ISS 2016 Statistics Paper-2 Solution: Question 63 (MLE vs Method of Moments for Uniform(0,a))
Continuing our question-by-question walk through the ISS Statistics Paper-2 archive, this post takes up Question 63 from the 2016 paper. It's a classic point-estimation question built around the uniform distribution, and it's a good one to slow down on because it tests two estimation methods side by side rather than just one formula.
Quick Summary
Topic: Point Estimation — Maximum Likelihood Estimator (MLE) vs. Method of Moments (MOM) for Uniform(0, a)
Question Reference: ISS 2016, Statistics Paper-2, Question 63
Correct Answer (as derived in this post): Option (a) — Statements 2, 3 and 4 are correct
The Question
If X1, X2, X3, ..., Xn is a random sample from the uniform distribution on [0, a] where a ∈ (0, ∞), then which of the following statements are correct?
Maximum likelihood estimator of 'a' is same as the moment estimator of 'a'
Maximum likelihood estimator of 'a' is X(n) = Max{X1, X2, X3, ..., Xn} = â
(n+1)/n · â is an unbiased estimator of 'a' where â is the maximum likelihood estimator of 'a'
Method of moments estimator is an unbiased estimator of 'a'
Select the correct answer using the code given below:
(a) 2, 3 and 4
(b) 1, 3 and 4
(c) 1, 2 and 4
(d) 1, 2 and 3
Step 1: Set Up the Likelihood Function
X follows Uniform(0, a), so its pdf is f(x; a) = 1/a for 0 ≤ x ≤ a, and 0 otherwise. For a random sample of size n, the joint density (the likelihood) is:
L(a) = ∏i=1n f(xi; a) = 1/an, provided every xi lies in [0, a], i.e. provided a ≥ max(x1, ..., xn).
If even one xi exceeds a, the likelihood is exactly 0, because that observation would be impossible under Uniform(0, a). So the likelihood is only ever defined (and positive) when a is at least as large as the sample maximum.
Step 2: Find the MLE by Reasoning, Not Calculus
Here's the part that trips up a lot of students coming from "take the derivative and set it to zero" habits: you cannot differentiate L(a) = 1/an with respect to a and set it to zero here, because that expression is strictly decreasing in a — the derivative is never zero on the valid range. Instead, the maximization has to be done by direct reasoning about the shape of the function.
Since L(a) = 1/an is a decreasing function of a, it is largest when a is as small as possible. But a cannot be made arbitrarily small — it is constrained by a ≥ max(x1, ..., xn). So the smallest permissible value of a, and therefore the value that maximizes L(a), is:
âMLE = X(n) = Max{X1, X2, ..., Xn}
This confirms Statement 2 is correct.
Step 3: Find the Method of Moments Estimator
The method of moments equates the first population moment (the mean) to the first sample moment. For X ~ Uniform(0, a):
E(X) = (0 + a)/2 = a/2
Setting the population mean equal to the sample mean X̄ and solving for a:
X̄ = a/2 ⇒ âMOM = 2X̄
Compare this with the MLE from Step 2: âMLE = X(n) (the maximum of the sample), while âMOM = 2X̄ (twice the sample mean). These are two structurally different statistics — one depends only on the single largest observation, the other depends on the average of all observations — and for almost any actual sample they will give different numerical values.The MLE and the method-of-moments estimator are therefore not the same estimator, which means Statement 1 is false.
Pro Tip: A quick sanity check that saves you from this exact trap in the exam hall — if an MLE is a function of just one order statistic (like the max or the min) while the MOM is built from a sample mean, they almost never coincide, except perhaps by numerical accident on a specific sample. Whenever a paper tempts you to call "MLE = MOM" a general rule, go back and write the two formulas side by side before answering. This is exactly the kind of habit that gets drilled in when students work through enough of these estimation problems systematically — it becomes an automatic first check rather than a conscious one.
Step 4: Check Whether the MLE Is Biased, and Fix It
A natural next question: is âMLE = X(n) itself an unbiased estimator of a? To check, we need E[X(n)].
The pdf of the maximum of n iid Uniform(0, a) variables is g(y) = n yn-1/an for 0 < y < a. So:
E[X(n)] = ∫0a y · n yn-1/an dy = (n/an) ∫0a yn dy = (n/an) · an+1/(n+1) = na/(n+1)
Since na/(n+1) ≠ a (it's strictly less than a for finite n), X(n) on its own is a biased estimator — it systematically underestimates a, because the sample maximum can never exceed a but is also, on average, a little short of it.
