ISS 2016 Statistics Paper-2 Solution: Question 37 (Sufficient but Not Complete Statistic)

Continuing our question-by-question walk through the ISS Statistics Paper-2 archive, today's post picks up right where Question 36 left off. This one moves us from estimator properties into a topic that trips up even well-prepared candidates: the gap between a statistic being sufficient and a statistic being complete.

Quick Summary

• Topic:Sufficiency and completeness of a joint statistic for the N(θ, θ²) family

• Question Reference:ISS 2016, Statistics Paper-2, Question 37

• Correct Answer:(b) Sufficient but not complete

The Question

Consider a random sample X1, X2, X3, ..., Xn of size n from N(θ, θ²), the statistic T = (Σi=1n Xi, Σi=1n Xi2) for θ is:

• (a) Sufficient and complete

• (b) Sufficient but not complete

• (c) Not sufficient but complete

• (d) Neither sufficient nor complete

Step 1: Understand What Makes N(θ, θ²) Unusual

A normal distribution is usually written N(μ, σ²) with two separate parameters — one for location and one for spread. Here, both roles are played by the same single parameter θ: the mean is θ and the variance is θ². So even though we are sampling from a two-parameter-looking family, there is really only one unknown, θ, to estimate.

This single fact — one parameter driving both the mean and the variance — is the whole key to the question, so keep it in mind as we go.

Step 2: Write Down the Likelihood

The pdf of a single observation is:

f(x; θ) = 1 / (θ√(2π)) · exp[ −(x − θ)² / (2θ²) ]

For the full sample of n independent observations, the likelihood is the product:

L(θ) = (2πθ²)−n/2 · exp[ −Σ(xi − θ)² / (2θ²) ]

Now expand the square in the exponent, term by term, exactly as you would in any algebra class:

(xi − θ)² = xi² − 2θxi + θ²

Summing over i = 1 to n:

Σ(xi − θ)² = Σxi² − 2θΣxi + nθ²

Substitute this back into the exponent of L(θ):

L(θ) = (2πθ²)−n/2 · exp[ −Σxi²/(2θ²) + Σxi/θ − n/2 ]

Step 3: Apply the Factorization Theorem to Check Sufficiency

The Fisher–Neyman factorization theorem says a statistic T(X) is sufficient for θ if the likelihood can be written as L(θ) = g(T(x), θ) · h(x), where h(x) does not depend on θ at all.

Look at the expression we derived above. Every place θ appears, it appears multiplying or dividing either Σxi or Σxi². There is no leftover piece of the data that escapes being summarised through these two sums. So we can write:

L(θ) = g( Σxi, Σxi² ; θ ) · 1

with h(x) = 1 (a constant, so trivially free of θ). This satisfies the factorization criterion exactly, so T = (Σxi, Σxi²) is sufficient for θ. That already rules out options (c) and (d), which claim T is not sufficient.

Step 4: Check Completeness — the Real Trap

A statistic T is complete if the only function g(T) satisfying Eθ[g(T)] = 0 for every θ is the zero function itself (g(T) = 0 with probability 1). If we can find even one non-zero function of T whose expectation is zero for all θ, completeness fails.

Let S1 = Σxi and S2 = Σxi². We compute their moments directly from the fact that each Xi has mean θ and variance θ²:

For S1:Since the Xi are independent, E[S1] = nθ and Var(S1) = nθ². Using Var(S1) = E[S1²] − (E[S1])², we get:

E[S1²] = Var(S1) + (E[S1])² = nθ² + n²θ² = nθ²(n + 1)

For S2:Each E[Xi²] = Var(Xi) + (E[Xi])² = θ² + θ² = 2θ². Summing over n independent terms:

E[S2] = n · 2θ² = 2nθ²

Now divide the first result by (n + 1) and the second by 2:

E[S1²] / (n + 1) = nθ² and E[S2] / 2 = nθ²

Both equal nθ². So if we define the function:

g(T) = S1² / (n + 1) − S2 / 2

then Eθ[g(T)] = nθ² − nθ² = 0 for every single value of θ— yet g(T) is clearly not the zero function (it depends on the actual sample values). We have just constructed a genuine counterexample, so T is not complete.

Pro Tip: The fastest way to spot this kind of question under exam pressure is to compare dimensions. Here we have a 2-dimensional statistic T = (S1, S2) carrying information about only a single scalar parameter θ. Whenever the sufficient statistic has strictly more components than the parameter has dimensions — this is called a

Step 5: Match with the Options and the Official Key

We have shown T is sufficient (Step 3) but not complete (Step 4). That corresponds exactly to option (b):Sufficient but not complete. This matches the official ISS 2016 Paper-2 answer key, which also records the answer as (b) for Question 37.

Why This Question Matters

This question is a favourite in ISS and similar exams because it tests something deeper than formula recall: whether you actually understand why the Lehmann–Scheffé theorem requires both sufficiency and completeness together to guarantee a UMVUE. If a statistic is sufficient but not complete — as here — you cannot blindly apply Lehmann–Scheffé, even though the statistic still summarises all the sample information. Students who work through several curved exponential family examples like N(θ, θ²) until the sufficiency-versus-completeness distinction becomes automatic tend to handle the trickier estimation-theory sections of Paper-2 far more comfortably.

Frequently Asked Questions

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

A sufficient statistic captures all the information in the sample about the parameter, so the original data gives no extra information about θ once you know T. A complete statistic is a separate property: it means no non-trivial function of T can have zero expectation for every value of θ. A statistic can be sufficient without being complete, exactly as in this question.

Why does the pair (ΣXi, ΣXi²) fail to be complete for N(θ, θ²)?

Because the parameter θ is one-dimensional while the sufficient statistic is two-dimensional, there is enough "extra room" for a non-zero function of the two sums — specifically S1²/(n+1) − S2/2 — to have expectation zero for every θ. That single counterexample is enough to disprove completeness.

What is a curved exponential family, and how does it apply here?

A curved exponential family is one where the natural parameters trace out a lower-dimensional curve inside a higher-dimensional natural parameter space, rather than filling it out completely. Here the natural parameters are 1/θ and −1/(2θ²), both functions of the single θ, so they trace a curve in a 2-dimensional space instead of covering it — a classic sign that completeness will fail even though sufficiency holds.

If T is not complete, does that mean we can never find a UMVUE of θ here?

Not necessarily — it just means the Lehmann–Scheffé theorem, which requires a complete sufficient statistic, cannot be applied directly to this T. A UMVUE might still exist and be found through other methods, or via a different, complete sufficient statistic if one can be identified, but the neat "any unbiased function of T is automatically the UMVUE" shortcut is no longer available.

How do you generally apply the factorization theorem to prove sufficiency?

You write the joint likelihood of the sample and check whether it can be split into two pieces: one, g(T(x), θ), which depends on the data only through your candidate statistic T and may depend on θ freely; and another, h(x), which depends on the data but not on θ at all. If such a split exists, T is sufficient — this is exactly the check we carried out in Step 3.

Is N(θ, θ²) a standard one-parameter exponential family?

Technically its likelihood can be written in exponential family form with two natural statistics, ΣXi and ΣXi², but since only one true parameter θ generates both natural parameters, it behaves as a curved exponential family rather than a full, standard one-parameter exponential family. That distinction is precisely why sufficiency and completeness part ways in this problem.

If any step above still feels shaky, especially the moment calculations in Step 4, drop a comment below and we'll work through it together. And if you know a fellow ISS aspirant grinding through Paper-2 estimation theory, sharing this post with them costs you a minute and might save them an hour of confusion.