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For a count $s>0$ from a Poisson distribution, equal-tail inversion uses the inclusive upper-tail event $S\ge s$ for the lower endpoint and the inclusive lower-tail event $S\le s$ for the upper endpoint. Their cumulative distribution indices differ by one, as displayed. If $s=0$, the lower endpoint is zero. For $n$ observations of mean $\mu$, apply this to their total of mean $\lambda =n\mu$ and divide endpoints by $n$. The construction has at least the nominal frequentist coverage; discreteness can make it conservative.

If host-specific equilibrium mean burden has a gamma distribution, conditional Poisson distributions produce a Poisson-gamma mixture, hence a negative binomial distribution. With mean $m$ and shape $k$, its zero-burden probability is $(1+m/k{)}^{-k}$. Decreasing $k$ increases aggregation: a smaller infected-host prevalence can accompany the same mean burden and more heavily infected individuals.

For a constant-rate radiation measurement through $n$ identical absorbing sheets, a simple Poisson distribution model has expected counts ${\mu }_n=t[N_0{e}^{-bn}+B]$. Here $N_0$ is the unattenuated signal rate, $b$ is attenuation per sheet, $B$ is an additive background rate and $t$ is exposure. At low attenuation the signal is approximately exponential; at strong attenuation the observed rate approaches $B$. Consequently a logarithmic plot of raw counts bends towards a plateau rather than remaining a straight line. Measure background independently with the source absent, or fit it with its uncertainty; subtracting it and then taking a logarithm is unreliable when the net count is small or negative.

For stable positive detector pixel means $N_i$ and uniform detector conversion gain $g$, high-count independent Poisson distribution noise gives $\mathrm{Var}(A_i/B_i)\simeq 2/(g{N}_i)$. Thus substituting a global mean in the equal-signal formula yields $g_{\mathrm{naive}}=g/[⟨N_i⟩⟨1/N_i⟩]\le g$, by the Jensen inequality. Equality holds for uniform means. The flat-field correction pattern cancels from the ratio mean but not from the signal-dependent noise. Local signal bins, or a variance fit after normalization of differences by the square root of their local mean, avoid this leading bias.

For independent factories with accident probability $p$, and independent Poisson distribution claim counts of mean $\lambda$ per accident, the accident count is binomial and each accident supplies a cluster. The random-sum transform identity gives the displayed probability generating function. In particular the no-claim chance is $(1-p+p{e}^{-\lambda }{)}^m$; a factory with an accident can still have zero claims. The binomial accident law has Panjer parameters $a=-p/(1-p)$ and $b=(m+1)p/(1-p)$.

Independent arrivals of charges $q$ at rate $\lambda$ give a Poisson distribution for the number $N$ in time $\tau$, with mean and variance $\lambda \tau$. Since ${\overline{I}}_{\tau }=qN/\tau$ and $I=q\lambda$, its variance is $qI/\tau$. This agrees with integrating the one-sided shot noise power spectral density $2qI$ through the averaging filter: its effective noise bandwidth is $1/(2\tau )$. A two-sided convention instead assigns density $qI$ to each sign of frequency.

The charge-transfer efficiency is the fraction of a packet delivered into its intended next site or time slot in one transfer, under stated conditions. It is distinct from quantum efficiency, which concerns conversion of incident photons into collected charge carriers. Define the complementary inefficiency as $ϵ=1-{\eta }_{\mathrm{tr}}$.

In an original model with constant $0\le {\eta }_{\mathrm{tr}}\le 1$, let $N_j$ denote the mean number of original packet electrons still delivered on time after $j$ transfers. Assuming no reinjection into that packet,Repeated substitution proves the result; for different efficiencies the fraction is ${\prod }_{j=1}^n{\eta }_j$. For $ϵ\ll 1$, a Taylor expansion of $\ln (1-ϵ)$ gives ${\eta }_{\mathrm{tr}}^n\simeq e^{-nϵ}$, while the simpler $1-nϵ$ requires $nϵ\ll 1$.

Choose $N_{\mathrm{in}}=10000$ and $n=2000$. For ${\eta }_{\mathrm{tr}}=0.9999$, the expected on-time output is $8187.23$ electrons; for ${\eta }_{\mathrm{tr}}=0.99999$, it is $9801.99$. These are model means, so fractional electron counts do not describe an individual packet. Even a small per-transfer imperfection can imprint distance-dependent image errors.

