Best Unbiased Estimators for Poisson Probabilities

Suppose we observe i.i.d. samples X₁, ..., Xₙ from a Poisson(λ) distribution. In this tutorial, we explore how to construct the best unbiased estimators (BUEs) for two key probabilities:

• e^(−λ): the probability that a Poisson variable equals 0
• λe^(−λ): the probability that a Poisson variable equals 1

When graduate homework solutions don’t match the rubric, the issue is often interpretation rather than computation. Online tutoring can help you align definitions, steps, and conclusions with graduate-level expectations.

This article presents the method in a graduate-level, definition-first way. It emphasizes assumptions, step-by-step reasoning, and common misunderstandings that cause errors. The discussion is designed to be accurate, self-contained, and useful for assignments, exams, and applied projects.

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(a) Best Unbiased Estimator of e^(−λ)

Define the sample proportion of zeros:

T = (1/n) ∑ I(Xᵢ = 0)

Since P(Xᵢ = 0) = e^(−λ) and I(Xᵢ = 0) is an unbiased indicator, we have:

E[T] = E[(1/n) ∑ I(Xᵢ = 0)] = (1/n) ∑ E[I(Xᵢ = 0)] = e^(−λ)

Thus, T is an unbiased estimator for e^(−λ). It is also the uniformly minimum variance unbiased estimator (UMVUE) by Lehmann–Scheffé, since it is a function of the sufficient statistic ∑Xᵢ.

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(b) Best Unbiased Estimator of λe^(−λ)

Similarly, let the sample proportion of ones be:

S = (1/n) ∑ I(Xᵢ = 1)

Since P(Xᵢ = 1) = λe^(−λ), we have:

E[S] = λe^(−λ)

So S is the BUE for λe^(−λ), and again qualifies as the UMVUE.

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(c) Asymptotic Relative Efficiency vs MLE

The maximum likelihood estimator (MLE) of λ is the sample mean:

λ̂ = x̄ = (1/n) ∑ Xᵢ

Plugging into the parametric forms gives:

MLE for e^(−λ): e^(−x̄)
MLE for λe^(−λ): x̄ · e^(−x̄)

These MLEs are biased but consistent. To compare efficiency, we examine asymptotic variances:

• Var(BUE for e^(−λ)) ≈ (1/n) e^(−λ)(1 − e^(−λ))
• Var(MLE for e^(−λ)) via delta method ≈ (1/n) λ e^(−2λ)

Relative efficiency depends on λ:

• For small λ, BUEs outperform MLEs.
• For large λ, MLEs converge faster.

Importantly, BUEs remain unbiased at all sample sizes, while MLEs provide smooth estimates even in sparse samples.

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(d) Example: Estimating from Mutation Data

Given counts from 15 petri dishes:

10, 7, 8, 13, 8, 9, 5, 7, 6, 8, 3, 6, 6, 3, 5

Let n = 15. Count zeros and ones:

Count of 0s = 0 ⇒ BUE for e^(−λ) = 0
Count of 1s = 0 ⇒ BUE for λe^(−λ) = 0

Sample mean:

x̄ = (∑Xᵢ)/15 = 104/15 ≈ 6.933

MLE estimates:

e^(−x̄) ≈ e^(−6.933) ≈ 0.00097
x̄ · e^(−x̄) ≈ 6.933 × 0.00097 ≈ 0.0067

If you’re preparing for a graduate exam and the method still feels mechanical, it usually means the assumptions haven’t been made explicit. Online tutoring can help you build intuition alongside the steps you need for problem-solving.

This post explains the topic at a graduate level, focusing on definitions, assumptions, and step-by-step reasoning. The goal is to clarify when the method applies, why it works, and how to avoid common errors. The discussion is written to be accurate, self-contained, and appropriate for coursework and exams.

Conclusion: The BUEs are zero due to no observed zeros or ones, while MLEs provide small positive estimates. In small samples, BUEs may be sparse, whereas MLEs offer smooth extrapolation.

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