ISS 2016 Statistics Paper-2 Solution: Question 25 (Sufficiency and Completeness for Bernoulli Sums)

Continuing our question-by-question walk through the ISS Statistics Paper-2 archive, today's post picks up right where we left off in the 2016 paper. This one sits in the estimation theory block, and it is a favourite examiner trick because it mixes three separate ideas — sufficiency, completeness, and unbiasedness — into a single multi-statement question.

Quick Summary

• Topic: Sufficiency, completeness and unbiasedness for a sum of i.i.d. Bernoulli variables

• Question Reference: ISS 2016, Statistics Paper-2, Question 25

• Correct Answer: Option (a) — Statements 1 and 2 only

Here is the question exactly as it appeared in the paper:

Q25.If X1, X2, X3, ..., Xn be independent random variables distributed as b(1, p), then which of the following statements are correct?

1. T = Σi=1n Xi is sufficient for p

2. T = Σi=1n Xi is complete for p

3. T2 is unbiased for p2

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: Understand the Setup

b(1, p) is just notation for the Bernoulli distribution — a Binomial distribution with a single trial. So each Xi takes the value 1 with probability p (success) and 0 with probability (1 − p) (failure), and the Xi's are independent of each other. The statistic in question is T = X1 + X2 + ... + Xn, the total number of successes in n independent Bernoulli trials. From basic distribution theory, this sum follows a Binomial distribution: T ~ Binomial(n, p).

Step 2: Check Statement 1 — Is T Sufficient for p?

A statistic T is sufficient for a parameter if, once you know T, the individual observations carry no further information about the parameter. The formal tool for checking this is the Factorization Theorem: T is sufficient for p if and only if the joint probability mass function of the sample can be written as a product of a function depending on the data only through T, and a function that does not involve p at all.

Write out the joint pmf of X1, ..., Xn:

P(X1 = x1, ..., Xn = xn) = ∏i=1n pxi (1 − p)1 − xi

Since exponents add when you multiply powers of the same base, this becomes:

= pΣxi (1 − p)n − Σxi = pt (1 − p)n − t, where t = Σxi

Notice that this entire expression depends on the sample only through t = Σxi, and there is no separate piece left over that depends on the individual xi values once t is fixed. So we can factor it as g(t, p) · h(x1, ..., xn) with h(x1, ..., xn) = 1. By the Factorization Theorem,T = ΣXi is sufficient for p. Statement 1 is TRUE.

Step 3: Check Statement 2 — Is T Complete for p?

Completeness is a stronger and less intuitive property than sufficiency, so let's build it up carefully. A statistic T is complete for p if the only function g(T) satisfying Ep[g(T)] = 0 for every possible value of p is the zero function (g(T) = 0 with probability 1).

Here T ~ Binomial(n, p), so its pmf is P(T = t) = C(n, t) pt (1 − p)n − t for t = 0, 1, ..., n. Suppose E[g(T)] = 0 for all p in (0, 1). Writing this out:

Σt=0n g(t) C(n, t) pt (1 − p)n − t = 0 for all p

Divide through by (1 − p)n (which is never zero for p in (0,1)) and let r = p / (1 − p):

Σt=0n g(t) C(n, t) rt = 0 for all r > 0

The left-hand side is a polynomial in r. A polynomial that equals zero for every value of r in an interval must have every coefficient equal to zero. Since r = p/(1 − p) ranges over all of (0, ∞) as p ranges over (0, 1), this forces g(t) · C(n, t) = 0 for every t = 0, 1, ..., n. Since C(n, t) is never zero, we must have g(t) = 0 for every t. This is exactly the definition of completeness being satisfied. So T is complete for p. Statement 2 is TRUE.

Pro Tip: The polynomial-in-r trick used above is not specific to this question — it is the standard method for proving completeness whenever the sufficient statistic follows a one-parameter exponential family distribution (Binomial, Poisson, Geometric, etc.). The way this argument is usually taught is to memorise the shape of it once — divide out, substitute to turn the parameter into a free variable, then invoke “a polynomial vanishing on an interval has all coefficients zero” — because once that shape is internalised, entire families of completeness questions across different distributions become the same five lines of algebra. Students who drill this particular substitution until it is automatic rarely lose marks on completeness questions again.

Step 4: Check Statement 3 — Is T² Unbiased for p²?

This is the step where most candidates trip up, because it feels natural to assume that if T behaves nicely with respect to p, then T² should behave nicely with respect to p². Let's actually compute E[T²] and see.