Now apply the correction suggested in Statement 3:
E[(n+1)/n · X(n)] = (n+1)/n · na/(n+1) = a
The (n+1)/n factor exactly cancels the shortfall, so (n+1)/n · â is unbiased for a.Statement 3 is correct. (This is the same idea behind the famous "German tank problem" — estimating a total count from the largest serial number observed.)
Step 5: Check Whether the Method of Moments Estimator Is Unbiased
From Step 3, âMOM = 2X̄. Taking expectation:
E[âMOM] = E[2X̄] = 2E[X̄] = 2 · E(X) = 2 · (a/2) = a
Since E[X̄] = E(X) = a/2 always holds for an iid sample (the sample mean is an unbiased estimator of the population mean, regardless of the distribution), âMOM = 2X̄ is unbiased for a with no correction needed.Statement 4 is correct.
Final Answer and a Flagged Discrepancy
Putting the four statements together:
Statement 1 (MLE = MOM):False
Statement 2 (MLE = X(n)):True
Statement 3 ((n+1)/n · â unbiased):True
Statement 4 (MOM unbiased):True
So the correct statements are 2, 3 and 4, which corresponds to Option (a).
Important note on the official answer key:the official ISS 2016 Statistics Paper-2 answer key (Series A) marks the answer to Question 63 as Option (d) — "1, 2 and 3." Having derived each statement independently from first principles above (and additionally verified the unbiasedness results by Monte Carlo simulation), this derivation is confident that Statement 1 is false (the MLE and the method-of-moments estimator are genuinely different formulas, X(n) versus 2X̄) and that Statement 4 is true (2X̄ is unbiased for a by direct expectation, with no approximation involved). This points to the published key likely containing a misprint or indexing slip for this item — a known, if infrequent, occurrence in these keys. Readers are encouraged to work through Steps 2–5 above themselves and compare; if you find a flaw in this reasoning, a correction is genuinely welcome in the comments.
Why This Question Matters
Question 63 packs three separate estimation-theory ideas — deriving an MLE on a bounded support, recognizing when MLE and MOM diverge, and checking/fixing bias via a correction factor — into a single four-way statement question. This combination shows up constantly across ISS, IIT JAM(MS), and similar exams, because the Uniform(0, a) case is the simplest setting where "take the derivative and set it to zero" fails and students are forced to actually reason about likelihood as a function. Getting comfortable with boundary-support MLEs here pays off directly on order-statistic and extreme-value questions elsewhere in Paper-2.
Frequently Asked Questions
Why can't I just differentiate the likelihood function to find the MLE here?
Because L(a) = 1/an is monotonically decreasing in a over its valid domain, its derivative is never zero — there is no interior stationary point. The maximum instead occurs at the boundary of the constraint set (a ≥ max of the sample), which is why you reason about monotonicity and the support condition rather than solving dL/da = 0.
Is the sample maximum X(n) always a biased estimator?
For Uniform(0, a), yes — E[X(n)] = na/(n+1), which is always less than a for finite n, so X(n) systematically underestimates a. The bias shrinks as n grows, since na/(n+1) approaches a as n approaches infinity, but it is never exactly zero for finite samples.
Why does multiplying by (n+1)/n fix the bias?
Because the bias has a known, exact multiplicative form: E[X(n)] equals a scaled by the factor n/(n+1). Multiplying by the reciprocal of that exact scaling factor, (n+1)/n, cancels it precisely, leaving an estimator whose expectation is exactly a.
Are MLE and method-of-moments estimators ever the same?
They coincide for some distributions (for example, both estimators equal the sample mean for a Normal distribution with known variance) but there is no general theorem guaranteeing this. Whenever the MLE depends on an order statistic (like a minimum or maximum) rather than a sum or mean, it is very unlikely to match the method-of-moments estimator, which is always built from sample moments.
What's an easy way to remember the method of moments for Uniform(0, a)?
Just remember that the mean of a Uniform(0, a) distribution is the midpoint, a/2. Setting the sample mean equal to this gives X̄ = a/2, so the method-of-moments estimator is simply twice the sample mean, â = 2X̄. No integration is needed beyond knowing this one mean formula.
Why does the exam test both MLE and MOM on the same distribution instead of just one?
Because comparing two different estimation philosophies on the same, simple distribution is an efficient way to test whether a candidate understands that "an estimator" is not unique — different estimation principles (likelihood-based vs. moment-matching) can produce different formulas with different bias and efficiency properties, even though both are reasonable ways to estimate the same parameter.
If any step above wasn't clear, or if you've spotted an issue with the discrepancy flagged against the official key, drop a comment below — and if this helped you untangle MLE versus MOM, consider sharing it with a fellow ISS aspirant working through the same paper.
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