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A Poisson observation is an integer-valued count modeled by a Poisson distribution of specified intensity, possibly conditionally on a latent field. Independent Poisson observations motivate Poisson data fidelity through their log-likelihood.

For independent photon detections with a fixed mean rate, counts in a fixed exposure have a Poisson distribution, so the variance equals the mean. Multiplying by a representative photon energy gives ${\sigma }_E=h\nu \sqrt{N}$. A source-only signal-to-noise threshold $q$ therefore needs an expected $N=q^2$ photons. Background, detector noise, variable illumination and photon correlations can invalidate that source-only formula.

A Homogeneous Poisson process models arrivals at a constant rate $\lambda >0$. Its counting process $N_t=N(t)$ counts arrivals in $(0,t]$, with $N_0=0$. Its paths are right-continuous nondecreasing functions, take nonnegative integer values, and have unit jumps almost surely. The defining Poisson distribution of each increment isIt has stationary increments: the probability distribution above depends on the duration $t$, not the starting time $s$. It also has independent increments: counts in any finite collection of disjoint time intervals are independent random variables. Together these conditions specify the joint count laws, not merely the Poisson distribution at a single time.^[ref]

The rate has units of arrivals per unit time; $\lambda t$ is dimensionless. The Poisson distribution givesFor a toy rate $\lambda =2$ per minute and a duration of $1.5$ minutes, the expected value and variance of the count are both $3$. The probability of no arrival is $e^{-3}\approx 0.049787$, and the probability of exactly three is $3^3{e}^{-3}/3!=\frac{9}{2}{e}^{-3}\approx 0.224042$. In two consecutive one-minute intervals, the probability of exactly two arrivals in the first and exactly one in the second is $(2e^{-2})(2e^{-2})=4e^{-4}\approx 0.073263$, using independent increments. These are exact model calculations with rounded decimals, not measured frequencies.

The diagram's nine hand-chosen arrivals illustrate a path; the line $\lambda t$ is an expected value, not a prescribed trajectory. Constant rate does not imply evenly spaced arrivals. Poisson interarrival times connect the count description to waiting, while Poisson thinning explains how independently routing arrivals produces separate streams. This makes the Poisson process useful as a simple arrival model for a queue, provided constant rate and independent increments are plausible. For a deterministic changing rate, use an Inhomogeneous Poisson process; bunching or history-dependent arrivals need additional assumptions. Conditional on $N_t=n$, the arrival times are the order statistics of $n$ independent uniform distributions on $[0,t]$, as developed in Poisson process conditional arrival times.

For independent Poisson distributions with log means $Z\beta +o$, where $o$ is a known offset, differentiation of the log-likelihood gives the score function $Z^T(y-\mu )$. An interior maximum-likelihood estimator therefore satisfies the displayed equations. An intercept and categorical indicator columns make fitted count totals equal observed count totals in each represented factor level. These equations match counts, rather than unweighted averages of rates when exposures differ.

// Target: probability-and-statistics.bigb

With ${\mu }_i=e^{\beta x_i}$, define $U={\sum }_i{x}_i(Y_i-{\mu }_i)$ and $I={\sum }_i{x}_i^2{\mu }_i$. The estimating equation and a Taylor expansion give $\hat{\beta }-{\beta }_0\approx I({\beta }_0{)}^{-1}U({\beta }_0)$. If responses are independent with correct means and $\mathrm{Var}(Y_i)=\varphi {\mu }_i$, then $EU=0$ and $\mathrm{Var}U=\varphi I$. Consequently the first-order estimator variance is $\varphi /I$, even if the working Poisson distribution is wrong. Regular asymptotics require growing information and control of leverage and the Taylor remainder.

For $N$ with Poisson distribution of mean $\lambda$, the equality $n\mathbb{P}(N=n)=\lambda \mathbb{P}(N=n-1)$ proves the identity by summation. For an independent random sum of independent claims $S$ with this count, exchangeability of the summands gives $\mathbb{E}[Sh(S)]=\lambda \mathbb{E}[Xh(S+X)]$, where $X$ on the right is independent of $S$ and has the severity law. These identities hold for nonnegative measurable functions, and for integrable signed functions. They obtain moments without assuming that a positive exponential moment exists.