We know two standard facts about T ~ Binomial(n, p):

E(T) = np and Var(T) = np(1 − p)

There is a basic identity connecting variance, mean, and the second moment: Var(T) = E(T2) − [E(T)]2. Rearranging this to solve for E(T2):

E(T2) = Var(T) + [E(T)]2

Now substitute the known values:

E(T2) = np(1 − p) + (np)2

Expand np(1 − p) = np − np2, and (np)2 = n2p2:

E(T2) = np − np2 + n2p2 = np + (n2 − n)p2 = np + n(n − 1)p2

For T2 to be an unbiased estimator of p2, we would need E(T2) = p2 to hold for every value of p. But what we actually got is:

E(T2) = np + n(n − 1)p2

This has an extra term np that does not vanish (since n ≥ 1), and even the coefficient of p2 is n(n − 1), not 1, unless n = 1. So E(T2) ≠ p2 in general. Therefore T2 is NOT unbiased for p2. Statement 3 is FALSE.

(As an aside, this is exactly why the correct unbiased estimator of p2 in this setting is actually T(T − 1) / [n(n − 1)], which removes the extra np term — a fact worth remembering since it often reappears in follow-up UMVUE questions.)

Step 5: Combine the Results

We found:

• Statement 1 (sufficiency of T): TRUE

• Statement 2 (completeness of T): TRUE

• Statement 3 (unbiasedness of T² for p²): FALSE

So only statements 1 and 2 are correct, which corresponds to option (a). This matches the official answer key for Series A of the ISS 2016 Statistics Paper-2 test booklet, which lists the answer to Question 25 as (a). Our independent derivation and the official key are in full agreement.

Why This Question Matters

This question is a compact test of three pillars of estimation theory that almost always show up together in ISS Paper-2: sufficiency (via the Factorization Theorem), completeness (via the polynomial-vanishing argument for exponential families), and the subtlety that unbiasedness does not automatically transfer through nonlinear transformations of a statistic. Together, sufficiency and completeness of T are exactly the two conditions required by the Lehmann-Scheffé theorem to identify a UMVUE (Uniformly Minimum Variance Unbiased Estimator) — so a strong grip on this question sets up almost every later estimation question in the paper.

Frequently Asked Questions

What does b(1, p) mean in this context?

b(1, p) denotes the Bernoulli distribution, which is a Binomial distribution with only one trial (n = 1). A random variable following b(1, p) takes the value 1 with probability p and 0 with probability (1 − p).

What is the difference between sufficiency and completeness of a statistic?

Sufficiency means the statistic captures all the information in the sample about the parameter, so nothing more can be learned from the raw data once the statistic is known. Completeness is a separate property about the family of distributions of the statistic: it means no non-trivial function of the statistic has zero expectation for every value of the parameter. A statistic can be sufficient without being complete, but when it is both, it becomes extremely useful for constructing UMVUEs.

Why is the sum of Bernoulli trials a sufficient statistic for p?

Because the joint probability of the sample can be factored so that the parameter p only ever multiplies the sample through the total count of successes, ΣXi. Once you know how many successes occurred, which specific trials were successes adds no further information about p, exactly as the Factorization Theorem requires.

Is every sufficient statistic also complete?

No. Sufficiency and completeness are independent properties in general, though they frequently coincide for statistics arising from one-parameter exponential families like the Binomial, Poisson, and Normal (with one parameter unknown). There exist sufficient statistics for certain distributions that fail to be complete, so each property must be checked on its own merits.

Why isn't T² an unbiased estimator of p² even though T is unbiased for np?

Unbiasedness does not carry through nonlinear transformations. Squaring introduces the variance of T as an extra additive term, since E(T²) = Var(T) + [E(T)]², and that variance term does not disappear just because you wanted an estimator of p². This is a classic exam trap, and recognising it quickly is a skill examiners specifically reward.

How do sufficiency and completeness relate to finding a UMVUE?

The Lehmann-Scheffé theorem states that if a statistic is both complete and sufficient for a parameter, then any unbiased function of that statistic is automatically the unique UMVUE for whatever it is estimating. This is precisely why examiners test sufficiency and completeness together — they are the two prerequisites you need before you can even start looking for a UMVUE.

What is the actual unbiased estimator of p² given T ~ Binomial(n, p)?

The correct unbiased estimator is T(T − 1) divided by n(n − 1). This adjusts for the extra np term that appears when you naively square T, and it is the kind of derivation that becomes second nature once you have worked through the Lehmann-Scheffé framework a few times.

If any step above felt rushed or you spotted a shortcut we didn't mention, drop a comment below — and if this helped clear up sufficiency and completeness for you, consider sharing it with a fellow ISS aspirant who might be stuck on the same question.