Poisson Statistics of Counting

Derivation

Poisson Statistics of Counting

D-116 Home PU-106 Threads chance Depends on The Binomial Distribution from Bernoulli Trials

Statement

Starting from the binomial distribution for N independent Bernoulli trials each of success probability p, we take the limit N → ∞, p → 0 with the mean μ = Np held fixed, and obtain the Poisson distribution P(k) = μ^ke^−μ/k!. Its variance equals its mean, σ² = μ, so a count of mean μ carries an absolute uncertainty √μ and a relative uncertainty 1/√μ — the √N rule of counting statistics.

Why it matters

Almost every experiment that counts discrete, independent events — photons striking a detector, radioactive decays in a time window, cosmic rays through a chamber, dark counts in a photomultiplier — obeys Poisson statistics. The distribution is the universal law for rare events drawn from a large reservoir of independent chances, and it fixes the intrinsic noise floor of any counting measurement before any instrumental imperfection is added.

The single most consequential corollary is the √N rule: to halve the fractional error on a rate you must quadruple the counts. This one fact governs integration times in astronomy, dose budgets in medical imaging, and the number of Monte Carlo samples needed for a target precision. Poisson statistics is the bridge between the microscopic randomness of individual chances and the macroscopic reliability of a measured number.

Assumptions

Trials are independent.If successes correlate (detector dead time, clustered arrivals, bunched photons), the counting variance no longer equals the mean and the distribution becomes super- or sub-Poissonian.

The per-trial probability p is constant across trials.A drifting rate (source decaying appreciably during the window, gain drift) makes the process inhomogeneous; the marginal count is then a mixture and is over-dispersed relative to Poisson.

The mean μ = Np is held finite while N → ∞ and p → 0.If p stays finite the limit is Gaussian (de Moivre–Laplace), not Poisson; if μ → ∞ the Poisson itself tends to a Gaussian of the same mean and variance.

Only the count matters, not which trials succeeded.Poisson describes the total number of events. If the labelling or timing of individual events is retained, the correct object is a Poisson point process; the count is its one-dimensional marginal.

Derivation

P_bin(k) = N!⁄[k!(N−k)!] · p^k(1−p)^N−k

Prior result: the binomial law for k successes in N independent Bernoulli trials. A

p = μ⁄N, μ = Np fixed

Impose the fixed-mean constraint that defines the limit; every p is now eliminated in favour of μ and N. A

P(k) = N!⁄[k!(N−k)!] · (μ⁄N)^k (1−μ⁄N)^N−k

Substitute step 2 into step 1. Purely algebraic. A

P(k) = μ^k⁄k! · N!⁄[(N−k)! N^k] · (1−μ⁄N)^N (1−μ⁄N)^−k

Regroup: pull the k-only factor μ^k/k! to the front and split the exponent N−k = N + (−k). No approximation yet. B

N!⁄[(N−k)! N^k] = N(N−1)…(N−k+1)⁄N^k = ∏_j=0^k−1(1 − j⁄N) → 1

The falling factorial has k factors; each ratio (1 − j/N) → 1 as N → ∞ with k fixed, and a finite product of terms each → 1 tends to 1. C

(1 − μ⁄N)^N → e^−μ

Standard limit: ln[(1−μ/N)^N] = N ln(1−μ/N) = N(−μ/N − μ²/2N² − …) → −μ, so the factor → e^−μ. C

(1 − μ⁄N)^−k → 1

Fixed exponent −k acting on a base → 1. B

P(k) = μ^k e^−μ ⁄ k!

Collect the three limits (steps 5–7: 1, e^−μ, 1) times the front factor of step 4. This is the Poisson probability mass function. A

∑_k=0^∞ P(k) = e^−μ ∑_k=0^∞ μ^k⁄k! = e^−μ e^μ = 1

Normalisation check using the exponential series ∑μ^k/k! = e^μ. Confirms a valid distribution. B

⟨k⟩ = μ, σ² = Var(k) = μ

Take the binomial moments ⟨k⟩ = Np = μ and Var = Np(1−p) = μ(1−p), then let p → 0 so (1−p) → 1. Mean and variance coincide. B

σ = √μ, σ⁄μ = 1⁄√μ

Square-root step 10 for the absolute spread and divide by the mean for the fractional spread. This is the √N rule. A

Result

P(k) = μ^ke^−μ⁄k! , ⟨k⟩ = μ , σ = √μ

Reading. The probability of observing exactly k counts when the expected number is μ is μ^ke^−μ/k!. The distribution is fixed by the single parameter μ; both its centre and its spread grow with μ, but the spread grows only as the square root, so larger counts are proportionally sharper. A raw count N is its own estimate of μ, with uncertainty √N.

