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A case chromosome is transmitted and a pseudo-control chromosome is untransmitted. Summing the paired table by columns for cases and by rows for controls givesFor the ordinary unpaired chi-squared test of independence, the expected cells are 30,30,70,70. Its uncorrected Pearson statistic isFor the paired McNemar's test, only the discordant transmissions matter, with counts 24 and 18. ThusBoth uncorrected statistics happen to coincide and have the same approximate one-degree-of-freedom chi-squared distribution reference, giving $p\approx 0.355$. The data do not show persuasive transmission imbalance at the conventional 5% level.

The equality is a numerical coincidence, not a justification for ignoring pairing. To see the equality criterion for paired and unpaired allele tests, write the original matched cells as $a,b,c,d$ and $n=a+b+c+d$. The unpaired marginal-table statistic iswhereas McNemar's is $(b-c{)}^2/(b+c)$. Here $n=100$, $b+c=42$, and $(2a+b+c)(2d+b+c)=60\times 140=2n(b+c)$, giving equality. Changing concordant counts can change the unpaired statistic without changing McNemar's. The transmission disequilibrium test should retain its conditional paired interpretation.

If a continuity correction is used, state it explicitly: McNemar's corrected statistic is $(|24-18|-1{)}^2/42=25/42\approx 0.5952$; the corresponding Yates-corrected marginal statistic also coincides here. An exact conditional test uses $\mathrm{Binomial}(42,1/2)$ and gives the two-sided probability $2\mathbb{P}(K\le 18)\approx 0.4408$.

The recombination fraction is the probability that a transmitted gamete is recombinant between the two loci in the ordinary two-locus genetic linkage model. Its conventional range isZero corresponds to complete linkage and $1/2$ to independent assortment. Multiple physical crossovers can restore parental phase, so the recombination fraction is not the expected number of crossovers or an unrestricted crossover probability.

The LOD score is the base-ten log likelihood ratio against independent assortment. For the two families,For a known-phase first family this becomeswith endpoint terms interpreted by limits, such as $0\log 0=0$ when the corresponding count is zero. Positive values favour genetic linkage, and the likelihood ratio is $10^{Z_{\max }}$. A conventional large positive threshold such as 3 represents a likelihood ratio of 1000, not automatically a posterior probability or a universal 5% test.

For a test calibrated to this study, use the distribution of the maximized statistic under $\theta =1/2$, for example by simulating transmissions conditional on the same parental genotype and ascertainment scheme. Reject for sufficiently large values with the chosen significance threshold; phase uncertainty and the boundary null prevent assuming an unqualified ordinary interior Wilks reference. The missing pedigrees prevent the numerical value or their explicit likelihood polynomial from being supplied, but the likelihood-ratio definition and testing procedure apply once those data are recovered.

The acceptance mass for an unnormalized rejection envelope follows by integrating the proposal-specific acceptance probability:The inequality $f\le Mg$ also gives $C\le M$, so this is a valid probability. It is the same at every independent trial.

If $J_i$ indicates acceptance of the $i$th proposal, then $J_1,\dots ,J_n$ are independent Bernoulli distribution variables with parameter $C/M$. Hence the accepted count $K={\sum }_i{J}_i$ satisfiesIts variance is $n{p}_{\mathrm{acc}}(1-p_{\mathrm{acc}})$. Conditional on any acceptance pattern, the accepted values have independent target probability density function $h$; mixing over patterns retains this product law conditional on their count.

Write $C={\int }_{\mathbb{R}}f(x)dx$, so $0<C<\infty$. The finite envelope condition means that $f\le Mg$ almost everywhere; in particular, the proposal must cover the support of the target.

For each independent proposal $Y\sim g$, generate an independent $U\sim \mathrm{Unif}(0,1)$ and accept $Y$ precisely whenDiscard a rejected proposal and continue. The ratio is in $[0,1]$ and is needed only where $g(Y)>0$. Neither evaluation of $C$ nor prior knowledge of the normalized target is required.

To prove the output law, for any measurable set $D$ the joint probability of acceptance and a proposal in $D$ isDividing by the total acceptance probability $C/M$ gives ${\int }_{D}f(y)/Cdy$, the target probability. Independent proposal-uniform pairs give independent accepted observations. With a finite proposal list there can be no accepted value; with continued independent trials, acceptance eventually occurs almost surely because $C/M>0$.

