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Write $A_k=N_{2k}-N_k$ and $B_k=N_k$. The stationary increments and independent increments of the Poisson process make $A_k$ and $B_k$ independent random variables, each with Poisson distribution of parameter $k$. ThusMultiplication of the two moment-generating functions givesThe odd terms cancel in Taylor's theorem, so $\log M_k(t)=t^2+O(k^{-1})$. The limit $e^{t^2}$ is the moment-generating function of a normal distribution with mean zero and variance two. The same continuity theorem for moment-generating functions therefore yieldsIn particular, the two equal-length increments must be separated before using independence: $N_{2k}$ and $N_k$ themselves are dependent.

The unit-time increments $D_j=N_j-N_{j-1}$ are independent and identically distributed random variables, by independent increments and stationary increments of the Poisson process. Each has the Poisson distribution of parameter one, hence expected value one and variance one. The central limit theorem therefore gives

A confidence interval is a pair of statistics $L(X),U(X)$ such that, for each parameter value $\mu$,In repeated sampling, at least the stated fraction of intervals constructed by this rule cover the fixed unknown parameter. It is not a posterior probability statement about the parameter after one interval has been observed. Discrete exact constructions can have coverage greater than the nominal value.

For independent observations from a Poisson distribution, $S={\sum }_i{X}_i$ has a Poisson distribution with mean $\lambda =n\mu$. Invert equal-tail tests of that total mean. At an observed integer $s>0$, the lower endpoint solvesand the upper endpoint solvesThen divide both endpoints by $n$. If $s=0$, use lower endpoint zero. Each inversion allocates at most $\alpha /2$ noncoverage probability to one side, giving at least $1-\alpha$ total coverage.

Here the single total count is 39. The supplied values with the correct adjacent count indices are $F_{27.7}(38)=0.975$ for the lower endpoint and $F_{53.3}(39)=0.025$ for the upper one. ThusThe values 28.6 and 52.2 use the other count index and are not the endpoints of this central interval. This is adjacent count tails in an exact Poisson confidence interval.

A Poisson distribution with parameter $\lambda \ge 0$ hasFor $\lambda =0$ it is concentrated at zero. For $\lambda >0$, shifting the index in the defining series gives $\mathbb{E}[X]=\lambda$ and $\mathbb{E}[X(X-1)]={\lambda }^2$, so $\mathrm{Var}X={\lambda }^2+\lambda -{\lambda }^2=\lambda$.

The full-day count has Poisson distribution with mean and variance $2\cdot 450=900$. Its normal approximation is $N(900,900)$, with standard deviation $30$. With the usual continuity correction,Without the half-unit correction the central-limit approximation gives $1-\Phi (1)\approx 0.159$.

The one-minute increment has Poisson distribution with parameter two, independently of where that interval occurs. Hence

The total $S$ under $H_0$ has Poisson distribution with mean $12$, and the observed total is $16$. Its one-sided p-value is thereforeThe supplied ${\varphi }_{12}(17)\approx 0.1$ also shows that the observed count does not enter the approximate 10% upper-tail rejection region. Thus there is insufficient evidence to reject $\mu =1$ at either the 10% or 5% level. The question specifies no testing level, so the p-value gives the general conclusion: rejection is justified only at levels at least about $0.1556$.

The printed ${\varphi }_1(16)\approx 0$ concerns a Poisson variable with mean one, not the twelve-month total. It is therefore not the null tail for this data; substituting it would incorrectly treat the yearly count as one month's count.

A Poisson process of rate $\lambda$ is a counting process starting at zero, with independent stationary increments, such that an increment over an interval of length $t$ has the Poisson distribution of parameter $\lambda t$. It has right-continuous paths with unit jumps almost surely.