Units check. P(k) is a probability, hence dimensionless: μ is a pure number (a count), μ^k and k! are pure numbers, and e^−μ is dimensionless. The mean μ and the standard deviation √μ are both counts, so σ/μ is dimensionless as a fractional uncertainty must be.

Limiting cases

μ → 0: P(0) → 1 and P(k≥1) → 0 — the "no events expected, none seen" limit.
μ ≫ 1 (large): Poisson → Gaussian of mean μ and variance μ; the central limit theorem applied to the sum of many rare sub-intervals. For μ ≳ 20 the Gaussian approximation is usually adequate.
Fixed p, large N (contrast): the limit is not Poisson but Gaussian with variance Np(1−p) < Np — the de Moivre–Laplace regime, where mean and variance differ.
Sum of independent Poissons: μ₁ + μ₂ gives a Poisson of mean μ₁+μ₂ — the distribution is closed under addition (infinite divisibility).

Breaks when

Dead time / saturation. When a detector cannot register events during a recovery window, high-rate counts are systematically lost, the effective p depends on recent history, independence fails, and the variance drops below the mean (sub-Poissonian).
Clustered or correlated arrivals. Photon bunching in thermal light, afterpulsing, or triggered showers make events arrive in groups; the count is then a compound (e.g. Neyman or negative-binomial) distribution with variance exceeding the mean (over-dispersed).
Non-stationary rate. If the source decays or the gain drifts appreciably during the observation, p is not constant; the marginal count is a mixture of Poissons and is over-dispersed relative to any single Poisson.
Finite population, non-negligible p. When p is not small (e.g. a dense target with high interaction probability) the binomial has not reached its Poisson limit; the (1−p) factor in the variance matters and σ² < μ.

Failure modes

Quoting √N on a rate, not a count. The √N rule applies to the raw number of events. If N = 400 counts in t = 10 s, the rate is 40 s⁻¹ with error √400/10 = 2 s⁻¹, not √40.
Adding errors on already-averaged data. √N is the error on a single realisation of the count, not on a mean of counts; students apply it to a per-second rate that has itself been averaged.
Subtracting background then using √(net). For a background-subtracted signal S = N−B, the error is √(N+B), not √S; variances of independent Poissons add even under subtraction.
Assuming σ = √μ when the data are over-dispersed. Applying Poisson errors to clustered or dead-time-affected data understates the true scatter and inflates χ² confidence.
Using the Gaussian approximation at small μ. For μ ≲ 5 a symmetric ±√μ band spills below zero and mis-states confidence intervals; the exact Poisson (or Feldman–Cousins) interval is required.

Discussion

The Poisson limit expresses a deep structural fact: when a large number of independent chances each carry a vanishingly small probability, only the product μ = Np survives into the answer. The individual N and p become invisible — a photon detector does not "know" how many quantum modes were available or how weakly each coupled, only the mean rate they conspire to produce. This is why the Poisson law is universal across wildly different microphysics: radioactive nuclei, incident photons, and clicks on a Geiger counter share the same statistics despite unrelated mechanisms.

The equality of mean and variance, σ² = μ, is the signature that distinguishes Poisson noise from every other kind. It is not an approximation of the counting process; it is an exact identity of the Poisson distribution, inherited from the binomial variance Np(1−p) in the p→0 limit. Because of it, the count itself is an unbiased estimator of its own uncertainty — a fact with no analogue in Gaussian measurement, where the spread must be estimated separately from repeated trials.

The √N rule then dictates the economics of precision. Fractional error 1/√N means diminishing returns are built into nature: the first thousand counts buy a 3% measurement, but reaching 1% demands ten thousand. This shapes experimental design everywhere — exposure times for faint stars, patient dose in a PET scan, and the sample budget of a Monte Carlo integral all trace back to this single scaling.

More abstractly, the Poisson distribution is the one-dimensional marginal of the Poisson point process, the unique simple point process with independent increments and no fixed atoms. Its infinite divisibility — a Poisson of mean μ is the sum of n independent Poissons of mean μ/n for any n — is precisely what allows the "large N, small p" derivation to be run at any granularity of sub-division and always return the same law. The same structure underlies shot noise in electronics (Schottky's formula, spectral density 2eI) and the quantum theory of coherent light, where a coherent state yields exactly Poissonian photon counts and sub-Poissonian statistics become a hallmark of non-classical light.