Independent Poisson distributions with a separate nuisance intercept in every stratum become multinomial distributions after conditioning on their totals. Profiling the intercept gives $e^{{\hat{\gamma }}_s}=N_s/{\sum }_l{e}^{x_s^T{\beta }_l}$. Substitution produces the same parameter-dependent log-likelihood as the conditional multinomial model. Both approaches fit ${\hat{\mu }}_{sl}=N_s{\hat{p}}_{sl}$. This exact equivalence differs from approximating a single binomial event count by a rare-event Poisson distribution.

For a known background rate $b$ observed for exposure $T$, independent events give a Poisson distribution with mean ${\mu }_b=bT$. An excess-count test of the background-only null hypothesis uses the upper-tail P-value in the title. The tail, rather than the probability of exactly $n$, counts all outcomes at least as discrepant in the chosen direction.

This probability is conditional on the null model; it is not the posterior probability that the null model is true. A stipulated discovery threshold is met only when the tail is below that threshold. Uncertainty in the background rate or selecting among many searches needs additional modelling rather than being absorbed into the same known-mean formula.

Poisson thinning splits a Poisson process by independently marking each arrival as retained or rejected. Let the original Homogeneous Poisson process have rate $\lambda >0$, and let $p(s)\in [0,1]$ be a deterministic measurable function of time. Attach independent uniform distributions $U_1,U_2,\dots$ on $(0,1)$, independent of the entire original Poisson process. At arrival time $T_i$, retain the arrival if $U_i\le p(T_i)$. Given all arrival times, these decisions are independent and have the specified retention probabilities. Define the two counting processes using an indicator function:The Poisson thinning theorem gives independent retained and rejected Inhomogeneous Poisson processes. Their rates are $\lambda p(s)$ and $\lambda (1-p(s))$, respectively, in the sense of the intensity function of a point process. In particular, for $0\le a<b$,The independence is of the two whole processes: every finite collection of retained counts is independent of every finite collection of rejected counts. It is not merely zero covariance for one interval. For constant $p$, the streams are Homogeneous Poisson processes of rates $\lambda p$ and $\lambda (1-p)$; the endpoints $p=0,1$ include the identically zero stream.^[ref]

For an exact toy calculation, take $\lambda =2$ per minute and retain with $p(s)=1/4$ on $(0,2]$ and $p(s)=3/4$ on $(2,4]$. In the first two minutes, the retained and rejected expected values are $2\cdot 2\cdot 1/4=1$ and $2\cdot 2\cdot 3/4=3$. In the next two minutes they are $3$ and $1$. Over all four minutes, $R_4,D_4$ are independent Poisson distributions with mean $4$ each, and their sum $N_4$ has mean $8$. ThusThese probabilities and expected values describe the model, not empirical measurements. In the diagram, the original nine synthetic arrivals split into five retained and four rejected arrivals; those realized counts need not equal their expected values.

The mechanism explains why mutually exclusive labels can yield independent counts: randomness of the original total offsets the dependence from allocating a fixed total. For constant $p$ and an interval with original mean $m$, the conditional retained count has a binomial distribution given $N=n$. The joint probability generating function isThis factorization proves the two independent Poisson distributions; disjoint original intervals give independent pairs. For varying $p(s)$, the Independent marking theorem for Poisson point processes gives the same conclusion with the two integrated intensities. The joint red and green Poisson count formula gives the corresponding direct calculation with probability mass functions.

Independence here is unconditional. After conditioning on $N=n$, the identity $D=n-R$ gives covariance $\mathrm{Cov}(R,D\mid N=n)=-np(1-p)$ for constant $p$. Thus the retained and rejected counts are dependent for $n>0$ and $0<p<1$, even though their unconditional covariance is zero. Poisson thinning can model independent routing of arrivals, or independent detection of arrivals with a known efficiency schedule; using only a retention fraction without checking the marking mechanism can give a wrong count law.

For a uniform prior distribution on $(l,u)$ with $0<l<u$, conditionally independent Poisson distribution observations have sufficient count $s={\sum }_i{x}_i$. Their Bayesian posterior density is proportional to $t^s{e}^{-nt}$ on that bounded interval. The displayed posterior mean is the Bayes estimator under squared error loss, and is generally not the affine Bühlmann credibility estimate.

