ISS 2016 Statistics Paper-2 Solution: Question 32 (MLE for a Decreasing Linear Density)
Continuing our question-by-question walk through the ISS 2016 Statistics Paper-2, we arrive at Question 32 — a compact but conceptually rich Maximum Likelihood Estimation (MLE) problem. As always in this series, the goal is not just to name the correct option but to see exactly how that option is derived from first principles.
Quick Summary
Topic: Maximum Likelihood Estimation (MLE) for a decreasing linear density
Question Reference: ISS 2016, Statistics Paper-2, Question 32
Correct Answer: Option (b) —the MLE of α is 2x
The Question, As Asked
Given a random sample from
f(x, α) = (2 / α2)(α − x), 0 < x < α.
What is the MLE for α?
(a) x
(b) 2x
(c) x/2
(d) None of the above
Step 1: Write Down the Likelihood Function
The density is given as
f(x; α) = (2/α2)(α − x), for 0 < x < α, and zero otherwise.
For a single observed value x, the likelihood function L(α) is simply the density evaluated at that observed x, but now viewed as a function of the unknown parameter α rather than as a function of x. So:
L(α) = (2/α2)(α − x), for α > x.
Notice the support restriction: since the original density needs 0 < x < α, once x is observed and fixed, α is only allowed to range over values strictly greater than x. This constraint on the domain of α will matter later when we check that our stationary point is actually a valid, attainable maximum.
Step 2: Expand the Likelihood into a Simpler Form
Before differentiating, it helps to expand the expression so each term's dependence on α is visible on its own:
L(α) = (2/α2) · α − (2/α2) · x = 2/α − 2x/α-2
Written more carefully:
L(α) = 2α-1 − 2xα-2
This is now a simple sum of two power-of-α terms, which is much easier to differentiate than the original product form.
Step 3: Differentiate With Respect to α
Since we are maximizing over α (not x), we differentiate L(α) with respect to α and set the derivative equal to zero. Using the power rule term by term:
d/dα [2α-1] = −2α-2
d/dα [−2xα-2] = −2x · (−2)α-3 = 4xα-3
Adding these together:
dL/dα = −2α-2 + 4xα-3
It is convenient to bring this to a single fraction with denominator α3:
dL/dα = (−2α + 4x) / α3
Step 4: Set the Derivative to Zero and Solve
Setting the numerator equal to zero (since α3 ≠ 0 for any valid α > 0):
−2α + 4x = 0
2α = 4x
α = 2x
So the stationary point of the likelihood occurs exactly at α = 2x.
Step 5: Confirm This Is a Genuine Maximum, Not Just Any Stationary Point
Pro Tip: Students often stop the moment they get a number out of dL/dα = 0 and circle an option — but with MLE problems on a restricted support like α > x, you must also confirm the candidate point actually lies inside the valid range and behaves like a maximum, not a minimum or an inflection. The habit of checking boundary behaviour before finalising an MLE answer is drilled repeatedly in how these problems are usually taught, precisely because exam options are deliberately designed to trap students who skip this check.
Check the boundary behaviour of L(α) over its valid domain α > x:
As α → x+ (from above), the factor (α − x) → 0, so L(α) → 0.
As α → ∞, both 2/α → 0 and 2x/α2 → 0, so L(α) → 0 as well.
Since L(α) is a smooth, positive function on (x, ∞) that approaches zero at both ends of its range, and it has exactly one interior stationary point at α = 2x, that stationary point must be the unique maximum. We can also verify this is indeed inside the valid region: since x > 0, we have 2x > x, so α = 2x does satisfy the required constraint α > x. Everything is consistent.
Step 6: State the Final Answer
The maximum likelihood estimator of α is:
α̂ = 2x
This corresponds exactly to option (b), which also matches the official ISS 2016 Statistics Paper-2 answer key for Question 32. Our independent derivation and the official key are in full agreement — no discrepancy to flag here.
Why This Question Matters
This question tests a small but essential skill set inside Maximum Likelihood Estimation: correctly writing the likelihood as a function of the parameter (not the data), differentiating a rational expression in α, and — critically — checking the support restriction imposed by the density itself. Many MLE problems on bounded or parameter-dependent supports (like Uniform(0, θ) or, as here, a decreasing linear density on (0, α)) cannot be solved by blindly setting a derivative to zero; the parameter's own domain restriction is often part of the answer. Building comfort with this pattern pays off directly in Paper-2, where such "supported-by-the-parameter" densities appear repeatedly across different distributional families. This kind of density is exactly why the domain check in Step 5 above is not optional.
Frequently Asked Questions
Why is the likelihood function treated as a function of α instead of x?
In MLE, the data point x is already observed and fixed; what is unknown is the parameter α. So we flip our perspective on the same formula — instead of asking how likely x is for a given α, we ask which α makes the x we actually observed most probable. This is exactly what defines the likelihood function.
Could this same method be extended to a sample of n observations instead of just one?
Yes. With observations x1, x2, ..., xn, the joint likelihood would be the product of the individual densities, and differentiating the log-likelihood with respect to α would typically involve the largest order statistic, x(n), because the support constraint 0 < xi < α must hold for every observation simultaneously. This single-observation version is a simplified stepping stone to that more general case.
Why does the answer key rule out option (a), x, and option (c), x/2?
Option (a), α = x, is actually impossible under the model's own constraint, since the density requires strictly 0 < x < α, meaning α can never equal x. Option (c), α = x/2, would make α smaller than x whenever x is positive, which directly violates the same constraint. Only α = 2x is both a valid stationary point of the likelihood and consistent with the support condition.
Is this density related to the Uniform distribution?
Not directly, though both are supported on an interval involving the unknown parameter. Here the density is not flat — it decreases linearly from a peak at x = 0 down to zero at x = α, making it a triangular-type or decreasing linear density rather than a Uniform one. The shape of the density is exactly why differentiating the likelihood (rather than reasoning purely from order statistics, as with Uniform(0, θ)) is the natural approach here.
What is the most common mistake students make on this type of question?
The most frequent error is treating the exponent algebra sloppily when differentiating −2xα-2, particularly mishandling the negative sign that appears twice — once from the power rule and once from the pre-existing minus sign in the term. Writing out each derivative term separately, as done in Step 3 above, rather than trying to differentiate the whole expression in one mental leap, avoids this error reliably.
How should I quickly verify an MLE answer under exam time pressure?
A fast sanity check is to confirm your candidate estimator satisfies any constraint implied by the original support. If your algebra gives an estimator that could violate the stated domain (for instance, an α smaller than the observed x here), that is an immediate red flag that a sign or algebra error crept in, and it is worth rechecking the derivative before marking your final option.
That wraps up Question 32 of the ISS 2016 Statistics Paper-2 series. If any step above felt unclear or you spotted something worth discussing further, drop a comment below — and if this walkthrough helped clarify the idea, consider sharing it with a fellow ISS aspirant who might be stuck on the same question.
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