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Independent increments of the Poisson process give $M\mid \lambda \sim \mathrm{Poisson}(\lambda )$ for a one-hour interval. The Poisson distribution therefore givesThis conditional probability still depends on the unknown rate.

A uniform prior on $(0,T)$ makes the posterior density proportional to the Poisson change-point posterior likelihood above. The zeros trick introduces an observed zero with Poisson distribution mean $K-\log L(\theta )$, making its likelihood function equal to $e^{-K}L(\theta )$. Choose $K=T+n\log 2+1$; since $\log L(\theta )\le n\log 2$, this mean is strictly positive. Rough BUGS code, with zero=0 supplied as data, isFor $n=0$, omit the array and set j <- 0. Monitor the sampled theta to obtain its posterior mean, credible interval and interval probabilities.

There is also an exact sampling method. On $(t_j,t_{j+1})$ the posterior density is proportional to $2^{-j}{e}^{\theta }$, so choose the interval with weightsThen draw $U$ from a uniform distribution on $(0,1)$ and set $\theta =\log (e^{t_j}+U(e^{t_{j+1}}-e^{t_j}))$. These weighted interval draws sample the posterior directly, without asking a local Markov chain Monte Carlo update to cross its discontinuities.

Construct a rate-$\lambda$ Poisson process, for $\lambda >0$, from independent waiting times $E_j$ with exponential distribution $\mathrm{Exp}(\lambda )$. Put $S_k=E_1+\cdots +E_k$, $S_0=0$, and $N_t=\max \{k:S_k\le t\}$. The strong law of large numbers gives $S_k/k\to 1/\lambda$, so there are only finitely many arrivals on each finite interval. Consequently the counting paths are càdlàg, and $N_0=0$ almost surely.

At any deterministic time $s$, the memorylessness of the exponential distribution says that the residual waiting time has again exponential distribution $\mathrm{Exp}(\lambda )$ and is independent of the observed history. Subsequent waiting times are fresh independent copies. Thus the Poisson process restarts independently at $s$, proving independent increments and stationary increments. Integrating the joint waiting-time densities over $0<s_1<\cdots <s_k\le t<s_{k+1}$ givesso its time values have the expected Poisson distribution.

For $h\downarrow 0$,Together with stationary increments, this gives stochastic continuity at every time, from either side where applicable. All the Lévy process requirements hold, andFor $\lambda =0$, the identically zero process gives the degenerate case.

For $\lambda >0$, write the Poisson distribution mass function in exponential dispersion family form:Thus the natural parameter of an exponential family is $\theta =\log \lambda$ and the dispersion parameter is fixed at $1$. The counting-measure support is $0,1,\dots$; there is no independent free dispersion parameter in an ordinary Poisson model. The degenerate boundary $\lambda =0$ is obtained as a limit, rather than a finite natural parameter.

The exponential dispersion family identities give $\mathbb{E}Y=b^{\prime }(\theta )$ and $\mathrm{Var}(Y)=a(\varphi )b^{\prime \prime }(\theta )$. Since both derivatives of $e^{\theta }$ equal $e^{\theta }$,

For the exponential distribution with expected value $\mu$,The Poisson distribution has both expected value and variance equal to $\lambda$. Substitution into the random sum of independent claims formulas gives the portfolio A moments

The expected values of early and late new diagnosis counts at step $k$ areEach summand in $n_k$ corresponds to an infection that avoids early diagnosis, enters the non-early state one step later, remains there for the intervening steps and is then diagnosed. Define $h_j=0$ for $j\le 0$, and take an empty sum as zero. The early share of diagnosis intensity isIt is defined only when $e_k+n_k>0$; in particular there are no diagnoses at step 1 under the initial condition. With infection counts following a Poisson distribution and independent marking, early and late diagnosis counts in a fixed interval are independent Poisson random variables. Conditional on a positive total, the early count has a binomial distribution given the total with probability $p_{Ek}$, so this ratio also equals the expectation of the observed early fraction conditional on a nonzero total. Without that conditioning a sample fraction at zero diagnoses is undefined.

First consider an M-M-K queue with $K$ servers, arrival rate $\lambda$, and service rate $\beta$ at each server. Assume $\lambda <K\beta$ so that a stationary distribution exists. Its population is a birth-death process with rates $q_{n,n+1}=\lambda$ and $q_{n,n-1}=\beta \min (n,K)$. The equilibrium probabilities satisfy detailed balanceThus the population process is a reversible Markov chain. In its time reversal of a continuous-time Markov chain, forward departures become upward jumps whose rate is the constant $\lambda$, independent of the state. They can be generated by a rate-$\lambda$ Poisson process independent of the initial reversed state. Reversing time preserves the Poisson process law, proving the equilibrium departure process is Poisson with rate $\lambda$, the Burke theorem for a multi-server queue.