Independence of the two processes makes increments of $A+B$ independent on disjoint intervals. The sum of independent Poisson random variables with parameters $\alpha t$ and $\beta t$ is Poisson with parameter $(\alpha +\beta )t$, as follows by convolution:Thus $A+B$ is a Poisson process of rate $\alpha +\beta$. Its first red and green waiting times are independent exponential random variables of rates $\alpha ,\beta$, so

For $j\ge 1$ the only entrance is from $j-1$, so stationarity requires ${\pi }_j={\pi }_{j-1}\lambda /j$. Induction gives ${\pi }_j={\pi }_0{\lambda }^j/j!$, and normalization using the exponential series gives ${\pi }_0=e^{-\lambda }$. HenceThe remaining zero-state equation also holds: ${\sum }_i{\pi }_i\lambda /(i+1)=e^{-\lambda }(e^{\lambda }-1)=1-{\pi }_0$, so ${\sum }_i{\pi }_i(1-\lambda /(i+1))={\pi }_0$. Thus this normalized Poisson law really is stationary, not just a solution of the positive-state equations. Irreducibility makes the stationary probability law unique.

Write $t=1-\rho \in (0,1)$. The target Poisson distribution and proposal geometric distribution have probability mass functions $p_k=e^{-\lambda }{\lambda }^k/k!$ and $q_k=(1-t)t^k$. Their ratio isThe successive ratio $R_{k+1}/R_k=\lambda /[t(k+1)]$ shows that a maximizing integer is $m=\lfloor \lambda /t\rfloor$, with a tie at $m-1$ when $\lambda /t$ is an integer. Thus the tight envelope isFor any fixed $t$, draw $Y\sim \mathrm{Geom}(1-t)$ using (ii)'s inversion formula with this new parameter, and draw an independent $U$ from the uniform distribution on $(0,1)$. Accept whenotherwise restart. The acceptance probability is $1/M(t)$ and the expected number of proposals is $M(t)$.

To minimize that expected count, set $K=\lceil \lambda \rceil$. The exact answer isHere is a proof that includes the changes of maximizing integer. Each $\log R_k(t)$ is a convex function of $t$, since its second derivative is $(1-t{)}^{-2}+k/t^2>0$. Their pointwise maximum $\log M(t)$ is also a convex function. When $m$ is fixed its derivative isIf $K-1<\lambda \le K^2/(K+1)$, take $t_{\ast }=\lambda /K$, where $m$ changes from $K$ on the left to $K-1$ on the right. The left derivative is nonpositive because $\lambda \le K^2/(K+1)$; the right derivative is nonnegative because $\lambda \ge K-1$. This is a minimum. If $K^2/(K+1)<\lambda \le K$, take $t_{\ast }=K/(K+1)$. There $K<\lambda /t_{\ast }\le K+1$ and the active branch has $m=K$ except for the endpoint tie; its derivative vanishes. Convexity again proves the global minimum. The formulas agree at the boundary.

For a fully specified optimal algorithm, compute $K,t_{\ast },{\rho }_{\ast }$ and $m_{\ast }=\lfloor \lambda /t_{\ast }\rfloor$ once, then use the preceding acceptance test with $t=t_{\ast }$ and $m=m_{\ast }$. At an integer tie either maximizing integer gives the same test. In code, factorials can be evaluated through the logarithm of the gamma function:and acceptance is $\log U\le \min (0,\log a_Y)$. This avoids overflowing factorials.

The optimum minimizes expected trials, the usual idealization of time when each proposal and test has constant cost. Actual runtime also depends on the implementation of discrete inversion and factorial evaluation. Simply minimizing an upper bound for $M(t)$ would not necessarily give the exact optimum above; see optimal geometric envelope for a Poisson distribution.

Let $s={\sum }_i{x}_i$, $A=a+s$ and $B=b+n$. Conditional independence gives the Poisson distribution's likelihoodMultiply by the prior gamma distribution's density and discard factors independent of $\theta$:Normalizing with the gamma function identifiesHere $A=a+s$ is the shape and $B=b+n$ is the rate, not the scale. This is conjugate prior updating for a Poisson distribution.

