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PU-106 / L11 · lecture

Discrete Distributions II: Poisson and Counting Statistics

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BEFORE YOU START

You should be comfortable with the binomial distribution and the concept of a probability mass function. Familiarity with basic calculus, specifically limits and factorials, is required to follow the transition from discrete trials to continuous limits.

Why this matters

Counting events is the fundamental act of experimental physics, and the Poisson distribution dictates the precision of every such measurement.

In experiments ranging from particle detector hits to radioactive decay counts, we rarely know the exact number of events in advance. Instead, we observe a random integer \( k \). The Poisson distribution describes the probability of observing \( k \) events when the average rate is known. Crucially, it provides a direct link between the mean count and the uncertainty, establishing the \( \sqrt{N} \) rule that underpins all counting statistics. Without this framework, we cannot distinguish a genuine signal from statistical fluctuation.

The Intuition of Rare Events

Imagine a vast number of independent opportunities for an event to occur, but each opportunity has a very small chance of success. This is the scenario of "rare events." For example, consider a large detector volume where particles are sparse, or a long time interval where radioactive decays are infrequent. If we divide the total time or space into many small bins, each bin has a low probability of containing an event. However, because there are so many bins, the total expected number of events remains finite. The Poisson distribution emerges as the limit of the binomial distribution under these conditions: many trials, low probability per trial, but a fixed average total. It is the natural distribution for counting processes where events occur independently and at a constant average rate.

From Binomial to Poisson

The derivation of the Poisson distribution relies on taking a specific limit of the binomial probability mass function. Recall that for \( N \) independent trials with success probability \( p \), the probability of \( k \) successes is given by the binomial formula. We are interested in the regime where \( N \) is very large and \( p \) is very small, such that their product \( \mu = Np \) remains constant and finite. This quantity \( \mu \) represents the expected number of events. By substituting \( p = \mu/N \) into the binomial expression and taking the limit as \( N \to \infty \), the combinatorial term simplifies significantly. The term \( (1 - p)^{N-k} \) approaches \( e^{-\mu} \), and the factorial terms converge to \( \mu^k / k! \). This limit process transforms the discrete binomial probabilities into the smooth Poisson form, which depends only on the mean \( \mu \) and the observed count \( k \).

\[ P(k) = \frac{\mu^k e^{-\mu}}{k!} \]

This expression is valid for \( k = 0, 1, 2, \dots \). It is important to note that the Poisson distribution is defined for non-negative integers only. It is not a continuous distribution, but a discrete one. The parameter \( \mu \) can be any positive real number, but the outcome \( k \) is always an integer. This distinction is crucial when applying the distribution to experimental data, where counts are inherently discrete.

Mean and Variance

A defining characteristic of the Poisson distribution is that its mean and variance are equal. To see this, we calculate the first and second moments of the distribution. The mean \( \langle k \rangle \) is the expected value of the count. Using the definition of expectation, we sum \( k P(k) \) over all possible values of \( k \). Through algebraic manipulation involving the properties of factorials and the exponential series, this sum evaluates exactly to \( \mu \). Similarly, the variance \( \sigma^2 \), which measures the spread of the distribution around the mean, is calculated as \( \langle k^2 \rangle - \langle k \rangle^2 \). The calculation reveals that \( \sigma^2 = \mu \) as well. This equality is a unique feature of the Poisson distribution and has profound implications for error analysis. It means that the uncertainty in a count is determined solely by the count itself. There is no separate parameter for the spread; the mean dictates both the central value and the width of the probability distribution.

\[ \langle k \rangle = \mu, \quad \sigma^2 = \mu \]

Consequently, the standard deviation, which represents the absolute uncertainty, is \( \sigma = \sqrt{\mu} \). This result is the foundation of counting statistics. It tells us that if we expect to see \( \mu \) events, the typical fluctuation around this mean is of order \( \sqrt{\mu} \). For large \( \mu \), the relative uncertainty \( \sigma / \mu = 1/\sqrt{\mu} \) becomes small, meaning the measurement becomes more precise. For small \( \mu \), the relative uncertainty is large, reflecting the inherent randomness of rare events.

