ISS 2016 Statistics Paper-2 Solution: Question 54 (Consistency vs Unbiasedness of Estimators)
Continuing our question-by-question walk through the ISS Statistics Paper-2 archive, today's post takes up Question 54 from the 2016 paper. This item packs four separate true/false claims about estimators into one question, which makes it a great one to slow down on and work through statement by statement.
Quick Summary
Topic: Unbiasedness vs. consistency of estimators (normal, exponential and uniform populations)
Reference: ISS 2016, Statistics Paper-2, Question 54
Correct Answer: Option (d) — Statements 1, 2 and 3 are correct
The Question
Consider the following statements:
If X1, X2, X3, ..., Xn is a random sample from a normal population, then the sample variance S2 is a consistent estimator of σ2.
If X1, X2, X3, ..., Xn is a random sample from a normal population with mean 0, then ΣXi2/n is an unbiased estimator of σ2.
If X1, X2, X3, ..., Xn is a random sample from an exponential population with probability density function f(x, θ) = (1/θ)e-x/θ; x > 0, 0 otherwise, then X̄ is a consistent estimator for parameter θ.
If X1, X2, X3, ..., Xn is a random sample from U(0, β), then the nth order statistic Yn is an unbiased estimator of β.
Which of the following statements are correct?
(a) 2, 3 and 4
(b) 1, 2 and 4
(c) 1, 3 and 4
(d) 1, 2 and 3
Before We Start: Unbiasedness vs. Consistency
These two properties get mixed up constantly, so let's pin them down first. An estimator T is unbiased for a parameter θ if E(T) = θ exactly, for every sample size n, even n = 1. An estimator T is consistent if it converges in probability to θ as n → ∞ — in practice, for most estimators we meet in this syllabus, that means its mean squared error (bias squared plus variance) shrinks to zero as n grows. A crucial consequence: an estimator can be biased for every finite n and still be perfectly consistent, because the bias simply has to vanish in the limit. Statement 4 below is built entirely around this distinction, so keep it in mind.
Statement 1: Is the Sample Variance S² Consistent for σ²?
Let X1, ..., Xn be a random sample from N(μ, σ2), and let S2 denote the sample variance (whether you define it with divisor n or n − 1, the argument below works the same way asymptotically).
We know the standard result: (n − 1)S2/σ2 follows a chi-square distribution with (n − 1) degrees of freedom, when S2 is defined with divisor (n − 1). From the properties of the chi-square distribution:
E[(n − 1)S2/σ2] = (n − 1), so E(S2) = σ2 — S2 is unbiased.
Var[(n − 1)S2/σ2] = 2(n − 1), so Var(S2) = 2σ4/(n − 1).
Now check what happens to the mean squared error as n → ∞. Since S2 is unbiased, MSE(S2) = Var(S2) = 2σ4/(n − 1), and this clearly tends to 0 as n → ∞. An estimator whose MSE shrinks to zero converges in probability to the parameter it is estimating. So S2 is both unbiased and consistent for σ2.
Statement 1 is TRUE.
Statement 2: Is ΣXᵢ²/n Unbiased for σ² When the Mean Is Known to Be 0?
Here the population is N(0, σ2) — note carefully that the mean is given as 0, not estimated from the data. This is the detail that makes the statement work.
Since each Xi ~ N(0, σ2), standardising gives Xi/σ ~ N(0, 1), and therefore (Xi/σ)2 ~ χ²1 for each i. Summing n independent such terms:
ΣXi2/σ2 ~ χ²n
A chi-square distribution with n degrees of freedom has mean exactly n, so:
E(ΣXi2/σ2) = n, which gives E(ΣXi2) = nσ2
Dividing both sides by n:
E(ΣXi2/n) = σ2
That is exactly the definition of unbiasedness for σ2. Notice this only works because the mean was given as 0 — if the mean were unknown, you would have to replace it with X̄ and lose one degree of freedom, which is precisely why the usual sample variance divides by (n − 1) instead of n.
Statement 2 is TRUE.
Pro Tip: Whenever a question explicitly tells you the population mean is a known constant (here, 0), that is almost always a signal to use ΣXᵢ²/n rather than the usual S² formula — dividing by n instead of (n − 1) is only valid once you're not "spending" a degree of freedom estimating the mean. Drilling this one distinction until it becomes automatic — the way it's typically taught in a structured estimation-theory course — saves a lot of silent marks lost on this exact trap.
Statement 3: Is X̄ Consistent for θ in the Exponential Distribution?
The population here has pdf f(x, θ) = (1/θ)e-x/θ for x > 0. This is the exponential distribution with mean θ (this particular parametrisation has E(X) = θ, not 1/θ — always check which parametrisation a question is using before writing down moments).
For this distribution:
E(X) = θ, so E(X̄) = θ — X̄ is unbiased for θ.
Var(X) = θ2, so Var(X̄) = θ2/n.
