ISS 2016 Statistics Paper-2 Solution: Question 21 (UMVUE via Complete Sufficient Statistic)

Continuing our question-by-question walk through the ISS 2016 Statistics Paper-2 previous year paper, today we tackle Question 21 — a favourite examiner's trick that quietly tests three different ideas about estimation theory in a single MCQ. If you've ever felt shaky on the difference between a sufficient statistic, a complete statistic, and a UMVUE, this question is exactly the kind of practice that clears it up for good.

Quick Summary

• Topic:Complete sufficient statistics and UMVUE for the Poisson distribution

• Question Reference:ISS 2016, Statistics Paper-2, Question 21

• Correct Answer:Option (d) — 1, 2 and 3

The Question As It Appeared in the Paper

Let X1, X2, X3, ..., Xn be i.i.d. P(λ) variables. Which of the following statements are correct?

1. T = Σi=1n Xi is a complete sufficient statistic for λ 2. X̄ = T/n is the unique UMVUE for λ 3. E(X1 | T) is the unique UMVUE for λ

Select the correct answer using the code given below:

(a) 1 and 2 only (b) 2 and 3 only (c) 1 and 3 only (d) 1, 2 and 3

Step 1: Set Up the Poisson Model Carefully

We are told X1, X2, ..., Xn are independent and identically distributed Poisson(λ) random variables. This means each Xi takes values 0, 1, 2, 3, ... with probability mass function

P(Xi = x) = e−λ λx / x! , for x = 0, 1, 2, ...

Since the observations are independent, the joint probability (the likelihood) of observing X1 = x1, X2 = x2, ..., Xn = xn is simply the product of the individual probabilities:

L(λ) = ∏i=1n [e−λ λxi / xi!] = e−nλ λΣxi / (x1! x2! ... xn!)

Notice something important here: the whole expression depends on the data only through the sum Σxi. That single fact is the seed from which every part of this question grows.

Step 2: Show T = ΣXi Is Sufficient (Factorization Theorem)

The Factorization Theorem says a statistic T is sufficient for λ if the likelihood can be written as

L(λ) = g(T, λ) × h(x1, x2, ..., xn)

where g depends on the data only through T, and h does not depend on λ at all. From Step 1, we already have exactly this split:

L(λ) = [e−nλ λT] × [1 / (x1! x2! ... xn!)]

Here g(T, λ) = e−nλ λT depends on the sample only through T = Σxi, and h(x1,...,xn) = 1/(x1! x2! ... xn!) has no λ in it. So by the Factorization Theorem,T = ΣXi is a sufficient statistic for λ. That settles half of Statement 1 — now we need completeness.

Step 3: Show T Is Also Complete

First, recall a standard fact: the sum of n independent Poisson(λ) variables is itself Poisson(nλ). So T ~ Poisson(nλ).

A statistic T is called complete if, whenever a function g(T) satisfies Eλ[g(T)] = 0 for every possible value of λ > 0, it forces g(t) = 0 for (almost) every value t that T can take. In plain words: no non-trivial function of T can have zero expectation for every λ — if it does, that function must itself be identically zero.

Let's verify this directly for T ~ Poisson(nλ). Suppose

Eλ[g(T)] = Σt=0∞ g(t) · e−nλ (nλ)t / t! = 0 for all λ > 0

Since e−nλ is never zero, we can divide it out, leaving

Σt=0∞ [g(t) nt / t!] λt = 0 for all λ > 0

The left-hand side is a power series in λ. A power series that equals zero for every value of λ must have every one of its coefficients equal to zero (this is a basic uniqueness property of power series — two power series that agree everywhere must agree term by term). So we need

g(t) nt / t! = 0 for every t = 0, 1, 2, 3, ...

Since nt/t! is never zero, this forces g(t) = 0 for every t. That is exactly the definition of completeness being satisfied, so T is a complete statistic for λ.

Combining Steps 2 and 3: T is both sufficient and complete, so Statement 1 is correct.

Pro Tip: Most students can quote "T is sufficient" from the factorization theorem without much trouble, but completeness is where marks are lost — largely because textbooks state it as a fact for Poisson/Binomial/Normal families without showing the power-series argument. The way this derivation is usually taught in a structured classroom is to make you write out the power series step explicitly at least once by hand, so that "complete statistic" stops being a memorized label and becomes something you can actually verify from scratch under exam pressure — that habit alone is worth several marks a year in ISS Paper-2.

Step 4: Find the UMVUE Using the Lehmann–Scheffé Theorem

The Lehmann–Scheffé theorem is one of the most useful results in estimation theory. It says: if T is a complete sufficient statistic for a parameter, and φ(T) is any function of T that is an unbiased estimator of that parameter, then φ(T) is the unique Uniformly Minimum Variance Unbiased Estimator (UMVUE) of that parameter.