Use $A=a+{\sum }_i{x}_i$ and $B=b+n$ as above. The future observation is independent of the old observations conditional on $\theta$, so the posterior predictive probability is obtained by integrating its Poisson distribution against the posterior distribution of $\theta$:The last integral is the normalizing gamma function integral, equal to $\Gamma (A+k)/(B+1{)}^{A+k}$. This is a negative binomial distribution in the count-of-failures convention with shape $A$ and success probability $B/(B+1)$; the gamma-function expression remains valid for noninteger $A>0$. As checks, its mean is $A/B$, and the law of total variance gives variance $A/B+A/B^2$. The additional term is the contribution of posterior uncertainty in $\theta$.

Writing $s={\sum }_i{x}_i$, the Poisson distribution likelihood is proportional to ${\theta }^s{e}^{-n\theta }$. Multiplying by the gamma distribution prior givesThis is Poisson-gamma conjugacy, with the rate convention made explicit:Its normalized density is $(\lambda +n{)}^{r+s}{\theta }^{r+s-1}{e}^{-(\lambda +n)\theta }/\Gamma (r+s)$ for $\theta >0$.

Write $m=r+s$ and $\beta =\lambda +n$. Conditional independence of the next count given $\theta$ allows us to average its Poisson distribution over the posterior distribution:This Poisson-gamma mixture is the posterior predictive distribution. In the stated negative binomial distribution convention its parameters areIts expected value $m/\beta$ equals the posterior mean of $\theta$, while its variance $m/\beta +m/{\beta }^2$ includes both conditional Poisson variation and uncertainty in $\theta$.

Let $Z=(Z_n{)}_{n\ge 0}$ be a Markov chain with transition kernel $K$, and let $N$ be an independent Poisson process of constant rate $r>0$. A Pseudo-Poisson process is $X_t=Z_{N_t}$. On the Banach space $B_b(E)$ of bounded Borel measurable functions, with the supremum norm, writeIts infinitesimal generator, transition semigroup, and resolvent operator are respectivelyand, for $\lambda >0$,The transition semigroup averages $K^n$ against the Poisson distribution of the number of updates. Since $K$ is a Markov operator, $\Vert K\Vert \le 1$; both series converge in operator norm. Updates with $Z_{n+1}=Z_n$ are allowed and do not count as changes of the observed state. The degenerate rate $r=0$ gives $L=0$, $P_t=I$, and $R_{\lambda }=I/\lambda$.

Conditioning on the observed time. The native PDF defines ${\tau }_2$ as the first production time and then conditions on that production occurring at ${\tau }_2$. If the intended information is its numerical value ${\tau }_2=t$, the answer depends on $t$. We first compute this conditional probability using a conditional probability density; conditioning on an exact value of a continuous random variable means a density ratio, since $\mathbb{P}({\tau }_2=t)=0$.

Before ${\tau }_2$, no molecules of the second chemical species are present, so the joint-removal chemical reaction cannot occur. Molecules of the first chemical species arrive as a rate-$2$ Poisson process, and each independently starts a clock with a rate-$1/2$ exponential distribution for producing the first molecule of the second chemical species. Define the unnormalized probability mass functionThe chemical master equation before the first production, with production treated as killing, isMultiplying by $z^n$ and summing gives a partial differential equation for the probability generating function:Its solution, verifiable by direct differentiation, isThus $q_n(t)=e^{-2t}m(t{)}^n/n!$, the survival function is $S(t)=F(1,t)=e^{-2t+m(t)}$, and the conditional distribution of $A_1(t)$ given survival to $t$ is a Poisson distribution of mean $m(t)$.

Production in $(t,t+dt)$ from Markov chain state $n$ has joint probability $(n/2)q_n(t)dt+o(dt)$. Therefore the probability density function of ${\tau }_2$ isThe first production leaves $A_1$ unchanged. Taking the density ratio gives a size-biased distribution of the surviving count:In particular, under this reading the requested conditional probability isThe shift by one matters: an observation made exactly when production occurs preferentially selects larger counts, since their production reaction propensity function is larger.