Subtracting an exactly known mean sky level leaves source-plus-sky variance of a Poisson distribution $S+B$. If the sky per source pixel is estimated from $n_b$ equal-area, equally exposed independent sky pixels, its additional variance is $B/n_b$, so $R=S/\sqrt{S+B+B/n_b}$ before read and dark terms. The variance of removed background photons does not disappear with their mean.

For a mixed Poisson distribution with known exposure $n>0$, multiply the unnormalized posterior density ${\theta }^s{e}^{-n\theta }\pi (\theta )$ by $\theta$ and compare its integral with the next probability mass. The factors $n^s/s!$ give the displayed identity whenever $p_n(s)>0$. Both mass probabilities have the same exposure $n$. For a future conditionally independent unit-exposure Poisson distribution count, this is also its posterior predictive expected value. It holds for arbitrary proper mixing distributions, with integrals taken against the prior measure when there is no density.

For two independent samples of size $n$ with event probabilities $p_0,p_1$, put $\delta =p_0-p_1$, $\bar{p}=(p_0+p_1)/2$, $s_0^2=2\bar{p}(1-\bar{p})/n$ and $s_1^2=[p_0(1-p_0)+p_1(1-p_1)]/n$. The normal approximation to a two-sided level-$\alpha$ test has powerFor rare events, independent Poisson distributions approximate the counts. Conditional on their total, the null count allocation is binomial distribution with probability one half; under the alternative the allocation probability is $p_0/(p_0+p_1)$. Averaging its exact rejection probability over the alternative total supplies a discrete power calculation.

// Target: probability-theory.bigb

Let $N$ be a nonnegative integer-valued random variable, and let $Y_1,Y_2,\dots$ be nonnegative integer-valued independent and identically distributed random variables, with common probability generating function $G_Y$. Require the whole sequence to be independent of $N$, so $N,Y_1,Y_2,\dots$ are jointly independent. Set the random sumwith the empty sum equal to zero. Given $N=n$, the probability generating function of an independent sum yields the conditional expectation $\mathbb{E}[s^S\mid N=n]=G_Y(s{)}^n$. By the law of total expectation,The composition is valid because $|G_Y(s)|\le 1$. Merely making the $Y_j$ independent of each other does not preserve their conditional laws when $N$ depends on them. This is the probability generating function version of transform composition for an independent random sum.

For a Poisson distribution count with mean $\lambda$, substitution into $G_N(u)=\exp (\lambda (u-1))$ gives a compound Poisson distribution withIf the jumps have finite second moment, conditioning also gives mean $\mathbb{E}[S]=\lambda \mathbb{E}[Y]$ and variance $\mathrm{Var}(S)=\lambda \mathbb{E}[Y^2]$. The extra spread comes from randomness in the count as well as randomness in the jumps.^[ref]

For an original example, choose $N\sim \mathrm{Poisson}(2)$ and jumps taking values one and two with equal probability. Keep the independence assumptions above. ThenHere $\mathbb{E}[Y]=3/2$, $\mathrm{Var}(Y)=1/4$, and $\mathbb{E}[Y^2]=5/2$. The law of total expectation and law of total variance giveThus the aggregate is not itself a Poisson distribution, whose mean and variance would coincide.

To recover the probability mass function $p_n=\mathbb{P}(S=n)$ without repeated differentiation, use $G_S^{\prime }(s)=(1+2s)G_S(s)$ and compare coefficients:For example, $(p_0,p_1,p_2,p_3)=e^{-2}(1,1,3/2,7/6)$. As an independent check, expanding $e^{s+s^2-2}=e^{-2}{e}^s{e}^{s^2}$ gives

The plotted probability masses stop at 20, although the support of a probability distribution is unbounded. The omitted tail probability isThis model has bounded jumps and an unbounded count. Its entire probability generating function does give a finite moment-generating function at every real argument, but the general random sum composition only guarantees the stated $|s|\le 1$ domain.

An invariant distribution of a continuous-time Markov chain whose coordinates are independent Poisson distributions. This property needs verification from the reaction propensity functions; it does not follow merely from having an equilibrium for the deterministic reaction-rate equations.