The same argument gives the departure component of quasireversibility needed for the tandem calculation: the present upstream population is independent of its entire past departure process. Indeed, these departures become future arrivals in the reversed chain, independent of its initial population. First check this on finite time windows, then extend to the full past. A Poisson output alone, without this independence, would not justify multiplying queue-empty probabilities.

Now let $a=\nu \mu$, where $\mu$ denotes the mean service time at passport control, so the service rate at each of its two servers is $1/\mu$. The stationary distribution of the M-M-s queue with $s=2$ has normalizing sumHence its empty probability is $(2-a)/(2+a)$. Its output feeds a stable M-D-1 queue with rate $\nu$ and deterministic service time $\tau$. In a stationary work-conserving single-server queue, the busy fraction is the rate of completed work, $\nu \tau$, so the empty probability is $1-\nu \tau$.

The downstream stationary workload is a causal function of past upstream departures. For example, it is the limit of workloads constructed from arrivals in increasingly long finite past intervals; stability makes this stationary construction well defined. The backward independence just proved therefore makes the two present queue populations independent. Consequently

With walking time $\sigma$, the downstream arrival stream is the upstream departure stream shifted by $\sigma$. Its present workload depends only on departures up to time $-\sigma$. The number walking at time zero is the number of departures in $(-\sigma ,0]$, which has Poisson distribution of mean $\nu \sigma$ and is independent of the earlier departures. The upstream present population is independent of both parts of its past departure stream. Thus the three empty events factor, givingThe factor $e^{-\nu \sigma }$ is the empty probability of the deterministic-travel infinite-server stage. Both printed stability inequalities ensure the stationary calculations apply.

Writing $y_i$ for the observed counts, independence of the Poisson distributions gives the log-likelihoodIts score function and negative second derivative areUnless all $x_i$ vanish, $J(\beta )>0$, so the log-likelihood is strictly concave. Use Newton method, starting from a finite ${\beta }_0$, withStop when the score function or the change in $\beta$ is small; a reduced step can ensure that the log-likelihood increases. Any finite zero of the score function is the unique maximum-likelihood estimator.

Existence needs a qualification. If the nonzero $x_i$ include both signs, $U$ decreases from $+\infty$ to $-\infty$, so a finite root exists. If all nonzero $x_i$ are positive, a finite root exists exactly when ${\sum }_i{x}_i{y}_i>0$; otherwise the supremum is approached as $\beta \to -\infty$. The corresponding condition for all negative $x_i$ is ${\sum }_i{x}_i{y}_i<0$, with a supremum at $+\infty$ otherwise. When all $x_i=0$, the parameter is unidentifiable. These cases explain when the iterative algorithm actually has a finite estimate to find.

If $C=0$, every bin contains at most one ball, so the total number of balls is at most $m$. Since $n=dm>m$, the pigeonhole principle makes that event impossible:As $n\to \infty$ along these values, $m\to \infty$ and the mean occupancy $n/m$ is $d$. Apply part (b):Both powers tend to $e^{-d}$. HenceThe Poisson limit for occupancy fractions interprets the answer as the probability that a Poisson distribution variable with mean $d$ is at least two.

Write $S={\sum }_i{X}_i$. The likelihood function for the Poisson distribution is proportional to $e^{-n\theta }{\theta }^S$. For $S>0$, differentiation of the log-likelihood gives $-n+S/\theta =0$, and its second derivative is negative. For $S=0$, the likelihood decreases with $\theta$ and has its maximum at the boundary zero. HenceThe estimator is unbiased, since $\mathbb{E}\hat{\theta }=n^{-1}{\sum }_i\mathbb{E}{X}_i=\theta$. Its variance is $\theta /n$. If the parameter space excludes zero, an all-zero sample gives a supremum as $\theta \downarrow 0$ rather than an attained maximum.

For independent Poisson distributions, ignoring the factorial term only when taking likelihood differences,The saturated likelihood maximizes each mean separately at ${\tilde{\mu }}_i=y_i$, including its limiting value at zero. Therefore the Poisson deviance iswhere $0\log (0/{\hat{\mu }}_i)$ is defined to be zero and ${\hat{\mu }}_i=e^{3.24926+0.01898a_i}$. The expression is twice a saturated-model likelihood ratio difference, not twice a likelihood ratio itself.