For normal rejection sampling with an exponential envelope, complete the square in the probability density function ratio:The maximum occurs at $x=\lambda$, which lies in the support because $\lambda >0$. ThusThe PDF exponent is $e^{{\lambda }^2}$ inside the square root; this distinction is lost in the converted TeX.

The acceptance rule simplifies toAn exponential proposal can be generated as $Y=-{\lambda }^{-1}\log V$ from an independent uniform distribution value $V$. Since the target is normalized, the overall acceptance probability is $1/M(\lambda )$. To maximize it,Therefore the optimal rate is $\lambda =1$, with acceptance probability $\sqrt{\pi /(2e)}$.

Finally, give each accepted half-normal distribution value $Y$ an independent fair sign $R\in \{-1,1\}$ and output $Z=RY$. On each half-line its probability density function is half the reflected half-normal distribution probability density function:Consequently $Z\sim N(0,1)$. The sign must be drawn independently; rejection itself does not supply it.

Here the proposal is $Y=x+\eta$ with $\eta \sim N_2(0,I_2)$. Its probability density function depends only on $\Vert y-x{\Vert }^2$, so $q(y\mid x)=q(x\mid y)$. The determinant and normalizing constant of the target also cancel. With $Q$ as in the preceding entry,Equivalently, writing $u_j=(y_j-{\mu }_j)/{\sigma }_j$ and $z_j=(x_j-{\mu }_j)/{\sigma }_j$,One can also compute the exponent without subtracting two large quadratic forms:Generate two independent standard normal distribution proposal increments, compute this log ratio, and accept when $\log U\le \min (0,\log (\pi (Y)/\pi (x)))$. Retain the previous pair on rejection. Detailed balance gives the correct invariant distribution, and the everywhere-positive Gaussian proposal makes this random-walk Metropolis algorithm chain irreducible; a rejection probability provides a holding step. The resulting target-distributed pairs are dependent, with proposal scale and target anisotropy controlling how quickly they mix.

For the Metropolis–Hastings algorithm, start at a point $x$ with positive target probability density function. Draw $Y$ from a proposal probability density function $q(y\mid x)$, generate an independent uniform distribution value $U$, and set the next state to $Y$ ifOtherwise retain $x$, including that repeated state in the sample. Where a proposed forward move is possible but its reverse probability density function vanishes, its acceptance probability is zero. Unknown normalizing constants in $\pi$ cancel.

For distinct states, the accepted probability flow iswhich is symmetric in $x,y$. This detailed balance identity, together with the holding probability, proves that $\pi$ is invariant. Under appropriate irreducibility and aperiodicity conditions the chain converges to $\pi$; successive states are generally dependent.

The Gibbs sampler updates coordinates from their full conditional distributions. In a systematic sweep, draw successivelyEach update preserves the target joint law: integrating the old coordinate against its conditional probability density function and replacing it by an independent draw from that same conditional leaves the joint probability density function unchanged. The composition of the coordinate updates therefore also preserves $\pi$. A Gibbs sampler coordinate move can be regarded as a Metropolis–Hastings algorithm move with acceptance probability one. Systematic sweeps preserve stationarity but need not themselves be reversible. Initialization away from stationarity requires convergence before treating the draws as approximately target-distributed, and dependence must be accounted for in Monte Carlo method error estimates.

A microparasite typically multiplies within an infected host on a much shorter time scale than between-host transmission. A useful first model classifies hosts by infection status. A macroparasite, such as an adult parasitic worm, instead requires a count of parasites per host: two infected hosts can carry very different within-host parasite burdens. These are modelling distinctions rather than a rule determined only by an organism's physical size.

For a microparasite with no lasting immunity, a homogeneous SIS model isThe basic reproduction number is $\beta /\gamma$, and the prevalence $I/N$ describes the infected-host fraction. A SIR model adds an immune class, and a SEIR model adds a latent class. Contact structure, age, susceptibility, recovery rate and infectiousness can require several such classes. Then a next-generation matrix or operator, rather than a single averaged contact rate, sets the invasion threshold. Stochastic epidemic models are relevant to early extinction and finite populations; network or spatial models represent nonuniform contact opportunities.

