Poisson Samples and Their Sufficient Statistics

Why the Total Count Can Summarize a Poisson Sample

Suppose several clinics record event counts over observation periods of different lengths. If all clinics share a common event rate, a longer observation period is expected to produce more events. The individual counts therefore need not have the same mean, even though they are governed by one unknown rate. The key question is whether the full vector of counts contains information about that rate beyond its total.

The answer comes from the joint likelihood. With independent Poisson observations and known exposures, the likelihood separates into a part involving the data only through their sum and a part that does not involve the unknown rate. This remains true when the exposures differ.

Definition: Let \(X_1,\ldots,X_n\) be independent, with \(X_i\sim\operatorname{Poisson}(\lambda t_i)\), where the common rate \(\lambda>0\) is unknown and the exposures \(t_i>0\) are known and fixed. Define the total count \(S=\sum_{i=1}^n X_i\) and total exposure \(T=\sum_{i=1}^n t_i\). A statistic \(S\) is sufficient for \(\lambda\) if the conditional distribution of the data given \(S\) does not depend on \(\lambda\). Under the likelihood factorization criterion, it is sufficient if the joint probability mass function can be written as \(g_\lambda(S)h(x_1,\ldots,x_n)\), where \(h\) does not depend on \(\lambda\).

Here the exposure \(t_i\) might be a person-time quantity, such as the number of patient-years observed. The model says that the expected count in observation \(i\) is \(\lambda t_i\). The exposures are treated as fixed known values, and all counts are conditionally independent under the model. We take \(\lambda>0\), so every nonnegative integer count vector has positive probability.

Factor the Joint Likelihood

For observed counts \(x_1,\ldots,x_n\), independence gives the joint probability mass function, which is also the likelihood as a function of \(\lambda\).

L(\lambda;x_1,\ldots,x_n) =\prod_{i=1}^n \frac{e^{-\lambda t_i}(\lambda t_i)^{x_i}}{x_i!} =e^{-\lambda T}\lambda^{S} \prod_{i=1}^n\frac{t_i^{x_i}}{x_i!}.

The first two factors depend on the observations only through \(S\), while the final product contains no \(\lambda\). Consequently, the likelihood factorization criterion applies.

Theorem: If \(X_1,\ldots,X_n\) are independent with \(X_i\sim\operatorname{Poisson}(\lambda t_i)\), for known positive exposures \(t_i\) and common \(\lambda>0\), then \(S=\sum_i X_i\) is sufficient for \(\lambda\).

Proof. Write the joint mass function in the factored form above. In the notation of the factorization criterion, take \(g_\lambda(S)=e^{-\lambda T}\lambda^S\) and \(h(x_1,\ldots,x_n)=\prod_i t_i^{x_i}/x_i!\). The second function is independent of \(\lambda\), and the first depends on the data only through \(S\). The sample space is the nonnegative integer lattice and does not depend on \(\lambda>0\). Thus the criterion establishes that \(S\) is sufficient for \(\lambda\). \(\square\)

The total exposure \(T\) affects the likelihood and the distribution of \(S\), but it is known, not an additional statistic of the observed counts. The sufficient statistic is the total count \(S\), not an exposure-weighted sum such as \(\sum_i t_iX_i\).

Worked Example: Equal Exposures

Four independent observation periods each have exposure \(t_i=1\). The observed counts are \(2,0,5,3\). Find the sufficient statistic and write the likelihood as a function of \(\lambda\).

The total count and exposure are \(S=2+0+5+3=10\) and \(T=1+1+1+1=4\). Substituting into the factorization gives

L(\lambda) =\prod_{i=1}^4 \frac{e^{-\lambda}\lambda^{x_i}}{x_i!} =e^{-4\lambda}\lambda^{10} \frac{1}{2!\,0!\,5!\,3!}.

Since \(0!=1\), the data-only factor is \(1/(2\cdot120\cdot6)=1/1440\). Thus \(L(\lambda)=e^{-4\lambda}\lambda^{10}/1440\). For sufficiency, the important feature is that all dependence on \(\lambda\) uses the counts only through \(10\). The individual pattern of counts appears only in the factor \(1/1440\), which does not depend on \(\lambda\).

Unequal Exposures Do Not Change the Statistic

When observation periods differ, each count has a different mean, \(\lambda t_i\). The factorization still works: multiplying the likelihoods adds the exposures in the exponential term and adds the counts in the power of \(\lambda\). The known exposure values also appear in a data-only factor, but do not change which feature of the observed counts carries the rate dependence.

Worked Example: Different Observation Periods

Three clinics have known exposures \(t_1=0.5\), \(t_2=1.5\), and \(t_3=3\) patient-years, and record counts \(x_1=2\), \(x_2=4\), and \(x_3=7\). Identify the sufficient statistic and display the likelihood factorization.

The totals are \(S=2+4+7=13\) events and \(T=0.5+1.5+3=5\) patient-years. The joint likelihood is

\begin{aligned} L(\lambda) &=\frac{e^{-0.5\lambda}(0.5\lambda)^2}{2!} \frac{e^{-1.5\lambda}(1.5\lambda)^4}{4!} \frac{e^{-3\lambda}(3\lambda)^7}{7!}\\ &=e^{-5\lambda}\lambda^{13} \frac{(0.5)^2(1.5)^4(3)^7}{2!\,4!\,7!}. \end{aligned}

The unknown rate appears only in \(e^{-5\lambda}\lambda^{13}\), with the observations entering through their total \(13\). In particular, multiplying each observed count by its exposure would not be the sufficient statistic indicated by this likelihood. The unequal exposures change the known factor, but not the statistic.

