ISS 2016 Statistics Paper-2 Solution: Question 30 (Consistent Estimator of Gamma Variance)

This post continues our ongoing question-by-question walkthrough of the ISS Statistics Paper-2 previous year papers. Today we work through Question 30 from the 2016 paper, which sits inside the estimation-theory block of the syllabus and tests whether you can connect a probability distribution's moments to a consistent estimator. As always, every algebraic step is shown in full so that even a first-time reader can follow the logic without needing outside help.

Quick Summary

• Topic:Point estimation — consistent estimators built from sample moments of the Gamma distribution

• Question Reference:ISS 2016, Statistics Paper-2, Question 30

• Correct Answer:Option (c) — Σxi2 ⁄ [nα(α+1)]

The Question As Asked

Let X1, X2, X3, ..., Xn be a random variable with pdf

f(x; α, β) = [1 ⁄ (βα Γ(α))] e−x/β xα−1, for x, α, β > 0, and 0 otherwise.

What is the consistent estimator for β2?

• (a) Σxi2 ⁄ [nα(α+1)2]

• (b) Σxi2 ⁄ [nα2(α+1)]

• (c) Σxi2 ⁄ [nα(α+1)]

• (d) Σxi2 ⁄ [n2α(α+1)]

Step 1: Recognise the Distribution

The pdf given is the standard two-parameter Gamma distribution, written here with shape parameter α and scale parameter β. All four answer options are built purely from α and Σxi2, which is a strong hint: the question wants us to relate E(X2) of this Gamma distribution to β2, and then invert that relationship to build an estimator out of the sample sum of squares.

Step 2: Compute the k-th Raw Moment

Start from the definition of expectation for a continuous random variable:

E(Xk) = ∫0∞ xk · [1 ⁄ (βα Γ(α))] xα−1 e−x/β dx

Combine the powers of x first:

E(Xk) = [1 ⁄ (βα Γ(α))] ∫0∞ xα+k−1 e−x/β dx

Now substitute u = x/β, so x = βu and dx = β du. The limits stay 0 to ∞ since β > 0:

∫0∞ (βu)α+k−1 e−u β du = βα+k ∫0∞ uα+k−1 e−u du = βα+k Γ(α+k)

The last integral is just the Gamma function definition, Γ(α+k) = ∫0∞ uα+k−1 e−u du. Substituting this back:

E(Xk) = [1 ⁄ (βα Γ(α))] · βα+k Γ(α+k) = βk · [Γ(α+k) ⁄ Γ(α)]

This one formula gives every raw moment of the Gamma distribution just by plugging in k.

Step 3: Get E(X) and E(X2) From the General Formula

For k = 1:

E(X) = β · [Γ(α+1) ⁄ Γ(α)]

Using the standard Gamma function recurrence Γ(α+1) = α Γ(α), this simplifies to:

E(X) = β · [α Γ(α) ⁄ Γ(α)] = αβ

This is the familiar mean of the Gamma distribution — good, it confirms our formula is correct. Now for k = 2:

E(X2) = β2 · [Γ(α+2) ⁄ Γ(α)]

Apply the recurrence twice: Γ(α+2) = (α+1) Γ(α+1) = (α+1) · α Γ(α). So:

E(X2) = β2 · [(α+1) α Γ(α) ⁄ Γ(α)] = α(α+1) β2

Step 4: Turn E(X2) Into an Estimator

By the Weak Law of Large Numbers, the sample second moment converges in probability to the population second moment as n → ∞:

(1/n) Σ Xi2 →p E(X2) = α(α+1) β2

We want an estimator for β2 itself, not for α(α+1)β2. Since α is treated as known here (only β2 is being estimated), we simply divide both sides by the constant α(α+1):

[(1/n) Σ Xi2] ⁄ [α(α+1)] →p β2

Dividing a sequence that converges in probability by a fixed nonzero constant still converges in probability to the corresponding limit (this is a direct consequence of the continuous mapping / Slutsky results), so the left-hand side is itself a consistent estimator of β2. Rewriting it in the same form as the options:

Estimator of β2 = Σ Xi2 ⁄ [n α(α+1)]

This matches option (c) exactly.

Pro Tip: the part students most often get wrong here isn't the integration — it's forgetting that "consistent estimator" only requires convergence in probability to the parameter, not unbiasedness in finite samples. You don't need E(estimator) = β2 exactly for every n; you only need the estimator to settle down onto β2 as n grows. The way this derivation is usually taught is to first nail down the general raw-moment formula E(Xk) = βkΓ(α+k)/Γ(α) for the Gamma distribution once and for all, because it instantly answers questions about the 1st, 2nd, or even 3rd moment without repeating the substitution each time — students who drill this one derivation until it's automatic tend to solve an entire family of Gamma-based estimation questions in under a minute.

Step 5: Confirm Against the Official Answer Key

The official ISS 2016 Statistics Paper-2 answer key lists the correct response to Question 30 as option (c), which is exactly what our derivation produced. So the final answer is confirmed:

Final Answer: (c) Σxi2 ⁄ [nα(α+1)]

Why This Question Matters

This question is a compact test of three separate skills at once: recognising a named distribution from its pdf, deriving moments using the Gamma function's substitution trick, and correctly applying the definition of consistency via the Weak Law of Large Numbers. That combination shows up repeatedly across ISS and similar statistical services exams, not just for the Gamma distribution but for the Exponential, Chi-square, and Beta families too, since they all share the same integration pattern. Once you can derive E(Xk) for one member of this family, you can do it for all of them with almost no extra work.

Frequently Asked Questions

What exactly does "consistent estimator" mean here?

An estimator Tn is consistent for a parameter θ if, as the sample size n grows without bound, Tn converges in probability to θ. In plain terms, the estimator's values cluster closer and closer to the true parameter as you collect more data, even if the estimator is slightly biased for small samples.

Why did we treat α as a known constant in this problem?

The question asks specifically for the estimator of β2, and the answer choices are all expressed purely in terms of α and the sample data — none of them attempt to estimate α itself. That tells us α is being held fixed and known, so it is legitimate to divide by α(α+1) as if it were an ordinary constant.

How is the Gamma distribution's mean and variance related to this question?

For a Gamma(α, β) distribution, the mean is αβ and the variance is αβ2. Since E(X2) = Var(X) + [E(X)]2, you get E(X2) = αβ2 + (αβ)2 = αβ2(1+α), which is the same α(α+1)β2 we derived directly — a useful shortcut once you already know the mean and variance formulas by heart.

Could the sample variance be used instead of the sample second moment here?

No, not directly. The sample variance estimates Var(X) = αβ2, which would give a different-looking (but related) estimator for β2. The question specifically frames the options in terms of Σxi2, which points to the raw second moment E(X2) rather than the central second moment.

Is this estimator unique, or could other consistent estimators for β2 exist?

It is not unique — many other functions of the sample could also converge in probability to β2. This particular one is favoured because it is the simplest and most natural estimator arising directly from the method of moments applied to E(X2).

What's the most common mistake candidates make on this type of question?

The most frequent error is mixing up Γ(α+1) = αΓ(α) with Γ(α+2) = (α+1)Γ(α+1), and stopping one recurrence step too early, which leaves an extra or missing factor of α or (α+1) in the final answer — exactly the kind of slip that separates options (a), (b), and (d) from the correct option (c).

If any step above felt rushed or you worked through it differently, drop a comment below — questions like this are much easier to lock in once you've talked through the tricky substitution out loud. And if you know a fellow ISS aspirant working through the same paper, sharing this post with them costs you nothing but might save them a genuinely confusing half hour.