldsExistence and uniqueness are unchanged. The problem is therefore a well-posed Bayesian inverse problem in Hellinger distance.

By Bayes theorem, the frailty distribution among survivors has densityIt is an exponential distribution with rate $1+H_0(t)$.

For a fixed observed colour ${\tilde{c}}_s$, Bayes theorem givesCompleting the square shows that this is a truncated normal distribution with variance ${\sigma }_c^2$, lower endpoint zero, and untruncated locationHenceConditioning first on $E_s$ and using the law of total expectation yields the dusty colour-magnitude relationAs ${\tilde{c}}_s\to -\infty$, the Inverse Mills ratio implies ${\overline{E}}_s\to 0$ with vanishing derivative, so the asymptotic slope is $\beta$. As ${\tilde{c}}_s\to +\infty$, ${\overline{E}}_s\sim {\tilde{c}}_s-c_0-{\sigma }_c^2/\tau$, so the asymptotic slope is $R$.

Using Bayes theorem and the fact that the prior is proper,Thus $\hat{I}$ is an unbiased estimator of $Z^{-1}$, and the Harmonic mean estimator of Bayesian model evidence is $\hat{Z}=1/\hat{I}$. The reciprocal is not itself generally unbiased, though it is consistent when the strong law of large numbers applies.

Under $M_1$, Bayes theorem at the nested value $\psi =0$ givesSeparability of the prior and equality of the $\varphi$ priors implyRearranging proves the Savage-Dickey density ratiowhich is the Bayes factor in favor of the nested model.

The graph has no cause of $D$, so $D$ is independent of its potential outcomes. By consistency of potential outcomes, the G-formula therefore givesCondition on $M$ and use the structural zero from part ii:Consequently the causal risk ratio isApplying Bayes theorem to the second factor giveswhich proves the stated formula. The unmeasured common cause $U$ of $M$ and $Y$ does not obstruct this total-effect argument because no mediator effect is being identified.

The directed acyclic graph factorization isThe graph gives $U\perp X_1$ and $U\perp X_3\mid (X_1,X_2)$. Therefore Bayes theorem givesConsequentlywhich contains no $x_1$. This observed equality is a Verma constraint: it is implied by the latent-variable causal graph even though it is not a conditional independence.

Bayes theorem turns the categorical distribution prior probabilities and the component multivariate normal densities intoThese probabilities define the label update in the Gibbs sampler.

Define the prior-predictive nuisance integralsThenFor each individual analysis, Bayes theorem also gives

Bayes theorem givesTaking its posterior expectation after subtracting $\log \pi (\theta )$ yieldsThus the equality holds for every proper prior and valid likelihood for which the displayed expectations are well-defined.