If every original electron independently stays on time with probability $p={\eta }_{\mathrm{tr}}^n$, a fixed initial count gives a binomial distribution with variance $N_{\mathrm{in}}p(1-p)$. An initial Poisson distribution with mean $\mu$ instead gives an on-time Poisson distribution with mean and variance $\mu p$, by Poisson thinning. A deterministic attenuation formula alone does not specify noise. Real trapping can correlate packets and depends on timing, illumination and temperature; the independent constant-efficiency model then fails.

The number of insurance claims in a specified exposure period. Its probability generating function combines with the moment-generating function of an independent claim size through the random-sum transform identity. A Poisson distribution, geometric distribution, or negative binomial distribution provides a possible count model, with different dispersion and zero-claim probabilities.

The probability distribution of the nonnegative integer number of claims over a specified accounting period. Its probability generating function determines the count probabilities. A Poisson distribution gives equal count expected value and variance; a Poisson mixture allows heterogeneity in the underlying risk intensity.

A normalized nonnegative probability measure over coherent states produces a classical mixture. The measure may contain Dirac delta distributions, so a pure coherent state is included. Conditioning on $\alpha$ gives Poisson distribution photon statistics; hence $\mathrm{Var}n=⟨n⟩+\mathrm{Var}(|\alpha {|}^2)\ge ⟨n⟩$. Each optical quadrature has variance at least its vacuum value because mixing adds nonnegative variance of conditional means. These inequalities provide simple nonclassicality witnesses.

For a coherent state, write $a|\alpha ⟩=\alpha |\alpha ⟩$ and $|\alpha ⟩={\sum }_n{c}_n|n⟩$ in number states. The annihilation operator gives $c_{n+1}\sqrt{n+1}=\alpha c_n$, so $c_n=c_0{\alpha }^n/\sqrt{n!}$. Normalization yields $|c_0{|}^2=e^{-|\alpha {|}^2}$. Thus photon statistics follows a Poisson distribution with $\mu =|\alpha {|}^2$, and the normal ordering of every moment gives $⟨(a^{†}{)}^k{a}^k⟩={\mu }^k$. This factorization is the defining virtue of coherent states for optical calculations.^[ref]

As an original comparison, choose $\mu =2$. Then $p_0=e^{-2}$, $p_1=p_2=2e^{-2}$, $\mathrm{Var}N=2$, and $⟨N(N-1)⟩=4$. A two-photon Fock state also has mean two but has $\mathrm{Var}N=0$ and pair moment two. ConsequentlyIn particular, $|1⟩$ cannot supply a pair. Nevertheless, photon statistics alone does not determine a quantum state: averaging $|\alpha ⟩⟨\alpha |$ over a uniform quantum phase removes its off-diagonal entries while retaining exactly the same $p_n$ and these equal-time moments.

The modeled matched filter search found an arrival difference of about $7\mathrm{ms}$, within the roughly $10\mathrm{ms}$ intersite light travel time. Background events were obtained by shifting one detector's timestamps far beyond that time and repeating the coincidence analysis. The binary search bounded the false alarm rate below one event per $203000$ years. That is an empirically estimated background rate, not a probability that the source interpretation is false.^[ref]

The logic is to destroy a common astronomical arrival while retaining each instrument's real glitches. A Poisson distribution approximation would turn a background rate $r$ into $1-e^{-rT}$ for at least one equally strong noise event during live time $T$, only after accounting for the search's ranking and trials. Repeated time slides reuse data; their accumulated background duration is not an equally long series of independent new observations. Correlated disturbances also require separate controls because shifting timestamps can erase them along with a true signal.

The dedicated noise study inspected auxiliary channels and deliberately applied magnetic, acoustic and mechanical disturbances to measure their coupling into strain. For GW150914, checks included environmental coincidences, instrument state, and exclusion of an intentional hardware injection. It also examined potentially correlated global magnetic disturbances, which time slides alone cannot exclude.^[ref] The point of an injection test is quantitative: a quiet environmental monitor is informative only if its sensitivity and coupling bound the disturbance needed to explain the candidate.

In slotted ALOHA, let old packets retry independently with fixed probability $0<p<1$, and let independent Poisson random variables of mean $\nu >0$ transmit once on arrival. Write $q=1-p$ and $r=e^{-\nu p}$. A hypothetical process which never removes a packet has backlog $b+A_t$, where $A_t$ is the cumulative arrival count over $t$ slots. Couple its attempts to those of the actual process until the first noncollision. Its one-slot noncollision probability at backlog $n$ is $h(n)=e^{-\nu }[(1+\nu )q^n+np{q}^{n-1}]$. The union bound and the Poisson distribution giveIndeed $\mathbb{E}[q^{A_t}]=r^t$ and $\mathbb{E}[A_t{q}^{A_t}]=\nu tq{r}^t$. Thus $\epsilon (b)\to 0$. From every sufficiently large backlog, permanent collisions have probability at least $1/2$. Independent arrival batches reach that threshold almost surely; after each failed attempt the Strong Markov property permits another attempt. The probability of $k$ consecutive failures is at most $2^{-k}$, proving almost-sure eventual permanent collisions without assuming in advance that the backlog escapes.

