UMVU Estimation in Poisson Models

Why Condition on the Poisson Total?

Suppose a process produces counts over several comparable periods, and each period is modeled by the same Poisson rate. The sample mean estimates that rate, but other targets—such as a power of the rate or the probability of observing a specified count—need different unbiased starting estimators. In each case, conditioning on the sample total provides a systematic route to the uniformly minimum-variance unbiased estimator.

The key feature is that, once the total count is known, its allocation among equal-exposure observations has a simple multinomial distribution. That conditional distribution removes the unknown rate from the calculation.

Model and notation: Let \(X_1,\ldots,X_n\) be independent and identically distributed, with \(X_i\sim\operatorname{Poisson}(\lambda)\), where \(n\) is a known positive integer and \(\lambda>0\) is the common expected count per observation. Define \(T=\sum_{i=1}^n X_i\). The equal-rate assumption means that the observations have equal exposures; the results below do not directly apply to independent counts with unequal means.

The Complete Sufficient Total and Its Conditional Distribution

The joint probability mass function is

P(X_1=x_1,\ldots,X_n=x_n) =e^{-n\lambda}\frac{\lambda^{x_1+\cdots+x_n}}{x_1!\cdots x_n!}.

The data enter the part involving \(\lambda\) through \(T\), so the factorization theorem gives sufficiency of \(T\). Also, \(T\sim\operatorname{Poisson}(n\lambda)\). This statistic is complete: if \(E_\lambda[g(T)]=0\) for every \(\lambda>0\), with the expectations defined, then

e^{-n\lambda}\sum_{t=0}^{\infty}g(t)\frac{(n\lambda)^t}{t!}=0 \quad\text{for every }\lambda>0,

and uniqueness of the coefficients of this power series implies \(g(t)=0\) for every nonnegative integer \(t\). Thus \(T\) is complete sufficient for \(\lambda\).

Given \(T=t\), the factors involving \(\lambda\) cancel from the joint probability. The conditional allocation of the \(t\) counts is multinomial:

P(X_1=x_1,\ldots,X_n=x_n\mid T=t) =\frac{t!}{x_1!\cdots x_n!}\left(\frac{1}{n}\right)^t, \qquad \sum_{i=1}^n x_i=t.

In particular, \(X_1\mid T=t\sim\operatorname{Binomial}(t,1/n)\). This is the important conditional law: the probability that a count lands in the first observation is \(1/n\), not a quantity involving \(\lambda\).

Theorem: If \(U\) is an unbiased estimator of a target \(\tau(\lambda)\), then \(E[U\mid T]\) is unbiased for \(\tau(\lambda)\). When this conditional expectation is a function of the complete sufficient statistic \(T\), it is the unique UMVU estimator of that target, by the Lehmann–Scheffé theorem.

Estimating the Rate and Its Integer Powers

The sample mean \(\bar X=T/n\) is unbiased for \(\lambda\), since \(E[T]=n\lambda\). It is already a function of the complete sufficient statistic, so it is UMVU. For powers of the rate, use falling factorials rather than ordinary powers. For a nonnegative integer \(r\), define \((x)_r=x(x-1)\cdots(x-r+1)\), with \((x)_0=1\).

For \(r\geq 1\), an individual Poisson observation satisfies \(E[(X_1)_r]=\lambda^r\). Conditioning this unbiased starting estimator on \(T\) gives the general result.

\widehat{\lambda^r}_{\mathrm{UMVU}} =E[(X_1)_r\mid T] =\frac{(T)_r}{n^r}, \qquad r=0,1,2,\ldots.

To justify the conditional factorial-moment step, if \(B\sim\operatorname{Binomial}(t,p)\), then

$$ E[(B)_r]=(t)_r p^r. $$

One way to see this is to write \((B)_r\) as the number of ordered selections of \(r\) distinct successes among \(t\) Bernoulli trials. There are \((t)_r\) ordered selections, and each has probability \(p^r\) of being all successes. Here \(p=1/n\), which gives the displayed UMVU formula.

