ISS 2016 Statistics Paper-2 Solution: Question 57 (MVUE of Normal Variance)

Continuing our question-by-question walkthrough of the ISS 2016 Statistics Paper-2 objective paper, today we tackle Question 57. This one sits at the heart of estimation theory — a topic that shows up again and again across ISS, IIT JAM(MS) and GATE(ST) syllabi, so it's worth getting completely solid on.

Quick Summary

• Topic:Minimum Variance Unbiased Estimation (MVUE) for the variance of a Normal distribution with known mean

• Question Reference:ISS 2016, Statistics Paper-2, Question 57

• Correct Answer:Option (b) — 1⁄n ∑ Xi2

The Question, As Asked

Let X1, X2, X3, ..., Xn be independently and normally distributed with mean 0 and variance σ2. What is the minimum variance unbiased estimator of σ2?

• (a) 1⁄n−1 ∑i=1n Xi2

• (b) 1⁄n ∑i=1n Xi2

• (c) 1⁄n ∑i=1n (Xi − X̄)2

• (d) 1⁄n−1 ∑i=1n (Xi − X̄)2

Step 1: Notice What Makes This Question Different

At first glance this looks like the standard "estimate the variance" problem you've solved a dozen times, where the answer is always S2 with (n−1) in the denominator. But read the setup carefully: the mean is stated to be exactly 0, a known constant, not an unknown parameter we also have to estimate. That single fact changes the entire answer. Whenever a question tells you the population mean, your first move should be to ask: "do I actually need X̄ here, or does using it throw away information I already have for free?"

Step 2: Write Down the Natural Candidate

Since each Xi ~ N(0, σ2), consider T = 1⁄n ∑ Xi2. Let's check this is unbiased for σ2.

For any random variable, Var(Xi) = E[Xi2] − [E(Xi)]2. Here E(Xi) = 0, so:

E[Xi2] = Var(Xi) + 02 = σ2

Therefore:

E(T) = E[1⁄n ∑ Xi2] = 1⁄n ∑ E[Xi2] = 1⁄n · nσ2 = σ2

So T is unbiased. Good — but unbiasedness alone doesn't make it the minimum variance unbiased estimator. We need to show it beats every other unbiased estimator in terms of variance.

Step 3: Use the Complete Sufficient Statistic Argument

Write the joint density of X1, ..., Xn:

f(x1,...,xn; σ2) = (2πσ2)−n/2 exp[ −1⁄2σ2 ∑ xi2 ]

This is a one-parameter exponential family in σ2, and the statistic ∑ Xi2 is exactly the quantity appearing in the exponent. By the standard exponential-family result, ∑ Xi2 is a complete sufficient statistic for σ2.

Now invoke the Lehmann–Scheffé theorem: if you can write down any unbiased estimator that is a function of a complete sufficient statistic, that estimator is automatically the unique MVUE. Since T = 1⁄n ∑ Xi2 is (i) a function of the complete sufficient statistic ∑ Xi2, and (ii) unbiased as shown in Step 2, Lehmann–Scheffé tells us immediately that T is the MVUE of σ2.

Pro Tip: The trap in this question is option (d). Most students' reflex is to write S2 = 1⁄n−1∑(Xi−X̄)2 the moment they see "unbiased estimator of variance," because that's the formula drilled into memory from the unknown-mean case. The way this distinction is usually taught is to always ask first whether the mean is known or unknown before reaching for a formula — here it's known, so replacing it with X̄ in (d) throws away real information, which is exactly why (d) is unbiased but NOT minimum variance.

Step 4: Confirm by Directly Comparing Variances

It helps to see numerically, not just via the theorem, why (b) beats (d). Since each Xi/σ ~ N(0,1), we know ∑Xi2⁄σ2 follows a chi-square distribution with n degrees of freedom. A chi-square variable with k degrees of freedom has variance 2k, so:

Var(∑ Xi2) = σ4 · 2n  ⇒  Var(T) = Var(1⁄n∑Xi2) = 2σ4⁄n

Now for option (d), the usual sample variance S2: we know (n−1)S2⁄σ2 ~ χ2n−1, which has variance 2(n−1). So:

Var((n−1)S2) = σ4 · 2(n−1)  ⇒  Var(S2) = 2σ4⁄n−1

Comparing the two: 2σ4⁄n is strictly smaller than 2σ4⁄n−1 for every finite n. So T in option (b) genuinely has lower variance than S2 in option (d) — numbers confirm what Lehmann–Scheffé already told us.

