ISS 2016 Statistics Paper-2 Solution: Question 59 (Sample Mean and Variance Properties)

Continuing this ongoing question-by-question walkthrough of the ISS Statistics Paper-2 previous year papers, today's post takes up Question 59 from the 2016 paper. It sits in the estimation theory portion of the syllabus and tests three separate ideas about the sample mean and sample variance at once, which is exactly why it trips up students who know the individual facts but haven't drilled how they interact.

Quick Summary

• Topic:Properties of the sample mean and sample variance (unbiasedness, consistency, chi-square distribution) for a normal population

• Reference:ISS 2016, Statistics Paper-2, Question 59

• Correct Answer:(a) 1 and 2 only

The Question As Asked

If X̄and S²are the mean and variance of a random sample of size n from a normal population with mean μand standard deviation σ, then consider the following statements:

1. X̄is unbiased and consistent estimator of μ.

2. S²is consistent estimator of σ².

3. The random variable (n − 1)S²/σ² has a Chi-square distribution with (n − 1) degrees of freedom.

Which of the above statements are correct?

• (a) 1 and 2 only

• (b) 2 and 3 only

• (c) 1 and 3 only

• (d) 1, 2 and 3

Step 1: Fix the Definitions Before Touching Any Statement

Let X₁, X₂, ..., Xₙ be independent and identically distributed as N(μ, σ²). The sample mean is X̄ = (1/n) Σ Xᵢ. The question calls S² "the variance of the sample", which in this paper's convention means the directly computed sample variance with divisor n:

S² = (1/n) Σ (Xᵢ − X̄)²

This is different from the commonly seen "unbiased sample variance" that divides by (n − 1) instead of n. Getting this definition straight before evaluating the three statements is the single most important thing in this question — mixing the two conventions is exactly what makes statement 3 look true when it isn't.

Step 2: Check Statement 1 — Is X̄ Unbiased and Consistent for μ?

Unbiasedness: E(X̄) = E[(1/n) Σ Xᵢ] = (1/n) Σ E(Xᵢ) = (1/n)(nμ) = μ. So X̄ is unbiased for μ.

Consistency: Var(X̄) = σ²/n, which tends to 0 as n → ∞. An estimator that is unbiased (or asymptotically unbiased) with variance shrinking to zero is consistent, by Chebyshev's inequality. So X̄ is also consistent for μ.

Statement 1 is true.

Step 3: Check Statement 2 — Is S² Consistent for σ²?

This is where the exam-specific definition of S² (divisor n) actually matters. A standard distributional result (it will be re-derived properly in Step 4) gives:

nS²/σ² ~ χ²₍ₙ₋₁₎

Taking expectation of a chi-square variable with (n − 1) degrees of freedom, whose mean equals its degrees of freedom:

E(nS²/σ²) = n − 1 ⟹ E(S²) = [(n − 1)/n] σ²

So S² is slightly biased downward in finite samples — but notice the bias factor (n − 1)/n → 1 as n → ∞, so the bias vanishes asymptotically. For the variance, using Var(χ²₍ₙ₋₁₎) = 2(n − 1):

Var(nS²/σ²) = 2(n − 1) ⟹ Var(S²) = 2(n − 1)σ⁴/n²

As n → ∞, Var(S²) → 0 as well. An estimator whose bias and variance both go to zero is consistent — this follows from the mean-squared-error criterion, since MSE(S²) = Var(S²) + [Bias(S²)]² → 0, which implies convergence in probability to σ².

Statement 2 is true. The key insight non-novice students often miss: consistency does not require unbiasedness. A slightly biased estimator can still be perfectly consistent, as long as the bias disappears in the limit.

Step 4: Check Statement 3 — Does (n − 1)S²/σ² Really Follow a Chi-Square Distribution?

This is the statement that decides the whole question, so it deserves the most careful treatment.

