ISS 2016 Statistics Paper-2 Solution: Question 42 (Unbiased Estimator for Poisson Mean)

Continuing our question-by-question walk through the ISS Statistics Paper-2 archive, today we pick up right where we left off in the 2016 paper. This one lives in the estimation theory block and is a great example of how a seemingly short question can hide a subtle trap if you rush the algebra.

Quick Summary

• Topic: Point estimation — unbiasedness for the Poisson distribution

• Question reference: ISS 2016, Statistics Paper-2, Question 42

• Correct answer (derived and verified):(b) λ/n — note this differs from the officially printed key, explained below

The Question, Exactly As Asked

X has a Poisson λ distribution. Then (X̄ − S²), where X̄ is the sample mean based on a random sample of size n and S² = Σ(Xi − X̄)² / n, is an unbiased estimator of:

• (a) 0

• (b) λ/n

• (c) λ − λ²

• (d) (1 − 1/n)λ

Step 1: Write Down What We Actually Know About the Poisson Distribution

Let X1, X2, …, Xn be a random sample from a Poisson(λ) distribution. The single most important fact about the Poisson distribution — the one this entire question is built around — is that its mean and variance are both equal to the parameter λ:

E(Xi) = λ and Var(Xi) = λ

We are also told:

X̄ = (1/n) Σ Xi   and   S² = (1/n) Σ (Xi − X̄)²

Notice carefully: this S² divides by n, not (n − 1). That single detail is what the whole question hinges on, so don't skip past it.

Step 2: Find E(X̄)

Since X̄ is just the average of n independent observations each with mean λ, this one is immediate:

E(X̄) = (1/n) Σ E(Xi) = (1/n) · (nλ) = λ

Step 3: Find E(S²) — The Part Everyone Rushes

We need a general result that holds for any iid sample (not just normal data): the expected value of the sum of squared deviations from the sample mean. Start from the identity

Σ (Xi − X̄)² = Σ Xi² − n X̄²

Now take expectations term by term. For any random variable Y, E(Y²) = Var(Y) + [E(Y)]². Apply this twice:

E(Xi²) = Var(Xi) + [E(Xi)]² = λ + λ²

E(X̄²) = Var(X̄) + [E(X̄)]² = λ/n + λ²  (since Var(X̄) = Var(Xi)/n = λ/n)

Putting these into the identity:

E[Σ(Xi − X̄)²] = Σ E(Xi²) − n·E(X̄²) = n(λ + λ²) − n(λ/n + λ²)

E[Σ(Xi − X̄)²] = nλ + nλ² − λ − nλ² = (n − 1)λ

This is the well-known general result E[Σ(Xi − X̄)²] = (n − 1)σ², here with σ² = λ. Dividing by n to get S² itself:

E(S²) = (n − 1)λ / n

Pro Tip: whenever an MCQ gives you a formula involving X̄ and a "sample variance" dividing by

Step 4: Combine the Two Results

Now simply subtract:

E(X̄ − S²) = E(X̄) − E(S²) = λ − (n − 1)λ/n

E(X̄ − S²) = [nλ − (n − 1)λ] / n = λ[n − (n − 1)] / n = λ/n

So X̄ − S² is an unbiased estimator of λ/n, which is option (b).

Step 5: Verify With the n = 1 Sanity Check

This is the fastest cross-check, and it's worth doing even after a full derivation. If n = 1, there is only one observation X1, so X̄ = X1 and every deviation (Xi − X̄) is exactly zero. That forces S² = 0 identically, no matter what λ is. So:

X̄ − S² = X1 − 0 = X1, and E(X1) = λ

Now check which option equals λ when n = 1: option (b), λ/n, gives λ/1 = λ. ✓ Matches exactly. Option (d), (1 − 1/n)λ, gives (1 − 1)λ = 0, which is wrong. Option (c), λ − λ², would require λ − λ² = λ for all λ, i.e. λ² = 0 — false for any λ > 0. Option (a), 0, is also immediately ruled out. This confirms (b) is the only option consistent with even the simplest possible case.

Final Answer and a Note on the Official Key

Our full algebraic derivation gives E(X̄ − S²) = λ/n, which is option (b). We verified this two independent ways: the general expectation algebra in Steps 1–4, and the n = 1 edge-case check in Step 5, which rules out every other option outright. A numerical simulation (repeatedly drawing Poisson samples and averaging X̄ − S² across hundreds of thousands of trials) also converges to λ/n for every sample size tested, confirming the algebra.

The officially published answer key for this paper (Series A) lists the answer to Question 42 as (c) λ − λ². Based on the derivation above — and particularly the n = 1 check, which (c) fails outright since it would require λ² = 0 for every λ — we believe the printed key is in error for this item, and we are publishing our fully-justified derivation with this discrepancy explicitly flagged rather than simply reporting the key's letter.

Why This Question Matters

This question is a compact test of three ideas at once: knowing that Poisson mean equals Poisson variance, knowing the general (distribution-free) formula for E[Σ(Xi − X̄)²], and noticing whether a "variance" formula in a problem divides by n or (n − 1). That last distinction is one of the most commonly tested ideas in estimation theory across competitive statistics exams, because it directly separates students who have memorized a formula from students who understand why the (n − 1) correction exists in the first place. Building the habit of running a quick edge-case check (like n = 1 here) before locking in an MCQ answer is a small habit that pays off repeatedly in a time-limited exam.

Frequently Asked Questions

Why does the Poisson distribution have mean equal to variance?

This is a direct consequence of its moment generating function, which gives E(X) = λ and Var(X) = λ simultaneously for any λ > 0. It's one of the defining fingerprints of the Poisson model, and it's why count data showing mean ≈ variance is often modeled as Poisson in the first place.

What is the difference between S² here (dividing by n) and the usual sample variance (dividing by n − 1)?

Dividing by n gives the maximum likelihood estimator of variance, but it is biased downward by a factor of (n − 1)/n. Dividing by (n − 1) — the Bessel-corrected version — makes the estimator exactly unbiased for the population variance. Many exam questions, including this one, deliberately use the n-denominator version to test whether you notice the difference.

Is X̄ always an unbiased estimator of the Poisson parameter λ?

Yes. Since E(Xi) = λ for every observation, the sample mean X̄ is unbiased for λ regardless of sample size n, which is why it is also the maximum likelihood estimator in the Poisson case.

Could the question have intended S² with an (n − 1) denominator instead?

If S² were defined with (n − 1) in the denominator instead, it would itself be unbiased for λ, making E(X̄ − S²) = λ − λ = 0, which is option (a). This shows how sensitive the correct answer is to that one denominator — always read such definitions in an MCQ character by character before computing.

How should I quickly check an unbiasedness MCQ answer without doing the full algebra?

Plug in the smallest possible sample size, often n = 1 or n = 2, where quantities like S² collapse to something trivial (often zero). Compare the resulting expectation against each option; this usually eliminates two or three wrong answers within seconds, as shown in Step 5 above.

Does this result generalize beyond the Poisson distribution?

The formula E[Σ(Xi − X̄)²] = (n − 1)σ² holds for any distribution with finite variance, not just Poisson. What is specific to the Poisson case here is substituting σ² = λ, because Poisson variance equals its mean.

If any step above feels unclear, or if you've worked through this differently and arrived at another option, drop a comment below — working through these disagreements is often where the real learning happens. And if you found this useful, consider sharing it with a fellow ISS aspirant who's grinding through the same paper.