For a simple macroparasite burden model, let $p_j$ be the fraction of hosts carrying $j$ adult worms. Assume acquisitions at rate $\lambda =\beta L$, independent adult-worm losses at rate $\mu j$, and a number $L$ of infective stages in a well-mixed environmental reservoir. The immigration-death worm-burden model hasThis birth-death process needs the burden distribution or its probability generating function, not just an infected/not-infected partition. Its mean $m={\sum }_{j}j{p}_j$ satisfies $\dot{m}=\beta L-\mu m$. With host number $N$ fixed and each adult producing infective stages at rate $\sigma$, one possible environmental closure isFor this asexual, unsaturated transmission approximation, one worm's basic reproduction number is $(\sigma /\mu )\beta N/(d_L+\beta N)$. Growth above the invasion threshold requires additional density regulation if a finite endemic burden is sought. Sex-dependent mating would change this closure.

With constant external exposure $\lambda$, the equilibrium burden follows a Poisson distribution with mean $m=\lambda /\mu$, giving infected-host prevalence $1-e^{-m}$. If exposure rates vary across hosts as a gamma distribution, the Poisson-gamma mixture gives a negative binomial distribution with mean $m$ and aggregation parameter $k$:The same mean burden can therefore coexist with very different prevalence and highly infected subgroups. Aggregation of macroparasite burdens affects treatment coverage, burden-dependent host mortality, immunity and parasite fecundity. For sexually reproducing worms, hosts need both sexes for fertile output; a mean alone may not determine it. With independent equally likely sexes and Poisson burden, the probability of both sexes is $(1-e^{-m/2}{)}^2$, illustrating mating-limited macroparasite transmission and its low-burden bottleneck.

Both microparasite and macroparasite models need heterogeneous exposure and host responses. The special extra difficulty for macroparasites is translating a distributed worm burden, and possibly worm sex or developmental stage, into transmission and host harm. Conversely, complicated within-host microparasite processes can also require more than a binary host state when the simple fast-within-host approximation fails.

Use an assortatively mixed two-risk-group SIS model with equal per-person recovery rate $\gamma$. Let $x,y$ be infectious proportions in groups of sizes $pN,(1-p)N$. Let $c_H\gg c_L$ be their fixed contact rates and $\tau$ the transmission probability per contact. Suppose a small fixed fraction $ϵ<1/2$ of high-risk contacts is with the low-risk group. Contact reciprocity in a two-group epidemic requiresFor sufficiently small $p$, both groups make most contacts within their own group, and the total cross-group contacts are of order $pN$. Define $A=\tau c_H(1-ϵ)$, $B=\tau c_Hϵ$, $D=\tau c_L$ and $C(p)=Bp/(1-p)$. The forces of infection giveHere $C(p)<D$ and ${ϵ}_L<1/2$ hold in the small-$p$ regime. High-risk per-person cross contact is order one but its group size is order $p$; low-risk per-person cross contact is order $p$. These distinctions are needed to count each cross-group partnership consistently.

Linearize about the disease-free state. In infectious fractions the transmission matrix isIn infected-individual counts the next-generation matrix is insteadThe count and fraction formulations are related by a diagonal change of coordinates, so their eigenvalues coincide. The basic reproduction number is the spectral radius $R_0=\rho (M)/\gamma$, not an average of the two group reproduction numbers. The larger real eigenvalue isThis model makes a particular reciprocal mixing convention explicit; different defensible contact laws can change the first-order coefficients.

The payoff table is blank in the original PDF, so denote its four row-player payoffs by $R,S,T,P$ for $CC,CD,DC,DD$, with the prisoner's dilemma ordering $T>R>P>S$. No numerical entries are inferred. Let the two reactive strategies be $(p_1,q_1)$ and $(p_2,q_2)$, where $p_i$ is cooperation after the opponent's cooperation and $q_i$ cooperation after defection.