As n → ∞, Var(X̄) = θ2/n → 0. Since X̄ is unbiased, MSE(X̄) = Var(X̄) → 0, so X̄ converges in probability to θ. This is really just an application of the Weak Law of Large Numbers: the sample mean of iid random variables with finite variance always converges in probability to the population mean, and here the population mean is θ itself.
Statement 3 is TRUE.
Statement 4: Is the Largest Order Statistic Yₙ Unbiased for β in U(0, β)?
Now X1, ..., Xn is a random sample from U(0, β), and Yn = max(X1, ..., Xn) is the nth (largest) order statistic. To check unbiasedness we need E(Yn) and compare it to β.
The cdf of a single Xi is F(x) = x/β for 0 ≤ x ≤ β. The cdf of the maximum of n iid observations is:
F(n)(y) = [F(y)]n = (y/β)n, for 0 ≤ y ≤ β
Differentiating to get the pdf of Yn:
f(n)(y) = n yn-1/βn, for 0 ≤ y ≤ β
Now compute the expectation:
E(Yn) = ∫0β y · (n yn-1/βn) dy = (n/βn) ∫0β yn dy = (n/βn) · [βn+1/(n+1)] = nβ/(n + 1)
So E(Yn) = nβ/(n + 1), which is not equal to β for any finite n — it is strictly less than β. The bias is E(Yn) − β = −β/(n + 1), which only vanishes as n → ∞. So Yn is a biased estimator of β for every finite sample size, even though it happens to be asymptotically unbiased (and in fact consistent, since both its bias and its variance go to zero as n grows).
Statement 4 is FALSE — this is the one statement in the question that is not correct, and it is exactly the unbiased-vs-consistent trap flagged at the start of this post.
Final Answer
Statements 1, 2 and 3 are correct; statement 4 is false. That matches option (d) 1, 2 and 3.
This agrees with the official ISS 2016 Statistics Paper-2 answer key (Series A), which lists D for Question 54.
Why This Question Matters
Question 54 is really four mini-questions disguised as one, and together they cover almost every standard scenario examiners use to test the unbiasedness/consistency distinction: a textbook consistent-and-unbiased estimator (S2), a "spot the known mean" unbiasedness trick (ΣXi2/n), a Law of Large Numbers application (X̄ for the exponential mean), and the classic biased-but-consistent order-statistic case (Yn for the uniform maximum). If you can confidently re-derive all four of these from scratch, you have essentially covered the core toolkit that recurs across estimation-theory questions in every ISS Paper-2 since 2016.
Frequently Asked Questions
Why is the sample maximum Yₙ biased but the sample mean X̄ is unbiased in these two statements?
It comes down to how each statistic is built. X̄ is a simple average, and averaging preserves the population mean exactly regardless of sample size. Yn, on the other hand, is the maximum of n draws, and the maximum of several draws from U(0, β) always falls somewhat below β on average, so its expectation is systematically pulled down to nβ/(n + 1) rather than β.
Can a biased estimator still be a good estimator?
Yes. Yn here is biased for every finite n, but it is consistent, has much smaller variance than an unbiased alternative like 2X̄, and a simple bias correction — using (n + 1)Yn/n instead — turns it into an exactly unbiased estimator. In practice, estimators are often judged on mean squared error rather than bias alone.
Does every unbiased estimator automatically become consistent as n increases?
Not automatically — unbiasedness only guarantees that MSE equals variance, it says nothing about whether that variance shrinks to zero. However, for most standard estimators built from iid samples (like X̄ or S2), the variance does decay at rate 1/n, so an unbiased estimator with a finite, well-behaved variance will typically also be consistent. You still have to check the variance term explicitly rather than assume it.
Why does knowing the population mean change the divisor from (n − 1) to n in statement 2?
Degrees of freedom. When the mean is unknown and has to be estimated by X̄, one degree of freedom is "used up" in that estimation, which is why the usual sample variance divides the sum of squared deviations by (n − 1) to stay unbiased. When the mean is already known (here, 0), no degree of freedom is spent estimating it, so dividing by the full n keeps the estimator exactly unbiased.
Is this kind of "which statements are correct" format common in ISS Paper-2?
Very common. A large share of ISS Statistics Paper-2 questions are structured exactly this way — several numbered statements followed by a combination-coded answer choice. The only reliable way through them is to evaluate every statement completely independently before looking at the options, which is exactly the approach used question by question throughout this series.
What's the fastest way to remember the uniform maximum's expectation formula nβ/(n+1)?
Think of it as a weighted shrinkage of β: as n grows, n/(n+1) approaches 1, so the maximum's expectation approaches β — which makes intuitive sense, since with more draws from U(0, β) you are more likely to land close to the upper limit. For small n it pulls noticeably below β, which is exactly the bias captured in the derivation above.
If any step above wasn't clear, or you'd like a different approach to any of the four statements, drop a comment below — and if this helped, consider sharing it with a fellow ISS aspirant who's working through the same paper.
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