The correct classical result (a consequence of Cochran's theorem, applied to a normal sample) is about the sum of squared deviations, not S² directly:

Σ (Xᵢ − X̄)² / σ² ~ χ²₍ₙ₋₁₎

Now substitute the exam's definition S² = (1/n) Σ(Xᵢ − X̄)², which means Σ(Xᵢ − X̄)² = nS². Plugging this in:

nS²/σ² ~ χ²₍ₙ₋₁₎

That is the correct statement — the multiplier in front of S²/σ² should be n, not (n − 1). The quantity actually written in statement 3 is:

(n − 1)S²/σ² = [(n − 1)/n] × [nS²/σ²]

This is just the true chi-square variable nS²/σ² multiplied by the constant (n − 1)/n, which is strictly less than 1 for any finite n. Multiplying a chi-square random variable by a constant other than 1 produces a scaled chi-square variable, not a chi-square variable itself — the chi-square family is not closed under arbitrary rescaling. So (n − 1)S²/σ² does not follow a χ²₍ₙ₋₁₎ distribution under this exam's definition of S².

Pro Tip: Whenever a question states a multiplier in front of S²/σ² or s²/σ², the very first thing to check — before touching the degrees of freedom — is which divisor (n or n − 1) was used to define S² in that same question. The

Statement 3 is therefore false as written.

Step 5: Combine the Results

Statement 1: true. Statement 2: true. Statement 3: false. So the correct combination is 1 and 2 only, which is option (a).

This matches the official ISS 2016 Statistics Paper-2 answer key (Series A), which also lists option (a) for Question 59. No discrepancy here — the derivation and the official key agree.

Why This Question Matters

Questions like this one are less about memorising a formula and more about tracking definitions precisely through a multi-part statement. ISS Paper-2 repeatedly tests whether a candidate can distinguish unbiasedness from consistency, and whether a stated chi-square relationship uses the right scaling constant for the specific definition of S² given in that question. Getting comfortable with both the divisor-n and divisor-(n − 1) conventions for sample variance — and converting between them on the fly — pays off across many other questions on estimation theory, not just this one.

Frequently Asked Questions

Why is S² here defined with divisor n instead of the more common (n − 1)?

Different textbooks and exams use different conventions. When a question explicitly calls S² "the variance of the sample" without further qualification, it typically means the direct second moment about the sample mean, which uses divisor n. The (n − 1) version is usually introduced separately as the "unbiased estimator of variance". Always read the question's own wording carefully rather than assuming a convention.

Can an estimator be biased and still be consistent?

Yes. Bias and consistency are different properties. Bias is about whether the expected value of the estimator equals the true parameter for a fixed sample size, while consistency is about whether the estimator converges to the true parameter as the sample size grows without bound. S² here is biased for every finite n but is still consistent because the bias shrinks to zero as n increases.

What is Cochran's theorem and why does it matter for this question?

Cochran's theorem gives conditions under which quadratic forms in normal random variables follow chi-square distributions. For a simple random sample from a normal population, it is what guarantees that Σ(Xᵢ − X̄)²/σ² follows a chi-square distribution with (n − 1) degrees of freedom, which is the building block used to evaluate statement 3 in this question.

How do I quickly check whether a scaled chi-square claim is correct in an exam?

Write out the known correct relationship (Σ(Xᵢ − X̄)²/σ² ~ χ²₍ₙ₋₁₎) first, then substitute the sample variance definition given in that specific question to see what multiplier it actually produces. Compare that multiplier to the one stated in the question before deciding if the statement is true.

Is the sample mean always unbiased, regardless of the parent distribution?

Yes, as long as the population mean exists, E(X̄) = μ holds for any distribution, not just the normal distribution, because expectation is linear. What can change across distributions is the sampling variance and the exact shape of X̄'s distribution, not its unbiasedness.

Does losing one degree of freedom always mean subtracting exactly 1?

In this single-sample, single-mean-estimated setting, yes — estimating one parameter (μ) from the data before computing the sum of squared deviations costs exactly one degree of freedom, giving (n − 1). In more complex models, such as regression with several estimated parameters, the degrees of freedom lost equals the number of independent linear restrictions imposed by the fitted parameters, which can be more than one.

If any step above needs more explanation, or you spotted something worth double-checking, drop a comment below — and if a fellow ISS aspirant is working through the same 2016 Paper-2 set, feel free to share this with them.