ISS 2016 Statistics Paper-2 Solution: Question 62 (Minimal Sufficient Statistic for Normal Variance)

Continuing our question-by-question walkthrough of the ISS Statistics Paper-2 archive, today we pick up right where the series left off and tackle a core question from Point Estimation theory. If you've been following along, you know the drill: no shortcuts, every algebraic step shown, and a direct check against the official answer key at the end.

Quick Summary

• Topic:Minimal Sufficient Statistics for a Normal Distribution with Known Mean

• Question Reference:ISS 2016, Statistics Paper-2, Question 62

• Correct Answer:(a) ΣXi2 is the minimal sufficient statistic for σ²

The Question

If X1, X2, X3, ..., Xn is a random sample from N(0, σ²), then which one of the following statements is correct?

• (a) Σi=1n Xi2 is minimal sufficient statistic for σ²

• (b) Σi=1n Xi is minimal sufficient statistic for σ²

• (c) X̄ is minimal sufficient statistic for σ²

• (d) There is no minimal sufficient statistic for σ²

Step 1: Write Down the Joint Density

Each Xi follows N(0, σ²), so its individual density is:

f(xi; σ²) = (1 / √(2πσ²)) · exp(−xi2 / (2σ²))

Since X1, ..., Xn are independent, the joint density (likelihood) of the full sample is just the product of the n individual densities:

L(σ²; x1,...,xn) = ∏i=1n (1 / √(2πσ²)) · exp(−xi2 / (2σ²))

Pull the constant term out of the product (it appears n times) and combine the exponents, since exp(a)·exp(b) = exp(a+b):

L(σ²; x) = (2πσ²)−n/2 · exp(− (1/(2σ²)) Σi=1n xi2)

Notice something important here: the data x1, x2, ..., xn appear in this entire expression only through the single combined quantity Σxi2. Nowhere does the likelihood depend on, say, x1 and x2 separately, or on Σxi on its own.

Step 2: Apply the Factorization (Neyman-Fisher) Theorem

The Factorization Theorem says: a statistic T(X) is sufficient for a parameter θ if and only if the likelihood can be written as

L(θ; x) = g(T(x), θ) · h(x)

where g depends on the data only through T(x), and h(x) does not depend on θ at all.

Let's match our expression to this form, with T(x) = Σxi2:

• g(T(x), σ²) = (2πσ²)−n/2 · exp(−T(x)/(2σ²))

• h(x) = 1

This factorization works perfectly with T(x) = Σxi2. So ΣXi2 is a sufficient statistic for σ². That rules out option (d) immediately — a sufficient statistic clearly exists, so a minimal one must exist too (any sufficient statistic can always be reduced to a minimal one).

Step 3: Confirm Minimality Using the Likelihood Ratio Criterion

Sufficiency alone isn't enough to answer this question — we specifically need the minimal sufficient statistic, meaning the one that compresses the data as much as possible without losing any information about σ².

The standard tool here is the Lehmann-Scheffe criterion: T(X) is minimal sufficient if, for any two sample points x and y,

the ratio L(σ²; x) / L(σ²; y) is free of σ²   ⇔   T(x) = T(y)

Let's compute this ratio directly using our likelihood expression:

L(σ²; x) / L(σ²; y) = exp[ −(1/(2σ²)) (Σxi2 − Σyi2) ]

(the (2πσ²)−n/2 terms are identical top and bottom, so they cancel out). Now look at this ratio carefully:

• If Σxi2 = Σyi2, the exponent becomes zero, and the ratio equals exp(0) = 1 for every value of σ². The ratio is free of σ².

• If Σxi2 ≠ Σyi2, the exponent is a non-zero multiple of 1/σ², so the ratio genuinely changes as σ² changes. It is not free of σ².

This is exactly the "if and only if" condition the criterion asks for. So ΣXi2 is confirmed as the minimal sufficient statistic for σ².