When \(T<r\), \((T)_r=0\), as it should: there are not enough observed counts in the total to form \(r\) distinct successes. The estimator remains unbiased across repeated samples, even though it can be zero for a particular sample.

Worked Example: Estimating the Rate from Five Periods

Model and method. Five equal-exposure periods have independent counts \(X_i\sim\operatorname{Poisson}(\lambda)\). The observed counts are \(0,2,1,4,3\). Find the UMVU estimate of \(\lambda\), and explain why conditioning an estimate based on the first period leads to the same result.

Step 1: Calculate the total.

T=0+2+1+4+3=10,\qquad n=5.

Step 2: Use the conditional expectation. The starting estimator \(X_1\) is unbiased because \(E[X_1]=\lambda\). Given the total, \(X_1\mid T=10\sim\operatorname{Binomial}(10,1/5)\), so

E[X_1\mid T=10]=10\left(\frac15\right)=2.

Step 3: State the UMVU estimate. The conditional estimator is \(T/n\), so the estimate is \(10/5=2\) counts per period. It is UMVU because it is unbiased and is a function of the complete sufficient statistic \(T\).

Check and mistake avoided. The observed first count is \(X_1=0\), but using that value alone discards information in the other four periods. Conditioning on the total averages over the possible allocations of those ten counts, giving \(2\), not \(0\).

Worked Example: Estimating the Square of the Rate

Model and method. Three equal-exposure counts are independent \(\operatorname{Poisson}(\lambda)\) variables. The observations are \(1,3,4\). Find the UMVU estimate of \(\lambda^2\), starting from an unbiased estimator based on \(X_1\).

Step 1: Select an unbiased starting estimator. For a Poisson variable, \(E[X_1(X_1-1)]=\lambda^2\). Thus \(X_1(X_1-1)\) is unbiased for the target.

Step 2: Find the total and condition.

T=1+3+4=8,\qquad n=3,\qquad X_1\mid T=8\sim\operatorname{Binomial}\left(8,\frac13\right).

For a binomial variable, the conditional second falling factorial moment is

E[X_1(X_1-1)\mid T=8] =(8)_2\left(\frac13\right)^2 =(8)(7)\left(\frac19\right) =\frac{56}{9}.

Step 3: Report and check the estimate. The UMVU estimate is \(56/9\), or \(6.2222\) to four decimal places. Its unbiasedness follows because taking expectation of the conditional expectation returns \(E[X_1(X_1-1)]=\lambda^2\).

Common mistake avoided. Squaring the sample mean does not give an unbiased estimator: \(E[(T/n)^2]=\lambda^2+\lambda/n\). The falling-factorial construction removes this extra term before conditioning.

Estimating a Poisson Point Probability

For a fixed nonnegative integer \(k\), the target probability for one observation is

q_k(\lambda)=P(X_1=k)=e^{-\lambda}\frac{\lambda^k}{k!}.

The indicator \(I(X_1=k)\), equal to 1 when \(X_1=k\) and 0 otherwise, is an unbiased starting estimator of \(q_k(\lambda)\). Conditional on \(T=t\), its expectation is the conditional probability that \(X_1=k\). Therefore the UMVU estimator is

\widehat{q_k}_{\mathrm{UMVU}} =P(X_1=k\mid T) = \begin{cases} \displaystyle \binom{T}{k}\left(\frac1n\right)^k \left(1-\frac1n\right)^{T-k},&T\geq k,\\[6pt] 0,&T<k. \end{cases}

The estimator is unbiased by iterated expectation:

E_\lambda[\widehat{q_k}_{\mathrm{UMVU}}] =E_\lambda[E[I(X_1=k)\mid T]] =E_\lambda[I(X_1=k)] =q_k(\lambda).

It is a function of \(T\), and \(T\) is complete sufficient, so it is UMVU. The result estimates the probability for one observation from the common Poisson model; it is not a probability statement about the total \(T\).