For example, consider the cyclic chemical reaction network with stoichiometric vectors $(1,0)$, $(0,1)$ and $(-1,-1)$ and respective reaction propensity functions $\lambda$, $\beta n$, $\gamma nm$, where all three constants are positive. Put $a=\lambda /\beta$ and $b=\beta /\gamma$. For $\pi (n,m)=e^{-a-b}{a}^n{b}^m/(n!m!)$, the three incoming terms in the chemical master equation are respectively $\beta n\pi (n,m)$, $\gamma nm\pi (n,m)$ and $\lambda \pi (n,m)$. They exactly cancel the outgoing terms. Boundary terms vanish by the zero negative-index convention, and normalization follows from the two Poisson distributions. Establishing a limiting probability distribution additionally requires nonexplosion, an irreducible Markov chain and positive recurrence.

A Poisson distribution with mean $E\lambda$ and positive known exposure has Fisher information $E/\lambda$, so its Jeffreys prior kernel is ${\lambda }^{-1/2}$. The proper shape-rate gamma distribution $\mathrm{Gamma}(1/2,\epsilon )$ has kernel ${\lambda }^{-1/2}{e}^{-\epsilon \lambda }$, approximating the improper kernel where $\epsilon \lambda$ is small. Its posterior is $\mathrm{Gamma}(y+1/2,E+\epsilon )$, converging to $\mathrm{Gamma}(y+1/2,E)$. There is no normalized Jeffreys prior to which the proper priors converge weakly; this is kernel approximation and posterior convergence.

When the random intensity is $\alpha +Y$ with $Y$ having an exponential distribution of rate $\beta$, the mixed count is the sum of independent Poisson distribution and support-zero geometric distribution counts. The Poisson parameter is $\alpha$ and the geometric success probability is $\beta /(\beta +1)$. At $\alpha =0$ the law is geometric.

Shot noise is fluctuation associated with discrete transfer events. For an electric current, individual charge carriers contribute finite amounts of electric charge, even when a meter displays a smooth mean. The simplest model is a stream of equal, independently arriving charges with constant rate: its Poisson process predicts both the mean electric current and the variance around it. Discreteness alone does not imply a Poisson distribution; the timing and correlations matter.

The useful surprise is that noise can reveal something a mean hides. Two devices can carry the same mean electric current yet have different fluctuations because one transfers charges independently and the other groups or regulates them. A calibrated shot noise photocurrent measurement can connect a macroscopic voltage variance to the elementary charge. Thermal electrical noise instead exists in an unbiased resistor at thermal equilibrium; a measured background can contain both mechanisms.

Walter Schottky's 1918 paper Über spontane Stromschwankungen in verschiedenen Elektrizitätsleitern treated amplifier sensitivity limits. Part II considered independent discrete transfers, including thermionic emission through a vacuum. On p. 563 he proposed amplifying fluctuations from a separate discharge tube to determine the elementary charge; those experiments were prospective, not reported results.^[ref]

For an original model, set $q=1.602176634\times 10^{-19}\mathrm{C}$, $I=1\mu \mathrm{A}$, $R_F=100\mathrm{k}\Omega$, $G=10$, and the exponential low-pass filter's $\tau =100\mu \mathrm{s}$. Assume an ideal detector without charge multiplication and omit all additive noise for this calculation. ThenDoubling $I$ doubles the voltage variance and multiplies its standard deviation by $\sqrt{2}$. Halving $\tau$ also doubles the variance, provided the detector remains approximately white throughout the enlarged passband. Using $f_c$ in place of $B_{\mathrm{n}}$ underestimates the variance by the factor $2/\pi$. These are calculated predictions, not measured values or settings claimed for the laboratory below.

To read the dimensionless figure, set $a=\lambda \tau$ and $x=\tau i_{\mathrm{filtered}}/q$, with frequency $\nu =f\tau$ in cycles per unit $t/\tau$. A gate of duration $\tau$ has count mean and variance $a$. The continuously filtered $x$ instead hasThe count and the filtered electric current are different observables. The factor $1/2$ comes from the exponential weighting, not from non-Poisson arrivals. The dimensionless count illustration is a separate choice of rate; it does not assert that the $1\mu \mathrm{A}$ example has only four electrons per $100\mu \mathrm{s}$.

For source counts $S$ and independent background counts $B$ with Poisson distributions, the ideal counting standard deviation is $\sqrt{S+B}$, so the signal-to-noise ratio is $S/\sqrt{S+B}$. Detector noise and uncertainty in the estimated background add further variance. A larger sky aperture raises $B$, while better throughput raises $S$; both affect the faintness limit of a spectrograph.