Pro Tip: Don't let the general N(μ, σ²) case confuse you here. When both μ and σ² are unknown, the jointly minimal sufficient statistic is the pair (ΣXi, ΣXi2), which is why so many students instinctively reach for ΣXi or X̄ in this question. But read the question again: the mean is fixed at 0, not estimated. That collapses the problem to a single-parameter exponential family in σ² alone, and a one-parameter exponential family always has exactly one minimal sufficient statistic, tied to the single term sitting next to the parameter in the exponent. The way this distinction is usually drilled into students is to always rewrite the density in exponential-family form first — f(x;θ) = h(x)c(θ)exp(w(θ)T(x)) — and read T(x) straight off it, rather than guessing from memory which statistic "usually" goes with the normal distribution.

Step 4: Rule Out the Other Options

With the exponential-family view from the Pro Tip, this becomes quick to check. Writing our density as f(x;σ²) = (2πσ²)−1/2 exp(−(1/(2σ²))x²), the term multiplying the parameter-dependent part of the exponent is x² and nothing else. So:

• (b) ΣXi:This statistic simply does not appear anywhere in the likelihood. Two samples can have the exact same Σxi but very different Σxi2 (e.g., one sample clustered near 0, another spread far out but still summing to the same total) — and those two samples would have genuinely different likelihoods. So ΣXi isn't even sufficient, let alone minimal sufficient.

• (c) X̄:Since X̄ = (1/n)ΣXi, it carries exactly the same information (or lack of it) as option (b) and fails for the identical reason.

• (d) No minimal sufficient statistic exists:We explicitly constructed one in Step 3, so this is false.

This confirms option (a)is correct, and it matches the official ISS 2016 Paper-2 answer key, which also lists (a) for Question 62.

Why This Question Matters

Sufficiency and minimal sufficiency form the backbone of estimation theory in the ISS syllabus — they show up again and again in questions on UMVUE, the Rao-Blackwell theorem, and completeness. A question like this one is really testing whether you can mechanically derive the minimal sufficient statistic from the likelihood itself, rather than pattern-matching to a formula you memorized for the "usual" N(μ,σ²) case. Exams frequently tweak a standard setup (here, by fixing the mean at a known value) specifically to see whether your understanding is procedural or just memorized.

Frequently Asked Questions

What exactly is a minimal sufficient statistic?

A sufficient statistic T(X) already captures all the information in the sample about a parameter. A statistic is minimal sufficient if it is also a function of every other sufficient statistic — in other words, it achieves the maximum possible data reduction while still being sufficient. There can be many sufficient statistics for a parameter, but the minimal one is essentially unique (up to a one-to-one transformation).

Why does knowing the mean change the answer compared to the standard N(μ, σ²) case?

When μ is unknown, you need two pieces of information from the sample to pin down both μ and σ², so the jointly minimal sufficient statistic is the pair (ΣXi, ΣXi2). Here μ = 0 is already known and fixed, so there's only one unknown parameter left, and the likelihood reduces to depending on just ΣXi2.

Is the Lehmann-Scheffe likelihood ratio method the only way to find a minimal sufficient statistic?

No. For distributions belonging to a regular exponential family, you can read the (minimal) sufficient statistic directly off the exponential form of the density without computing any ratio. The likelihood ratio method is more general and works even outside the exponential family, which is why it's worth knowing both approaches.

Is ΣXi2 also sufficient for the standard deviation σ, not just σ²?

Yes. Sufficiency for σ² automatically implies sufficiency for any one-to-one function of σ², including σ itself, since estimating σ² and then taking a square root loses no further information.

Does this mean the sample mean is never sufficient when estimating variance?

Not in general, no. In this specific problem, the sample mean happens to be useless for σ² precisely because the true mean is already known to be exactly 0, so X̄ only reflects random noise around that known value, not anything about the spread. If μ were unknown, X̄ would become an essential part of the (joint) sufficient statistic for estimating μ, though still not by itself sufficient for σ².

How should I practice this topic for the exam?

Take the standard distributions one by one — normal, exponential, Poisson, uniform — and for each, write the density in exponential-family form and read off the sufficient statistic yourself, rather than memorizing a table of results. Doing this by hand a few dozen times across different distributions is what makes the pattern click, and it is a habit that is worth building deliberately rather than leaving to exam-week revision.

If anything in the derivation above felt unclear, drop a comment below and we'll work through it together — and if you found this useful, consider sharing it with a fellow ISS aspirant who might be stuck on the same question.