The joint action states form a four-state Markov chain. In state order $CC,CD,DC,DD$, its transition matrix isThe prescribed initial memories give independent first actions with cooperation probabilities $p_1,p_2$. Same-round actions remain independent: each player's next action uses the other player's previous action and its own independent random draw. Thus their marginal cooperation probabilities obeyPut $a_i=p_i-q_i$. If $|a_1{a}_2|<1$, the two-step recurrences contract and the long-run payoff of reactive strategies follows fromThe joint stationary distribution is $(x_{\ast }{y}_{\ast },x_{\ast }(1-y_{\ast }),(1-x_{\ast })y_{\ast },(1-x_{\ast })(1-y_{\ast }))$. Player 1's mean payoff is $R{x}_{\ast }{y}_{\ast }+S{x}_{\ast }(1-y_{\ast })+T(1-x_{\ast })y_{\ast }+P(1-x_{\ast })(1-y_{\ast })$; exchange $S,T$ for player 2.

The deterministic boundaries must retain the initial memories rather than divide by zero. Two Tit for tat players cooperate forever and earn $R$. Two players who always do the opposite of the opponent's previous action alternate $DD,CC$ and average $(P+R)/2$. One of each cycles through all four states and each averages $(R+S+T+P)/4$. These are the only cases with $|a_1{a}_2|=1$. A periodic Markov chain need not have convergent state probabilities, but its long-time average payoff is defined by its cycle frequencies.

Choose a real unitary matrix and undo it at the second stage:After the first rotation the photon qubit is $\sqrt{q}|0⟩+\sqrt{p}|1⟩$. A dud leaves this quantum superposition unchanged, so the inverse rotation restores $|0⟩$. Its final projective measurement can never return $1$.

For a live bomb the absorbing component explodes with probability $p$. If the photon survives, the normalized state is $|0⟩$, and the second rotation makes it $\sqrt{q}|0⟩-\sqrt{p}|1⟩$. The unconditional probabilities of the three possible live-bomb outcomes are consequentlyThey sum to one. The nonexplosive result $1$ certifies a live bomb because its dud probability is exactly zero. The inconclusive result $0$ resets the qubit to its original state, allowing another identical trial.

The repeated weak-rotation bomb-test efficiency follows by summing the geometric series over the first successful trial:This is conditional on testing a live bomb, not an average over an unspecified prior fraction of duds. To make the finite-run success probability exceed $1/2-\epsilon$, choose positive $p$ so that $p/[2(2-p)]<\epsilon /2$, then choose $N$ so that $q^{2N}<\epsilon /2$. The first condition controls the limiting deficit and the second controls the remaining tail. Exactly $p=0$ yields no explosion but also no information, so the limit must be approached with nonzero weak rotations.

The half-efficiency limit is consistent with the structure of this Elitzur-Vaidman bomb tester. Suppose generally $U|0⟩=a|0⟩+b|1⟩$. A final outcome that certifies a live bomb must be orthogonal to the dud output $U^{\prime }U|0⟩$. Pulling this outcome backwards through $U^{\prime }$ gives the normalized vector orthogonal to $(a,b)$. Its squared overlap with $|0⟩$ is $|b{|}^2$. Thus the safe-certification probability is $|a{|}^2|b{|}^2$, whereas the explosion probability is $|b{|}^2$. The former never exceeds the latter. Summing these mutually exclusive terminal events over repeated, possibly adaptive trials gives $P_{\mathrm{safe}}\le P_{\mathrm{explosion}}$ and hence $P_{\mathrm{safe}}\le 1/2$. This concerns the specified rotate–interact–rotate–measure protocol.

A differential form of degree $k$ is integrated over an oriented $k$-dimensional smooth manifold by integrating its coefficient in orientation-preserving coordinates. For an oriented parametrization $F:U\to M$, this means integrating the pullback $F^{\ast }\alpha$ over $U$; a partition of unity combines charts. The change of variables formula makes the result independent of the charts. Reversing the orientation of a smooth manifold reverses the integral. The Generalized Stokes theorem says, for a compact oriented $k$-dimensional smooth manifold with boundary and a smooth $(k-1)$-form $\beta$,where the right-hand side includes the boundary pullback, and the boundary has the outward-normal-first boundary orientation. Compact support suffices on a noncompact manifold. This unifies the fundamental theorem of calculus, circulation and flux identities.

