ISS 2016 Statistics Paper-2 Solution: Question 24 (Unbiased Estimators from Uniform Order Statistics)

This post continues our ongoing, question-by-question walk-through of the ISS 2016 Statistics Paper-2 objective paper. Today's question takes us into the world of order statistics drawn from a Uniform distribution, and it tests something that trips up a lot of first-time learners:an estimator can look perfectly natural and still be biased unless you actually work out its expectation.

Quick Summary

• Topic:Order statistics of the Uniform distribution and unbiased estimation

• Question Reference:ISS 2016, Statistics Paper-2, Question 24

• Correct Answer:(b) 2 only

The Question

Let X1, X2, X3, …, Xn be independent random variables distributed uniformly on the interval (0, θ). If n is odd, which of the following is/are the unbiased estimator(s) of θ?

1. X(n)

2. 2X((n+1)/2)

Select the correct answer using the code given below:

• (a) 1 only

• (b) 2 only

• (c) Both 1 and 2

• (d) Neither 1 nor 2

Step 1: Understand What Is Being Asked

We have a random sample X1, …, Xn drawn independently from a Uniform(0, θ) distribution, where θ is the unknown parameter we want to estimate. If we arrange the sample in increasing order, we get the order statistics X(1) ≤ X(2) ≤ … ≤ X(n). X(n) is simply the largest observation in the sample, and — because n is odd — X((n+1)/2) is the exact middle observation, i.e. the sample median.

An estimator T is called unbiased for θ if E(T) = θ for every value of θ. So to check statements 1 and 2, we need to actually compute E[X(n)] and E[X((n+1)/2)], and see whether they equal θ.

Step 2: Derive the General Formula for E[X(k)]

Rather than deriving the expectation separately for each order statistic, it is far more efficient to derive one general formula for the expectation of the k-th order statistic, X(k), and then just plug in the value of k we need.

For a Uniform(0, θ) parent distribution, the cdf is F(x) = x/θ and the pdf is f(x) = 1/θ for 0 < x < θ. The pdf of the k-th order statistic out of n observations is the standard order-statistic density:

f(k)(x) = [n! / ((k−1)! (n−k)!)] · [F(x)]k−1 · [1−F(x)]n−k · f(x)

Substituting F(x) = x/θ and f(x) = 1/θ:

f(k)(x) = [n! / ((k−1)! (n−k)!)] · (x/θ)k−1 · (1 − x/θ)n−k · (1/θ), for 0 < x < θ

Now compute the expectation:

E[X(k)] = ∫0θ x · f(k)(x) dx

To simplify this integral, substitute u = x/θ, so x = θu and dx = θ du. As x runs from 0 to θ, u runs from 0 to 1:

E[X(k)] = ∫01 (θu) · [n! / ((k−1)! (n−k)!)] · uk−1 (1−u)n−k · θ du

E[X(k)] = θ · [n! / ((k−1)! (n−k)!)] · ∫01 uk (1−u)n−k du

The remaining integral is a standard Beta integral: ∫01 uk(1−u)n−k du = B(k+1, n−k+1) = k! (n−k)! / (n+1)!

Substituting this back:

E[X(k)] = θ · [n! / ((k−1)! (n−k)!)] · [k! (n−k)! / (n+1)!]

The (n−k)! terms cancel, and n!/(n+1)! = 1/(n+1), and k!/(k−1)! = k, leaving the clean result:

E[X(k)] = kθ / (n+1)

This single formula tells us the expectation of every order statistic of a Uniform(0, θ) sample — we just substitute the value of k that matches the order statistic we care about.

Step 3: Check Statement 1 — Is X(n) Unbiased for θ?

For the maximum, k = n. Plugging into our formula:

E[X(n)] = nθ / (n+1)

Since n/(n+1) < 1 for every finite n, we get E[X(n)] < θ. So X(n) systematically underestimates θ on average — it is biased, not unbiased. Statement 1 is therefore false.