With sufficiently large fitted counts and regularity, $D$ is approximately ${\chi }_{30-2}^2$ under a correctly specified Poisson regression. Here $D=83.351$ on 28 residual degrees of freedom, with upper-tail probability $2.0935\times 10^{-7}$. The Poisson model fits poorly on this diagnostic. Its deviance is nearly three times its degrees of freedom, consistent with overdispersion, although a wrong mean function, dependence, or influential islands could also cause the discrepancy. The smaller Akaike information criterion than its competitors does not make this an adequate absolute fit.

For fixed $k$ and sufficiently large $N$,The finite product and last factor tend to one, while $(1-\lambda /N{)}^N\to e^{-\lambda }$. Thus the Poisson limit theorem givesThese numbers form a probability mass function because they are nonnegative and their sum is $e^{-\lambda }{\sum }_{k\ge 0}{\lambda }^k/k!=1$, by the exponential series. They define the Poisson distribution of parameter $\lambda$. Only $N\ge \lambda$ is needed to make the binomial parameter valid, which is sufficient for the limit.

For fixed known $r>0$, take the natural parameter of an exponential familyThe negative binomial distribution has log-likelihoodThus this is a negative binomial exponential family, with dispersion parameter $\varphi =1$ in the convention of part (a). The carrier contains the Gamma function and depends on the fixed size $r$, which need not be an integer. Differentiating the cumulant function of an exponential family givesThereforeThe size $r$ controls overdispersion relative to a Poisson distribution; it is different from the unit dispersion parameter used by this fixed-size generalized linear model. Another common convention applies to the scaled response $Z=Y/r$: then $b(\theta )=-\log (1-e^{\theta })$ and $\varphi =1/r$. Keeping track of the response scale prevents confusing these two representations. At $\lambda =0$ the law is concentrated at zero and is obtained as a boundary limit, rather than by a finite natural parameter of an exponential family.

The negative binomial regression assumes independent random variables conditional on their recorded days, withThe logarithmic link function applies to the conditional expectation, not to the observed count. The displayed maximum-likelihood estimates areand the dispersion parameter for the fitted family is one. Thus the fitted variance is ${\hat{\lambda }}_i+{\hat{\lambda }}_i^2/\hat{r}$, practically the Poisson distribution variance at these means.

Increasing the day by one multiplies the conditional expectation by $e^{{\hat{\beta }}_1}\simeq 0.996836$, a decrease of about $0.3164\%$. Comparing days 365 and 1 gives $e^{364{\hat{\beta }}_1}\simeq 0.3155$. The negative slope has Wald statistic approximately $-35.79$, so under the stated model assumptions there is very strong evidence of a decreasing conditional mean:This is an association with calendar day, not a causal effect established by the regression. The printed negative-binomial residual deviance is $7544.3$ on $1298$ residual degrees of freedom, which is a reason to investigate the adequacy of the assumed count model and dependence structure. In particular, the small reported slope standard error cannot make a misspecified model reliable.

Use the shape-rate convention for every gamma distribution, with $a,c>0$. Let $A_i=a+y_i$ and $s={\sum }_i{\theta }_i$. The Poisson distribution likelihood and the Gamma–Poisson hierarchical model giveAll factors involving the coordinate being updated must be retained, including $b^{na}$ from the conditional gamma distributions. The full conditional distributions areFor a blocked Gibbs sampler, draw $b$ from the second full conditional distribution, then redraw all the ${\theta }_i$ independently from the first ones. Repeating these two steps preserves the joint posterior distribution; observing $\theta$ after each complete sweep gives the Markov chain used in the next part. Standard gamma distribution sampling works for noninteger as well as integer shapes. The positive observation-period convention $t_i>0$ is used for the subsequent geometric drift condition.

Take observation periods to have positive lengths $t_i>0$, and let $P$ be the Markov kernel of a sweep which first draws $b$ given the current $\theta$ and then draws the new vector ${\Theta }^{\prime }$. Write $A_i=a+y_i$. Conditional on $b$, independence and the gamma distribution mean and variance giveAveraging over the full conditional distribution of $b$ preserves this bound, uniformly in the current state. Therefore $PV(\theta )\le M:=1+D<\infty$. This is the mechanism in bounded conditional moments imply a geometric drift.