Worked Example: Estimating the Probability of Exactly Two Events

Model and method. Four equal-exposure monitoring periods have independent \(\operatorname{Poisson}(\lambda)\) counts, and the observed counts are \(0,1,2,3\). Estimate \(P(X_1=2)\) by conditioning the unbiased indicator \(I(X_1=2)\) on the total.

Step 1: Calculate the total and conditional model.

T=0+1+2+3=6,\qquad X_1\mid T=6\sim\operatorname{Binomial}\left(6,\frac14\right).

Step 2: Evaluate the conditional probability.

\begin{aligned} P(X_1=2\mid T=6) &=\binom62\left(\frac14\right)^2 \left(\frac34\right)^4\\ &=15\cdot\frac1{16}\cdot\frac{81}{256}\\ &=\frac{1215}{4096}\\ &=0.296630859375. \end{aligned}

The UMVU estimate is \(0.2966\) to four decimal places. The conditioning argument also verifies the estimator’s unbiasedness over repeated samples.

Common mistake avoided. Substituting the sample mean \(T/n=1.5\) into the Poisson point-probability formula gives \(e^{-1.5}(1.5)^2/2\approx0.2510\). That plug-in calculation is not the conditional expectation of the unbiased indicator. The correct conditional law is binomial with success probability \(1/4\), giving \(0.2966\).

Choosing and Applying the Estimator

The target determines the unbiased starting estimator, while the common conditional distribution does the Rao–Blackwellization. The resulting estimator depends on the data only through \(T\).

Target	Unbiased starting estimator	UMVU estimator
\(\lambda\)	\(X_1\)	\(T/n\)
\(\lambda^r\), integer \(r\geq 0\)	\((X_1)_r\)	\((T)_r/n^r\)
\(P(X_1=k)\), integer \(k\geq 0\)	\(I(X_1=k)\)	\(\binom{T}{k}(1/n)^k(1-1/n)^{T-k}\) if \(T\geq k\), and \(0\) otherwise

The point-probability formula also handles boundary cases. If \(n=1\), then \(T=X_1\), and the conditional estimate is the indicator \(I(T=k)\). If \(T<k\), the estimate is zero because the conditional binomial variable cannot exceed its total number of trials.

Check the exposure assumption. The probability \(1/n\) comes from equal Poisson means. With unequal exposures, the conditional allocation probabilities are proportional to those exposures rather than all being \(1/n\).
Match the estimator to the target. A point probability is estimated by conditioning an indicator, not by inserting an estimated rate into a Poisson probability formula.
Use falling factorials for powers. The estimator \((T)_r/n^r\) is the unbiased conditional form for \(\lambda^r\); ordinary powers of the sample mean generally include extra terms.

Key takeaway: In an independent equal-rate Poisson sample, condition an unbiased starting estimator on \(T=\sum X_i\). The conditional allocation is binomial for one observation, and completeness of \(T\) makes the resulting unbiased function the unique UMVU estimator.

Check Your Understanding

For each problem, assume independent Poisson counts with the same rate and equal exposure. Use the total as the conditioning statistic.

Four periods have total count \(T=12\). What is the UMVU estimate of \(\lambda\)? (Answer: \(12/4=3\).)
Five periods have total count \(T=7\). Find the UMVU estimate of \(\lambda^2\). (Answer: \((7)_2/5^2=42/25=1.68\).)
Three periods have total count \(T=5\). Estimate \(P(X_1=1)\) and round to four decimal places. (Answer: \(\binom51(1/3)(2/3)^4=80/243\approx0.3292\).)
Two periods have total count \(T=0\). Estimate \(P(X_1=0)\). (Answer: \(\binom00(1/2)^0(1/2)^0=1\).)
Four periods have total count \(T=3\). Find the UMVU estimate of \(\lambda^3\). (Answer: \((3)_3/4^3=6/64=3/32=0.09375\).)

What You'll Learn