(As a side note, this is exactly why the classic unbiased estimator built from the sample maximum is [(n+1)/n]·X(n) — the correction factor (n+1)/n exists precisely to cancel this n/(n+1) shrinkage.)

Step 4: Check Statement 2 — Is 2X((n+1)/2) Unbiased for θ?

Since n is odd, the middle order statistic corresponds to k = (n+1)/2, which is a whole number (this is exactly why the question insists n be odd — otherwise there is no single "middle" observation). Plugging k = (n+1)/2 into our formula:

E[X((n+1)/2)] = [(n+1)/2] · θ / (n+1) = θ/2

Now consider the proposed estimator 2X((n+1)/2):

E[2X((n+1)/2)] = 2 · E[X((n+1)/2)] = 2 · (θ/2) = θ

This comes out to exactly θ, for every value of θ. So 2X((n+1)/2) — twice the sample median —is an unbiased estimator of θ. Statement 2 is true.

Step 5: Match with the Given Options

We have found that statement 1 (X(n)) is biased, while statement 2 (2X((n+1)/2)) is unbiased. That matches option (b) 2 only.

Pro Tip:

So the correct answer is (b) 2 only, and this matches the official ISS 2016 Statistics Paper-2 answer key, which also lists option B as correct for Question 24.

Why This Question Matters

This question is a compact test of three ideas at once: knowing the density of an order statistic, being comfortable evaluating a Beta-type integral, and — most importantly — understanding that "unbiased" is a precise mathematical statement (E(T) = θ) rather than a vague notion of an estimator "seeming reasonable." The sample maximum feels like an obvious candidate for estimating θ, since it can never exceed θ, but that same property (X(n) ≤ θ always) is exactly why its average falls short of θ. Meanwhile, the median-based estimator, which looks less intuitive at first glance, turns out to be exactly right on average. Questions like this appear repeatedly across ISS and similar exams because they reward candidates who verify claims with algebra rather than intuition alone.

Frequently Asked Questions

Why isn't the sample maximum X(n) an unbiased estimator of θ?

Because X(n) can never exceed θ (every observation lies in (0, θ)), its average value across repeated samples is always somewhat less than θ. The exact calculation gives E[X(n)] = nθ/(n+1), which is strictly smaller than θ for any finite sample size n, so it systematically underestimates the parameter.

How can X(n) be corrected to make it unbiased?

Multiplying by the reciprocal of the shrinkage factor fixes it: since E[X(n)] = nθ/(n+1), the estimator T = [(n+1)/n]·X(n) has E(T) = θ exactly, making it unbiased. This is a very common follow-up twist on this type of question.

Why does the question specifically require n to be odd?

The middle order statistic only has a clean, single index when the sample size is odd — with n odd, (n+1)/2 is a whole number, so X((n+1)/2) is unambiguously "the" median observation. If n were even, there would be two middle observations, and you'd typically average them instead, which changes the calculation.

What is the general formula for the expectation of the k-th order statistic of a Uniform(0, θ) sample?

For X1, …, Xn iid Uniform(0, θ), the k-th order statistic satisfies E[X(k)] = kθ/(n+1). This single formula covers the minimum (k=1), any intermediate order statistic, and the maximum (k=n) — you just substitute the appropriate k.

Is the minimum order statistic X(1) unbiased for θ?

No. Using the same formula with k = 1 gives E[X(1)] = θ/(n+1), which is far smaller than θ for any reasonably sized sample, so X(1) is heavily biased downward. Like the maximum, it needs a correction factor — here, (n+1)·X(1) would be unbiased for θ.

Are order statistics of the Uniform distribution commonly tested in ISS Paper-2?

Yes, very much so — Uniform(0, θ) order statistics are a recurring favorite in ISS and similar statistics exams because they combine order statistic theory, Beta integrals, and estimation theory in a single compact question, and because the results (like the kθ/(n+1) formula) are clean enough to test in an MCQ format.

If any step above felt unclear, or you'd like to see the even-n version of this question worked out too, drop a comment below — and if this explanation helped, consider sharing it with a fellow ISS aspirant who might be stuck on the same topic.