The Conditional Distribution Given the Total

Sufficiency can also be seen by asking how the total is divided among the observations once its value is known. For a fixed total count, the conditional allocation has a multinomial distribution whose probabilities are proportional to the exposures. This gives a direct interpretation: conditional on the number of events overall, a larger exposure receives a larger share on average, but the conditional allocation does not involve the unknown rate.

Theorem: Under the independent Poisson model with known positive exposures \(t_1,\ldots,t_n\), conditional on \(S=s\), the count vector has the multinomial distribution with total \(s\) and cell probabilities \(p_i=t_i/T\). That is, for nonnegative integers \(x_1,\ldots,x_n\) summing to \(s\), \(P(X_1=x_1,\ldots,X_n=x_n\mid S=s)=\frac{s!}{\prod_i x_i!}\prod_i(t_i/T)^{x_i}\).

Proof. Independence gives the joint mass function \(e^{-\lambda T}\lambda^s\prod_i t_i^{x_i}/x_i!\) for any allocation with total \(s\). Summing this expression over all nonnegative allocations with total \(s\), the multinomial theorem yields

\begin{aligned} P(S=s) &=e^{-\lambda T}\lambda^s \sum_{\sum_i x_i=s}\prod_{i=1}^n\frac{t_i^{x_i}}{x_i!}\\ &=e^{-\lambda T}\lambda^s\frac{(t_1+\cdots+t_n)^s}{s!} =e^{-\lambda T}\frac{(\lambda T)^s}{s!}. \end{aligned}

The sum identity follows by expanding \((t_1+\cdots+t_n)^s\): the term for allocation \((x_1,\ldots,x_n)\) has coefficient \(s!/\prod_i x_i!\). Dividing the joint mass by \(P(S=s)\) gives

\begin{aligned} P(X_1=x_1,\ldots,X_n=x_n\mid S=s) &=\frac{e^{-\lambda T}\lambda^s\prod_i t_i^{x_i}/x_i!} {e^{-\lambda T}(\lambda T)^s/s!}\\ &=\frac{s!}{\prod_i x_i!}\prod_{i=1}^n\left(\frac{t_i}{T}\right)^{x_i}. \end{aligned}

The final expression contains no \(\lambda\), proving the stated conditional distribution and confirming sufficiency by the conditional-distribution definition. \(\square\)

Worked Example: Conditional Allocation Across Exposures

Suppose three independent counts have exposures \(1,2,3\), respectively, and their total is known to be \(6\). Find the conditional probability that the observed allocation is \((1,2,3)\), without specifying the common rate.

The total exposure is \(T=1+2+3=6\), so the conditional cell probabilities are \(1/6\), \(2/6=1/3\), and \(3/6=1/2\). The multinomial formula gives

\begin{aligned} P(X_1=1,X_2=2,X_3=3\mid S=6) &=\frac{6!}{1!\,2!\,3!} \left(\frac16\right)^1 \left(\frac13\right)^2 \left(\frac12\right)^3\\ &=60\cdot\frac{1}{6}\cdot\frac{1}{9}\cdot\frac{1}{8} =\frac{60}{432} =\frac{5}{36}. \end{aligned}

The rate does not enter this calculation. Conditional on six events overall, the chance of this particular allocation depends on the known relative exposures, not on how large the common rate is.

Common Mistakes and Why the Result Matters

The factorization shows exactly what the data retain about a common Poisson rate: the total number of events, together with the known total exposure that sets the scale of the likelihood. It also explains why counts from unequal observation periods can be combined for inference about one common rate without pretending that the individual counts have identical means.

Wrong reasoning: “The exposures differ, so the total count cannot be sufficient.” Equal means are not required. The likelihood has rate-dependent factor \(e^{-\lambda T}\lambda^S\) even when the \(t_i\) differ. The correction is to check the factorization for the specified common-rate model.
Wrong reasoning: “The sufficient statistic is \(\sum_i t_iX_i\), because exposure appears in each Poisson mean.” Exposure appears as \(t_i^{x_i}\), a factor independent of \(\lambda\), while the power of \(\lambda\) is \(\sum_i x_i\). The correction is to separate parameter-dependent and parameter-free terms before choosing a statistic.
Wrong reasoning: “The result applies to any collection of Poisson counts.” It assumes a shared rate, known fixed exposures, and independence. If the rates differ across observations, the joint likelihood generally has separate rate-dependent terms, and this argument does not establish that the total is sufficient for all those rates.

In applications, the assumption of a shared rate is substantive: it says that after accounting for exposure, the expected event count per exposure unit is the same across observations. If this assumption is appropriate, the total count is a concise summary for the rate. The conditional allocation result is useful as a check on the same logic: once the total is fixed, the remaining pattern follows the relative exposures and contains no further dependence on the common rate.

Check Your Understanding

Use the common-rate Poisson model and the factorization or conditional distribution to answer these questions.

For independent counts with known exposures \(2,4,5\), what is the sufficient statistic for the common rate, and what is the total exposure?
Explain why unequal Poisson means do not by themselves prevent the total count from being sufficient.
For exposures \(1\) and \(3\), calculate the conditional probability of allocation \((2,1)\) given that the total count is \(3\).
Which assumption fails if each observation has its own unknown rate rather than one common rate?
In the likelihood factorization, which terms depend on the unknown rate, and which terms depend on the known exposures and observed allocation?