A compound mixed Poisson distribution is a random sum of independent claims whose count has a Poisson mixture law: conditional on a nonnegative random intensity $\Lambda$, $N$ has Poisson distribution with parameter $\Lambda$. The claim sizes are independent and identically distributed random variables, independent of the intensity and count mechanism. If $M_X$ is their moment-generating function, then $M_S(t)=\mathbb{E}\exp (\Lambda (M_X(t)-1))$ wherever finite. A gamma distribution intensity of shape $r$ and rate $\alpha$ gives a negative binomial distribution count with shape $r$ and success probability $\alpha /(\alpha +1)$. Independent intensities with common gamma rate add their shapes by additivity of independent gamma distributions with a common rate.

Conditioning on the independent Poisson distribution count in a random sum gives $\mathbb{E}{e}^{tS}=\exp (\lambda (M_X(t)-1))$ wherever finite. Differentiation at zero gives the displayed cumulants when the relevant derivatives exist. The aggregate variance is the count rate times the raw second claim moment, not just the claim variance. Positive moment-generating function arguments need an exponential moment; finite first and second claim moments suffice for the aggregate mean and variance by conditioning or right derivatives at zero of the Laplace transform of a nonnegative random variable.

The law of a sum of independent identically distributed claims with an independent Poisson distribution count of parameter $\lambda$. If the claim moment-generating function is $M_X$, the aggregate transform is $\exp (\lambda (M_X(r)-1))$ wherever finite. It has an atom at zero when the claims are positive. This is the fixed-time law of a Compound Poisson process.

Consider an approximate conditional likelihood in which a paired total $t_i$ leads to a structural zero with probability $\theta$, and otherwise a binomial distribution with success probability $p$. For an observed zero, the posterior cure weight is $q_i=\theta /[\theta +(1-\theta )(1-p{)}^{t_i}]$; for a positive count it is zero. Differentiating the log-likelihood gives the displayed equations at an interior maximum-likelihood estimator, with the weights evaluated at that estimator. Recomputing the weights and then updating the parameters is an expectation-maximization algorithm. Boundary fits must also be considered. These equations describe a conditional mixture: if cure is instead specified before two Poisson distributions are conditioned on their total, its conditional probability generally depends on the total and on the baseline nuisance rate. A zero observed in a short interval is not by itself proof of permanent biological cure.

The same intensity measure can belong to very different point processes. A Poisson point process with a locally finite intensity measure $\mu$ adds two requirements: each finite-intensity window count has the Poisson distribution of mean $\mu (W)$, and counts in pairwise disjoint observation windows are mutually independent. These requirements determine its probability law; they do not follow from the formula for $\mu$.^[ref]

A randomly shifted lattice point process, a unit-rate Poisson point process, and the shared random rate gives dependent point counts construction all satisfy $\mu (A)=|A|$ on the real line. Nevertheless, their counts in $[0,1)$ have respectively variance zero, one, and two. Counts in the disjoint observation windows $[0,1/2)$ and $[1/2,1)$ have respectively covariance $-1/4$, zero, and $1/4$. These exact calculations show why measuring mean density cannot identify a spacing or statistical dependence model.

For cell means ${\mu }_j$ with ${\sum }_j{\mu }_j=n$, the independent Poisson distribution likelihood and the multinomial distribution likelihood with probabilities ${\mu }_j/n$ differ by a parameter-independent factor. Thus a maximum-likelihood estimator for the unrestricted Poisson model whose fitted total is $n$ is also an estimator for the constrained multinomial model. Equality of named parameter estimates presumes identifiability and uniqueness; without these, the conclusion concerns the sets of maximizers. A free intercept in a log-linear Poisson model enforces the fitted-total equation.

For fixed real $c$, the number of isolated vertices in a binomial random graph with $p=(\log n+c)/n$ converges to a Poisson distribution of mean $e^{-c}$. Its fixed factorial moments tend to $(e^{-c}{)}^k$. A union bound for small connected components, together with exclusion of intermediate components, shows that isolated vertices are asymptotically the only obstruction to connectivity. Thus the displayed limit is the zero-mass of that Poisson distribution.