advanced29 min studyLesson 961 in Physics

Counting statistics and Poisson likelihoods

Build a count-generating model, infer an efficiency-corrected rate, and propagate independent background fluctuations.

On this page

What you’ll understand

  • Separate event counts, expected counts and physical rates.
  • Keep exposure and efficiency in a Poisson likelihood.
  • Distinguish a background-subtracted estimator from a Poisson observation.

A detector records integers, while the physical quantity of interest may be an activity, flux or cross section. Connecting them requires a probability model and a measurement chain. Counting uncertainty does not arise because an integer was read imprecisely; it describes how a repeated acquisition would fluctuate. We will use hypothetical counts and exactly known acquisition parameters, then introduce a separately measured background. This exposes which uncertainty belongs to the counting process and which would require an additional calibration model.

Write the expectation in detector space

Let r be a nonnegative source rate before detection, epsilon a known detection probability, b a known detected background rate, and T the live acquisition time. Independent thinning of a constant-rate Poisson source gives detected signal rate epsilon times r. Independent Poisson background adds its mean. The observed random count N then has mean mu, which is dimensionless; r and b have inverse-time units. Exposure refers to live time, so unmodeled dead time cannot simply be ignored.

The Poisson assumption requires more than rare individual events. Independent increments and a stable intensity are central. Afterpulsing, pileup, bursty sources, or fluctuating efficiencies can produce correlations or overdispersion. A fixed number of trials with a success probability instead leads to a binomial model. Conditioning independent Poisson bins on their total yields a multinomial shape model, discarding information about the total rate. These are different sampling experiments even when their histograms look similar.

P(N=n∣r)=μ(r)ne−μ(r)n!,μ(r)=T(ϵr+b),E⁡[N]=Var⁡(N)=μ(r)P(N=n\mid r)=\frac{\mu(r)^n e^{-\mu(r)}}{n!},\qquad \mu(r)=T(\epsilon r+b),\qquad \operatorname E[N]=\operatorname{Var}(N)=\mu(r)
Counts and their expectations are dimensionless. The equality of mean and variance concerns repeated Poisson acquisitions with fixed physical parameters.↔ Scroll wide equations horizontally.

A likelihood is a function of a parameter

After observing n, hold it fixed and vary r in the likelihood. The factorial can be omitted when maximizing because it has no parameter dependence. The expected-count normalization term cannot: it penalizes predicting many events when few appeared. Differentiating the log likelihood gives an unconstrained estimate from the observed total rate minus background, divided by efficiency. If this is negative, the maximum over the physical parameter space lies at zero; that boundary changes its sampling behavior.

For an interior solution, the variance of the unconstrained rate estimator follows directly from the Poisson variance through its linear transformation. Substituting n for the unknown mu gives an estimated standard error, not an exact confidence interval. At zero observed counts this plug-in error vanishes even though positive rates remain compatible with the observation. Small-count inference therefore needs an interval construction or explicit posterior rather than a symmetric error bar justified only by a local curvature approximation.

ℓ(r)=nln⁡μ(r)−μ(r)+const,r^unc=n/T−bϵ,Var⁡(r^unc)=μ(r)T2ϵ2\ell(r)=n\ln\mu(r)-\mu(r)+\mathrm{const},\qquad \widehat r_{\mathrm{unc}}=\frac{n/T-b}{\epsilon},\qquad \operatorname{Var}(\widehat r_{\mathrm{unc}})=\frac{\mu(r)}{T^2\epsilon^2}
The variance is exact for the unconstrained linear estimator with known T, epsilon and b. Replacing mu by n estimates that variance; clipping the estimator changes its distribution.↔ Scroll wide equations horizontally.

Subtraction does not remove background noise

Suppose a second acquisition measures background alone. Its count fluctuates independently if it uses a separate Poisson process and the detector conditions are stable. Subtracting a time-scaled background count removes its expectation, while adding its variance to that of the signal-region count. The resulting difference need not be nonnegative and is not itself a Poisson count. Modeling the two original observations jointly preserves their discrete information and allows the common background rate to become a nuisance parameter.

Efficiency uncertainty, exposure uncertainty, and contamination from the source in the control region require further terms. A shared calibration can correlate acquisitions that would otherwise have independent counting noise. Before adding variances, state whether independence concerns events, runs, or calibration information. Inspecting count-versus-time and control-region distributions helps test the assumptions. Increasing exposure reduces independent counting error, but it does not automatically remove a common efficiency bias or repair a wrong background model.

Convert a detected count into a source rate

Observe n = 25 in T = 5 s, with known efficiency 0.5 and known detected background rate b=1 s−1b=1\,\mathrm{s^{-1}}. Find the interior rate estimate and its plug-in counting standard error.

  1. The observed detected rate is 25/5=5 s−125/5=5\,\mathrm{s^{-1}}. Subtract the known background expectation per second to obtain detected signal rate four per second.

  2. Divide by efficiency: r^=(5−1)/0.5=8 s−1\widehat r=(5-1)/0.5=8\,\mathrm{s^{-1}}. This positive value is inside the allowed parameter region.

  3. Use Var⁡^(r^)=25/(52×0.52)=4 s−2\widehat{\operatorname{Var}}(\widehat r)=25/(5^2\times0.5^2)=4\,\mathrm{s^{-2}}. Its square root is two per second; this is an estimated standard error, not an exact interval.

The result

The estimated source rate is 8 s⁻¹ with plug-in counting standard error 2 s⁻¹ under the stated known calibrations.

Explore the model

Interactive lab

Interactive lab · probability

Mean and variance belong to a specified count model

Reset and set BOTH Success probability p = 0.5 and Trials n = 10; predict mean 5 and variance 2.5. For a fresh comparison, Reset and set p = 0.25 and n = 20; predict the same mean 5 but variance 3.75.

Mean and variance belong to a specified count modelThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe lab computes exact binomial probabilities. Equal means need not imply equal variances. It does not implement a Poisson likelihood, background subtraction or random acquisitions.

Check your understanding

What happens to independent counting noise when a measured background count is subtracted?

Try it yourself

Independent on-source and background-only runs each last 2 s with unit efficiency. Their counts are 18 and 8. Find the unconstrained net rate estimate and its plug-in standard error, treating the background rate as measured rather than known exactly.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Subtract counts for the central estimate, but add counts in the estimated variance before dividing by exposure squared.

Compare with the solution

The rate estimate is (18−8)/2=5 s−1(18-8)/2=5\,\mathrm{s^{-1}}. Independence gives estimated variance (18+8)/22=6.5 s−2(18+8)/2^2=6.5\,\mathrm{s^{-2}}, hence standard error 26/2≈2.55 s−1\sqrt{26}/2\approx2.55\,\mathrm{s^{-1}}. The difference is not a Poisson count. A joint two-count likelihood would retain both observations and the unknown background rate; treating eight as an exact background expectation would incorrectly omit its fluctuation.

Take it with you

  • Observed counts and physical rate parameters have different roles and units.
  • Poisson normalization and experimental selection belong in the likelihood.
  • Independent background subtraction adds variance and can produce a non-Poisson difference.

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