Counting Statistics and the \( \sqrt{N} \) Rule

In experimental physics, we often measure a signal by counting events. Suppose we measure a count \( N \) in a given time interval. According to the Poisson model, the best estimate of the true mean rate is \( N \) itself. However, this measurement carries an uncertainty. The standard deviation of the Poisson distribution tells us that the absolute uncertainty is \( \sqrt{N} \). This is known as the \( \sqrt{N} \) rule of counting statistics. It is a powerful and simple result that allows physicists to assign error bars to count data immediately. For example, if a detector records 100 photons, the uncertainty is \( \sqrt{100} = 10 \) photons. If it records 10,000 photons, the uncertainty is \( \sqrt{10,000} = 100 \) photons. Notice that while the absolute uncertainty grows with the count, the relative uncertainty decreases. This is why high-statistics measurements are more reliable: the signal-to-noise ratio improves as \( \sqrt{N} \).

When combining multiple independent measurements, the variances add. If we have two independent counts \( N_1 \) and \( N_2 \), the total count is \( N_{\text{tot}} = N_1 + N_2 \). The variance of the total is \( \sigma_{\text{tot}}^2 = \sigma_1^2 + \sigma_2^2 = N_1 + N_2 \). Thus, the uncertainty in the total count is \( \sqrt{N_1 + N_2} \). This rule of error propagation is essential for analyzing complex experiments where signals are summed from different channels or time bins. It ensures that the statistical uncertainty is correctly accounted for in the final result.

The derivation

◆ DERIVATION EMBED · D-116 · not re-derived here

Poisson Statistics of Counting

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) = μke−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.

P(k) = μke−μ⁄k! ,   ⟨k⟩ = μ ,   σ = √μ
Open full derivation →
Tier 3 ยท watch out

A common misconception is that the Poisson distribution is an approximation to the normal distribution. While it is true that for large \( \mu \), the Poisson distribution approaches a normal distribution with mean \( \mu \) and variance \( \mu \), this is only valid in the limit \( \mu \gg 1 \). For small \( \mu \) (e.g., \( \mu < 10 \)), the Poisson distribution is significantly skewed and asymmetric. Using a normal approximation in this regime leads to incorrect error bars and invalid statistical inferences. Always use the exact Poisson probabilities for low-count data. C

Worked example

Problem. A Geiger counter is used to measure the activity of a radioactive source. In a 60-second interval, the counter records 400 counts. Assuming the background is negligible and the counts follow Poisson statistics, determine the mean activity and the absolute uncertainty in the count. Also, calculate the relative uncertainty.

  1. Observed count: \( N = 400 \) in \( t = 60\ \mathrm{s} \). For Poisson statistics the mean \( \mu \) is estimated by \( N \).
  2. Absolute uncertainty in the count: \( \sigma = \sqrt{\mu} = \sqrt{400} = 20 \) counts.
  3. Relative uncertainty: \( \sigma/\mu = 20/400 = 0.05 \), i.e. 5%.
  4. Mean count rate (activity as recorded): \( R = N/t = 400/60 \approx 6.67\ \mathrm{s^{-1}} \), with uncertainty \( 20/60 \approx 0.33\ \mathrm{s^{-1}} \) (the same 5%).
\[ N = 400 \pm 20, \quad \sigma/\mu = 0.05, \quad R \approx 6.67 \pm 0.33\ \mathrm{s^{-1}} \]

Units check. The count is a dimensionless integer. The uncertainty is also a dimensionless integer representing the standard deviation of the count. The relative uncertainty is a dimensionless ratio. The result is consistent with the \( \sqrt{N} \) rule.

CHECK YOURSELF

Three quick tests

  1. If the mean count is \( \mu = 25 \), what is the absolute uncertainty?
    AnswerThe absolute uncertainty is \( \sqrt{25} = 5 \).
  2. Why is the variance of a Poisson distribution equal to its mean?
    AnswerThis is a mathematical property derived from the moments of the distribution, reflecting the independence of events and the constant rate assumption.
  3. Can the Poisson distribution be used for continuous variables?
    AnswerNo, it is a discrete distribution defined only for non-negative integers \( k \).
PROBLEMS

Problem set for PU-106 →

Each derivation in this unit ends with five graded problems with full solutions.

NOTES

Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.

Derivations cited: D-116 · Threads: chance