For example, take the unit disk $D$ oriented by $dx\wedge dy$ and the differential one-form $\beta =\frac{1}{2}(xdy-ydx)$. Its exterior derivative is $d\beta =dx\wedge dy$, so the area integral is $\pi$. Parametrizing the positively oriented circle by $(x,y)=(\cos t,\sin t)$ gives the pullback $\beta =\frac{1}{2}dt$ and the boundary integral ${\int }_0^{2\pi }\frac{1}{2}dt=\pi$, directly illustrating Generalized Stokes theorem.

For the distance calculation, write $u=|{\mathbf{r}}_1|$, $v=|{\mathbf{r}}_2|$, $x=u+v$ and $y=u-v$; these are scalar radii, not position vectors. The triangle inequality and its reverse give $r\le u+v=x$ and $|y|=|u-v|\le r$. Also,Thus $r\le x\le 2R-|y|$ and $|y|\le r$. These conditions, with $u,v\ge 0$, are also sufficient for a triangle with side lengths $u,v,r$. Endpoints correspond to collinear configurations.

Assume the two positions are independent random variables, each having constant volume probability density inside the radius-$R$ ball. Independence is needed: uniform marginal distributions alone do not determine the distance distribution. Put $C=3/(4\pi R^3)$ and $\mathbf{s}={\mathbf{r}}_2-{\mathbf{r}}_1$. The change of variables $({\mathbf{r}}_1,{\mathbf{r}}_2)\mapsto ({\mathbf{r}}_1,\mathbf{s})$ has unit Jacobian determinant, so the joint probability volume differential form is the product of the two normalized volume forms. In spherical coordinates for ${\mathbf{r}}_1$ and for $\mathbf{s}$ relative to its axis, it isThe polar coordinates fail on axes and at zero radii, which are sets of zero volume and do not affect the probability. Here $\theta$ is the angle from ${\mathbf{r}}_1$ to $\mathbf{s}$. The law of cosines becomesDifferentiating and taking the wedge product with $du\wedge dr$ eliminates the terms involving $du$ and $dr$:Consequently the positive integration density transforms by $|\sin \theta d\theta |=vdv/(ur)$ at fixed $u,r$. The minus sign is accounted for by reversal of limits: $v$ decreases as $\theta$ increases. The radial variable retained is $r$, while the polar angle is replaced by $v$; equivalently one can start with the angle between the two position vectors and replace that angle by $r$. The wording of the printed hint conflates these two coordinate choices, but its volume form gives the stated transformation directly.

Integrating $\sin {\theta }_{1}d{\theta }_{1}d{\varphi }_1$ over the first orientation gives $4\pi$, and integrating $d\chi$ gives $2\pi$. The probability density function of $r$ is thereforewhere $D_r$ has $0\le u,v\le R$ and $|u-v|\le r\le u+v$. There are no such configurations for $r>2R$.

To evaluate this integral explicitly, the Jacobian determinant of $(u,v)=((x+y)/2,(x-y)/2)$ has absolute value $1/2$. Hence $uvdudv=(x^2-y^2)dxdy/8$ as a positive density. For $0\le r\le 2R$, set $a=\min (r,2R-r)$. The domain is $-a\le y\le a$, $r\le x\le 2R-|y|$, and evenness in $y$ givesAn antiderivative, zero at $y=0$, yieldsSubstituting $a=r$ for $r\le R$ and $a=2R-r$ for $r\ge R$ gives the same polynomial in both intervals:Thus the distance between two uniform points in a three-dimensional ball has the final densityIt is zero outside this interval. Its nonnegativity also follows from $f_R(r)=3r^2(4R+r)(2R-r{)}^2/(16R^6)$, and direct integration gives ${\int }_0^{2R}{f}_R(r)dr=1$. The derivation uses the full six-dimensional probability volume differential form, rather than treating the three scalar distances as independent.

A Poisson distribution with parameter $\mu$ has probability generating function $G_N(s)=e^{\mu (s-1)}$. The preceding part gives $G_F(s)=e^{\mu p(s-1)}$, the probability generating function of a Poisson distribution with parameter $\mu p$. To prove independence, retain both counts:Consequently the joint probability generating function factors asComparing the coefficients of $s^f{t}^k$ proves that the joint probabilities are the products of the marginal probabilities. Hence $F$ and $S$ are independent, with distributions $\mathrm{Poisson}(\mu p)$ and $\mathrm{Poisson}(\mu (1-p))$. This is Poisson thinning, including the degenerate counts when $p=0$ or $p=1$.