Step 5: Eliminate the Remaining Options

Option (a), 1⁄n−1∑Xi2, is not even unbiased: its expectation is E[1⁄n−1∑Xi2] = n⁄n−1σ2 ≠ σ2, so it's disqualified immediately — an estimator must first be unbiased before we even ask whether it has minimum variance.

Option (c), 1⁄n∑(Xi−X̄)2, has expectation n−1⁄nσ2, which is also biased (this is the familiar "divide by n instead of n−1" bias everyone meets when first learning sample variance).

That leaves only (b) as both unbiased and minimum variance.

Final Answer

The minimum variance unbiased estimator of σ2 is (b) 1⁄n ∑i=1n Xi2, which matches the official ISS 2016 Statistics Paper-2 answer key (Series A: Q57 → B).

Why This Question Matters

This question is a compact test of three separate skills at once: recognizing when extra information (a known mean) should actually change your formula, proving unbiasedness by direct expectation calculation, and applying the Lehmann–Scheffé theorem to jump from "unbiased + function of complete sufficient statistic" straight to "this is the MVUE" without having to compare variances of every competing estimator by hand. ISS Paper-2 repeatedly tests exactly this combination — known-mean vs unknown-mean variance estimation is one of the most recycled ideas in the estimation theory section, so a student who drills this distinction until it's automatic will recognize it instantly in whatever disguise it reappears under.

Frequently Asked Questions

Why doesn't option (d), the usual sample variance formula, count as the MVUE here?

Option (d) is unbiased, but it is not the minimum variance unbiased estimator in this problem because it uses the sample mean X̄ to center the data even though the true mean 0 is already known. Using X̄ instead of the known mean discards information and inflates the variance, as shown directly in Step 4.

Is S2 = 1⁄n−1∑(Xi−X̄)2 ever the MVUE of σ2?

Yes — when the population mean is unknown. In that standard case, (X̄, S2) forms a jointly complete sufficient statistic, and S2 is the MVUE of σ2. The moment the mean becomes a known constant, the complete sufficient statistic changes to ∑Xi2, and the MVUE changes along with it.

What exactly does the Lehmann–Scheffé theorem say, in plain words?

It says that if you find any estimator that is both unbiased for your parameter and can be written purely as a function of a complete sufficient statistic, that estimator is guaranteed to be the unique minimum variance unbiased estimator. You don't need to compare it against every other possible unbiased estimator one by one.

How do I quickly tell whether a statistic is "complete and sufficient" in an exam setting?

For standard one-parameter exponential family distributions (Normal with known mean, Exponential, Poisson, Bernoulli, etc.), the natural sufficient statistic that appears in the exponent of the likelihood is automatically complete. Recognizing the exponential family form of the density, as done in Step 3, is usually the fastest route to identifying it under exam time pressure.

Why is Var(S2) = 2σ4/(n−1) and not 2σ4/n?

Because (n−1)S2/σ2 follows a chi-square distribution with (n−1) degrees of freedom, not n — one degree of freedom is "used up" in estimating the unknown mean via X̄. Since a chi-square with k degrees of freedom has variance 2k, substituting k = n−1 and rescaling gives Var(S2) = 2σ4/(n−1), which is exactly why S2 has higher variance than ∑Xi2/n.

Does this result generalize to other distributions with a known mean?

Yes. Whenever the mean (or any other nuisance parameter) is known rather than estimated, you should always re-derive the sufficient statistic from scratch rather than assuming the usual formula still applies — the same logic used here applies equally to estimating variance in other known-mean settings.

If any step above felt rushed or you'd solve it differently, drop a comment below with your doubt — and if this walkthrough helped, consider sharing it with a fellow ISS aspirant working through the same paper.