Choose $0<\rho <1$, $R>M/\rho$, and $C=\{\theta :V(\theta )\le R\}$. Outside $C$, $PV\le M<\rho V$; inside $C$, $PV\le M\le \rho V+M$. HenceFor completeness, $C$ is a small set, even though it approaches the boundary of the positive orthant. Choose $\delta >0$ small enough that $B=[\delta ,2\delta {]}^n\subset C$. On $C$, $s={\sum }_i{\theta }_i\le \sqrt{R-1}$. The conditional gamma distribution of $b$ has fixed shape $na+c$ and rate $1+s$ in a compact positive interval. Its probability density function therefore has a positive common lower bound on $b\in [1,2]$. For those $b$, the product probability density function of ${\Theta }^{\prime }$ similarly has a positive common lower bound on $B$, since every shape is positive and every rate $b+t_i$ lies in a compact positive interval. Integrating over $b\in [1,2]$ gives $P(\theta ,\cdot )\ge \epsilon \eta (\cdot )$ on $C$, where $\eta$ is the uniform distribution on $B$ and $\epsilon >0$.

The transition probability density function is positive throughout the positive orthant, giving an irreducible Markov chain. Since $B\subset C$, the box minorization includes starts in the same box, which gives an aperiodic Markov chain. The proper prior distributions and bounded Poisson distribution likelihood have positive finite evidence, so the invariant posterior distribution is proper. Together with the geometric drift condition, these hypotheses imply geometric ergodicity. In particular, the argument proves the required drift rather than assuming that every Gibbs sampler is geometrically ergodic.

The positive-period convention matters. If zero lengths are allowed, the printed assertion can fail: take $n=1$, $a=c=1$, $t_1=y_1=0$. Then $b\mid \theta \sim \mathrm{Gamma}(2,1+\theta )$ and ${\Theta }^{\prime }\mid b\sim \mathrm{Gamma}(1,b)$, soThis example has a proper posterior distribution but cannot satisfy a finite quadratic geometric drift condition. Thus the proof establishes the intended claim for positive observation periods, not an unrestricted extension to zero exposure.

Use the ideal detection convention of one collected Electron per detected photon; otherwise replace photon totals by detected Electron totals using the quantum efficiency. Let $A$ be the summed measurement on the star patch and $B^{\prime }$ the independent background-only patch. After additive offsets are removed, their means are $Q+B$ and $B$. Independent Poisson distribution counts have variance equal to their mean, and a single read of $n$ detector pixels contributes $n{r}^2$ to each patch's variance. ThereforeThe two factors of two arise from the independently measured background and the second patch's read noise, not from doubling the stellar signal. The formula assumes equal background means, independent detector pixels and patch measurements, one read per patch, and negligible dark current or other noise. Several exposure reads or correlated detector pixels require their own variance and covariance terms.

Let $k=m/2$. After offset subtraction, each summed image has mean $N$ analogue-to-digital units per detector pixel. For a constant detector conversion gain $g$ and independent Poisson distribution detected counts, $\mathrm{Var}(A)=\mathrm{Var}(B)=N/g$: summing $k$ independent frames adds both their means and variances. Here $N$ is the mean of the sum, not one frame.

Write $A=N+a$, $B=N+b$. At high counts, the delta method gives $A/B\simeq 1+a/N-b/N$. Independence therefore givesThus the formula is a high-signal gain estimation from flat-field ratios, not an exact identity for a ratio of random variables. A Poisson distribution denominator can even be zero at low counts. The exposures must be linear and unsaturated, offsets removed, and read noise negligible. Being close to saturation increases the signal but does not establish linearity; clipping or charge-induced correlations biases a variance estimate. A large ensemble of suitable detector pixels supplies the measured standard deviation. In normal photodetector terminology $g$ is collected Electrons per ADU, consistent with detected photons here only under the stated unit-yield convention; see the instrument gain definition.

If temporal variation is spatially uniform, summing all frames in each group still gives independent Poisson distribution totals, but their means may now be $N_A,N_B$ with ratio $c=N_A/N_B\ne 1$. At high counts and negligible read noise, direct propagation givesAfter dividing the ratio image by its measured mean $c$, the appropriate estimator isThe original formula is recovered only for $N_A=N_B=N$. Uniform brightness drift does not automatically create a spatial pattern or extra ratio scatter beyond these changed noise levels; its leading problem is unequal group normalization. Interleave the groups in time or pair exposures of comparable flux, measure their separate means, and propagate shot and read noise when rescaling. If the spatial pattern or photodetector sensitivity also evolves, the ratio develops systematic structure that must be modeled or those frames rejected. A normalization inferred from finitely many detector pixels introduces a small shared correlation, which vanishes in the large-pixel limit and can be included in a precise uncertainty analysis.