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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.

By the law of total probability,Bayes' theorem therefore gives the conditional probability mass functionThe six probabilities sum to one. A small red outcome is more likely after an immediate success because more white faces can meet its threshold.

Let $D$ denote having the disease and $+$ a positive result. Assume I am sampled uniformly from the stated population, so $\mathbb{P}(D)=n/N$. By the law of total probability, $\mathbb{P}(+)=an/N+b(N-n)/N$. Thus Bayes' theorem givesThe denominator must be positive; otherwise the observed event has zero probability in this model and the conditional probability is undefined. The answer depends on disease prevalence as well as the test's two error rates.

Let $G$ mean that the tested person is guilty and $M$ mean a positive match. Under the specified model, $P(G)=1/N$, $P(M\mid G)=1$ and $P(M\mid G^c)=p$. Bayes' theorem givesThe posterior probability for a unique source after a positive test depends on the number of alternative possible sources as well as the false-positive probability. In particular, the false-positive probability $p$ is not the probability of innocence conditional on a positive match.

The negative-test probability is $P(-)=1-0.05094=0.94906$. Applying Bayes' theorem again,

By Bayes' theorem, the required conditional probability isThis is about $1.94\%$; it is a different conditioning direction from the test's $99\%$ positive probability among carriers.

Let $D$, $C$, and $V_0$ denote the three disjoint categories, and $+$ or $-$ the test result. The law of total probability givesTo calculate the requested conditional probability, first combine the two categories with the virus. Their joint negative-test probability is $0.1(0.05)+0.2(0.1)=0.025$, whereas $\mathbb{P}(-)=1-0.31=0.69$. Bayes' theorem therefore givesThese numbers use the specified population proportions and test probabilities; the negative result reduces the initial combined proportion $0.3$ to about $3.62\%$ within this model.

The prior probability is $1/10$. Using Bayes' theorem, the posterior probability after two absences isThus the probability rises from 10% to about 73.5% within the assumed attendance model. Two absences are 25 times as likely under the lazy model, overcoming the prior odds of $1:9$; they still do not make the classification certain.