Without observing the numerical time. If the conditioning merely repeats the definition of ${\tau }_2$, the requested probability is instead the unconditional count at that stopping time. Before production, $A_1$ increases by one at rate $2$, while production occurs at rate $n/2$ in Markov chain state $n$. The competing exponential clocks implyThe first factor requires a second arrival before production from count one; the second requires production before a third arrival. The first arrival itself is certain. Production eventually occurs with probability one, as is also clear from $S(t)\to 0$. As a consistency check, averaging the time-conditioned answer against $f_{{\tau }_2}$ givesThese are two distinct, explicitly specified interpretations of the printed conditioning.

A product-Poisson stationary distribution is suggested by balancing the three reaction fluxes around the directed cycle of reaction complexes. If their mean counts are $a,b$, this complex balance condition iswhich gives $a=4$ and $b=1/2$. This suggests the joint probability mass functionIt is normalized because it is the product of two Poisson distributions. We verify it directly in the chemical master equation, rather than assuming a product-Poisson stationary distribution theorem. The three incoming terms satisfyWith negative-index values set to zero, these identities also hold on the axes. Their sum equals $(2+n/2+nm)\pi (n,m)$, precisely the outgoing probability flux. The stationary average total exit rate is $2+{\mathbb{E}}_{\pi }n/2+{\mathbb{E}}_{\pi }(nm)=6<\infty$, so these stationary balance equations do define an invariant distribution of a continuous-time Markov chain for the nonexplosive model verified below. This is complex balance; it does not assert pairwise detailed balance for the irreversible chemical reactions.

To identify this invariant distribution with the limiting probability distribution in the question, first check that the model is a nonexplosive continuous-time Markov chain. On any finite time interval, $A_1$ is bounded by its initial count plus the finite number of arrivals in a rate-$2$ Poisson process. Conditional on that bound, the rate of producing $A_2$ is bounded, so only finitely many such productions occur. The number of joint removals cannot exceed the available molecules. Thus only finitely many chemical reactions occur in finite time.

The continuous-time Markov chain is also an irreducible Markov chain. From a Markov chain state with both counts positive, repeated joint removals reduce both. On the $A_2=0$ axis, producing one $A_2$ and then removing a pair reduces $A_1$ by one. On the $A_1=0$ axis, adding one $A_1$ and then removing a pair reduces $A_2$ by one. Each prescribed finite sequence has positive probability, so every Markov chain state reaches $(0,0)$. Conversely, arrivals followed by productions reach any $(n,m)$ with $n>0$; to reach $(0,m)$, reach $(1,m+1)$ and remove one pair. The continuous-time Markov chain is therefore positive recurrent, and the continuous-time Markov chain convergence theorem gives $\varphi =\pi$ from the empty initial condition.

The two Poisson factors happen to have cancelling numerical coefficients at $(2,2)$:

The exponential input has norm squared $\exp (|{\alpha }_1{|}^2+|{\alpha }_2{|}^2)$. Its normalized form is a product of coherent states. Substitution of the inverse transformation givesSince the output creation operators commute, the normalized output factorizes as $|{\beta }_1⟩|{\beta }_2⟩$. For all nonnegative integers $n_1,n_2$,These are independent Poisson distributions, not a fixed-total-particle-number distribution. unitarity gives $|{\beta }_1{|}^2+|{\beta }_2{|}^2=|{\alpha }_1{|}^2+|{\alpha }_2{|}^2$. Equal complex inputs make ${\beta }_2=0$, so that port is certainly empty; opposite inputs empty port 1. This is amplitude interference of coherent inputs, distinct from the fixed-number two-particle interference in (c).

For the Poisson distribution with parameter $1$, the exponential series givesEvery derivative is $G^{(k)}(s)=e^{s-1}$, so the factorial moments areIn particular $G^{\prime }(1)=G^{\prime \prime }(1)=1$, and the variance is

Using the probability mass function of the Poisson distribution directly givesFor each fixed positive integer $n$, this series converges: the ratio of successive nonzero terms is $(1+1/j{)}^n/(j+1)$, which tends to zero. Equivalently, using the polynomial identity in (ii) and the factorial moments from (a)(iv) gives $\mathbb{E}[X^n]={\sum }_{k=1}^{n}A(n,k)$; the extension of that identity to $X=0,1$ is justified in (iv).