Common misconceptions. Poisson is not "the distribution of rare events" in the sense that the events must be improbable in absolute terms — μ can be large. What must be rare is each individual trial's success, with the trials numerous; the total count is then whatever μ demands. Nor does Poisson require a time axis: it governs any count of independent events over any domain (area of a photographic plate, volume of a bubble chamber), with μ the expected number in that domain.

Worked examples

A source gives N = 10 000 counts in a fixed window. Estimate the count rate uncertainty (fractional and, over t = 100 s, absolute rate).

The count is Poisson-distributed; use σ = √N. A

σ_N = √N = √10 000 = 100 counts

Apply the √N rule to the raw count. A

σ_N⁄N = 100⁄10 000 = 0.010 = 1.0%

Fractional uncertainty is 1/√N. A

R = N⁄t = 10 000⁄100 s = 100 s⁻¹; σ_R = √N⁄t = 100⁄100 s = 1 s⁻¹

Dividing a count by an exactly-known time t scales both value and error by 1/t. B

R = 100.0 ± 1.0 s⁻¹ (1.0%)

Reading. Ten thousand counts fix the rate to one percent; to reach 0.5% one would need 40 000 counts, i.e. four times the exposure. Units check. counts/s for both R and σ_R; the 1.0% is dimensionless.

In a signal region a detector records N = 2500 total counts; an off-source measurement of equal duration gives B = 1600 background counts. Find the net signal and its uncertainty.

Signal S = N − B; both N and B are independent Poisson counts. A

S = N − B = 2500 − 1600 = 900 counts

Background subtraction, symbols before numbers. A

σ_S² = σ_N² + σ_B² = N + B

Variances of independent quantities add, whether combined by sum or difference; each variance equals its own mean. B

σ_S = √(N+B) = √(2500+1600) = √4100 ≈ 64.0 counts

Square-root the summed variances — note it is not √900 = 30. B

S⁄σ_S = 900⁄64.0 ≈ 14.1

Significance of the excess in standard deviations. A

S = 900 ± 64 counts (14σ excess)

Reading. The net signal is a highly significant 14-sigma detection; the uncertainty is dominated by the total counts N+B, not by the net 900. Using √900 = 30 would overstate significance more than twofold. Units check. counts throughout; the significance 14 is dimensionless.

Problems

A radioactive sample gives a mean of μ = 3 decays per second. What is the probability of observing exactly 0 decays in a given one-second window?
Solution
P(0) = μ⁰e^−μ/0! = e⁻³ = 0.0498, i.e. about 5.0%. (Here μ⁰ = 1 and 0! = 1.)
Cosmic-ray muons pass through a detector at a mean rate giving μ = 4 per minute. Find the probability of exactly 2 in a one-minute interval.
Solution
P(2) = μ²e^−μ/2! = 4²e⁻⁴/2 = 16 × 0.01832 / 2 = 0.1465, i.e. about 14.7%.
You wish to measure a count rate to 0.5% fractional precision. How many counts N must you accumulate, and how does the required exposure compare with a 1.0% measurement?
Solution
Fractional error = 1/√N = 0.005 ⇒ √N = 200 ⇒ N = 40 000 counts. A 1.0% measurement needs 1/√N = 0.010 ⇒ N = 10 000. Since counts accumulate linearly with time at fixed rate, the exposure is 4× longer — halving the error costs four times the time.
Starting from the binomial mean Np and variance Np(1−p), show explicitly that in the Poisson limit the variance-to-mean ratio (Fano factor) equals 1, and state the value of the Fano factor for a binomial with p = 0.2.
Solution
Fano factor F = Var/mean = Np(1−p)/(Np) = 1−p. As p → 0, F → 1 — the Poisson signature. For p = 0.2, F = 1 − 0.2 = 0.8 < 1, i.e. the binomial is sub-Poissonian; its counts are less scattered than a Poisson of the same mean.
A telescope collects 400 photons from a star and 100 photons from the adjacent (equal-area, equal-time) sky background. Report the background-subtracted stellar signal with its uncertainty and significance.
Solution
Signal S = N − B = 400 − 100 = 300 photons. Variance σ_S² = N + B = 500, so σ_S = √500 ≈ 22.4 photons. Result: S = 300 ± 22 photons, a significance S/σ_S = 300/22.4 ≈ 13.4σ. (Using √300 = 17.3 would wrongly give 17σ.)