Three positive lengths form a nondegenerate triangle precisely when each is less than the sum of the other two. Necessity is the triangle inequality. For sufficiency, two circles centered at the ends of one side intersect off that side whenever the remaining lengths satisfy these strict inequalities, giving the third vertex.

Failure thus occurs when $X\ge Y+Z$, $Y\ge X+Z$, or $Z\ge X+Y$. Two such inequalities cannot hold together for strictly positive lengths. Zero lengths and equality cases have probability zero because the variables have continuous probability density functions. By symmetry each failure event has the probability $1/6$ just calculated. Hence the probability of forming a triangle is

Another convolution of probability densities gives, for $0\le s\le 1$,Only this part of the density is needed because the fourth length is at most one.

Four positive lengths $a,b,c,d$ can form a nondegenerate quadrilateral if and only if every length is less than the sum of the other three. Necessity follows by following the other three sides between the endpoints of the longest side and applying the triangle inequality. For sufficiency, select a diagonal length $r$ in the intersectionBoth intervals are nonempty, and their intersection is nonempty exactly when $|a-b|<c+d$ and $|c-d|<a+b$. These are precisely the four strict length inequalities. Construct triangles with sides $(a,b,r)$ and $(c,d,r)$ on opposite sides of their common diagonal. Their union has the required simple, nondegenerate quadrilateral as boundary.

The failure events, in which one length is at least the sum of the others, are disjoint except on sets of probability zero. For the specified length $W$, independence and the density above giveThere are four symmetric choices of the overly long side, so the probability isThis is the four-sided case of the polygon probability for independent uniform rods.

The binomial series givesThus $p_0=1-\alpha >0$ and, for $k\ge 1$,with the factors after $\beta$ omitted for $k=1$. The signs follow from $0<\beta <1$. As $s\uparrow 1$, the series has nonnegative terms and its sum increases to $G(1)=1$, so monotone convergence gives ${\sum }_{k\ge 0}{p}_k=1$. These are therefore genuine probabilities on the nonnegative integers, proving that $G$ is a probability generating function. Its expected value is infinite, since $G^{\prime }(s)=\alpha \beta (1-s{)}^{\beta -1}\to \infty$; finiteness of the mean is not required for this branching process.

Write $G_n(s)=1-A_n(1-s{)}^{{\beta }^n}$. The composition formula gives $A_{n+1}=A_n{\alpha }^{{\beta }^n}$, with $A_0=1$. Consequently $A_n={\alpha }^{1+\beta +\cdots +{\beta }^{n-1}}$, and summing the geometric series yields the explicit generation generating functionThis is the fractional-power offspring generating function and its iterates.

Evaluation of the probability generating function at zero selects the constant coefficient, soTo identify this limit with ultimate extinction, let $E_n=\{X_n=0\}$. The absorbing-zero property of the branching process gives $E_n\subseteq E_{n+1}$. Ultimate extinction means that some finite generation is empty, namely $E={\bigcup }_{n\ge 0}{E}_n$. Continuity of probability on increasing events therefore givesThis set argument is essential: it uses absorption, rather than inferring extinction merely from a numerical limit of generation probabilities.

Every urn contains $n-1$ balls. Let $k=n-r$ be its number of blue balls; choosing an urn uniformly makes $k$ uniform on $0,\dots ,n-1$. Conditional on that urn, the probability of a blue first draw is $k/(n-1)$, so the law of total probability givesDrawing twice without replacement requires $n\ge 3$. Given $k$, the probability that both draws are blue is $k(k-1)/((n-1)(n-2))$, and henceTaking the ratio defining conditional probability gives the second requested probabilityAlthough a blue first draw removes a blue ball from its urn, it also makes an urn rich in blue balls more likely. Averaging with this changed conditional distribution is essential.

Two independent events satisfy $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)$. If $\mathbb{P}(B)>0$, this is equivalent to $\mathbb{P}(A\mid B)=\mathbb{P}(A)$: learning that $B$ occurred does not change the probability of $A$. The product definition also covers zero-probability events without dividing by zero.