We already know T = ΣXi is complete and sufficient (Steps 2–3). Now check X̄ = T/n:

E(X̄) = E(T/n) = E(T)/n = (nλ)/n = λ

So X̄ is unbiased for λ, and it is a function of the complete sufficient statistic T. By Lehmann–Scheffé,X̄ = T/n is the unique UMVUE of λ. Statement 2 is correct.

Step 5: Evaluate E(X1 | T) Using a Symmetry Argument

This is the step that trips up most students, because at first glance E(X1 | T) looks like it needs a fresh, separate derivation. It doesn't — a short symmetry argument gets us there quickly.

Since X1, X2, ..., Xn are i.i.d., none of them is "special" relative to the others. This means, given that the total T = ΣXi is fixed, every individual Xi has exactly the same conditional expectation. In symbols:

E(X1 | T) = E(X2 | T) = ... = E(Xn | T)

Now sum both sides over i = 1 to n. On the left, summing n identical copies of E(X1 | T) gives n · E(X1 | T). On the right, we get

Σi=1n E(Xi | T) = E(Σ Xi | T) = E(T | T) = T

(because Σ Xi is just T itself, and the expectation of T given T is T). So we have

n · E(X1 | T) = T, which gives E(X1 | T) = T/n = X̄

So E(X1 | T) is literally the same random variable as X̄. Since we already proved in Step 4 that X̄ is the UMVUE of λ, it immediately follows that E(X1 | T) is also the UMVUE of λ— they are not two different estimators, they are the same estimator written two different ways. Statement 3 is correct.

Final Answer

All three statements — 1, 2, and 3 — are correct. This matches option (d) 1, 2 and 3, which is exactly what the official answer key gives for Question 21 (Series A) of ISS 2016 Statistics Paper-2. Our derivation and the official key are in full agreement.

Why This Question Matters

This question is a compact tour of three pillars of point estimation theory: sufficiency, completeness, and the Rao–Blackwell/Lehmann–Scheffé machinery for finding UMVUEs. In the ISS syllabus these ideas resurface constantly — in questions about exponential families, uniform distributions, normal distributions, and beyond — so a candidate who can reproduce this exact style of argument (factorization theorem, power-series completeness check, symmetry trick for conditional expectations) is well prepared for a large share of the estimation theory block, not just this one item. Students who drill this kind of three-step reasoning until it becomes automatic tend to solve similar-looking MCQs in under a minute during the actual exam, which matters a great deal given the strict two-hour time limit for 80 questions.

Frequently Asked Questions

What exactly is the difference between a sufficient statistic and a complete statistic?

A sufficient statistic captures all the information in the sample about the parameter — once you know T, the rest of the data tells you nothing more about λ. Completeness is a separate, stronger property: it guarantees that no non-zero function of T can have zero expectation for every parameter value. You need both properties together before you can invoke the Lehmann–Scheffé theorem to identify a UMVUE.

Why is the sum of independent Poisson variables also Poisson?

This follows from moment generating functions. The MGF of a single Poisson(λ) variable is exp(λ(es − 1)). Since the Xi are independent, the MGF of their sum is the product of the individual MGFs, which works out to exp(nλ(es − 1)) — exactly the MGF of a Poisson(nλ) variable. Because the MGF uniquely determines the distribution, T must be Poisson(nλ).

Can I always assume X̄ is the UMVUE whenever it's unbiased?

No — unbiasedness alone is not enough. X̄ is guaranteed to be the UMVUE only when it is also a function of a complete sufficient statistic for that parameter, which is exactly what the Lehmann–Scheffé theorem requires. For distributions where the sufficient statistic isn't complete, unbiased estimators are not unique, and X̄ may not have the minimum variance among them.

Is there a quicker way to see that E(X1 | T) = T/n without the summation argument?

Yes — you can also argue it through the conditional distribution directly. Given T = t, the joint distribution of (X1, ..., Xn) is Multinomial(t; 1/n, 1/n, ..., 1/n), because each of the t "successes" is equally likely to have come from any of the n Poisson variables. The mean of a single multinomial category with t trials and probability 1/n is t/n, giving the same answer, E(X1 | T) = T/n.

Does this same logic apply to other exponential family distributions like Binomial or Exponential?

Yes, this is a very general pattern. Any distribution belonging to a one-parameter exponential family (Poisson, Binomial, Exponential, Normal with known variance, and others) has a natural sufficient statistic that also turns out to be complete over an open parameter space, so the same three-step approach — sufficiency, completeness, then Lehmann–Scheffé — applies almost unchanged.

How much weight do estimation theory questions like this carry in ISS Statistics Paper-2?

Estimation theory (sufficiency, completeness, unbiasedness, UMVUE, Cramer-Rao bound, and related ideas) is one of the heaviest-weighted topics across ISS Statistics Paper-2 papers, often accounting for a meaningful chunk of the 80 questions in any given year. Building genuine fluency with theorems like Lehmann–Scheffé, rather than memorizing isolated results, tends to pay off across many questions rather than just one.

If anything in this derivation still feels unclear, drop a comment below and we'll work through it together — and if you found this useful, do share it with a fellow ISS aspirant who might be stuck on the same concept.