ISS 2016 Statistics Paper-2 Solution: Question 28 (Minimum Variance Linear Combination)

Continuing our question-by-question walk through the ISS Statistics Paper-2 archive, today we land on Question 28 from the 2016 paper. This one sits in the estimation theory portion of the syllabus and is a favourite type of question in ISS because it tests whether you actually understand what "minimum variance" means, rather than just knowing the definition of unbiasedness.

Quick Summary

• Topic: Minimum variance unbiased linear combination of two independent estimators

• Question Reference: ISS 2016, Statistics Paper-2, Question 28

• Correct Answer: (a) 2/3

The Question

If T1 and T2 are independent and unbiased estimators of parameter θ with V(T1) = σ12 and V(T2) = 2σ12, then the unbiased estimator of θ given by T = λT1 + (1 − λ)T2 has minimum variance if λ equals :

• (a) 2/3

• (b) 1/2

• (c) 1/3

• (d) 1/4

Step 1: Check That T Is Already Unbiased

Before touching variance, it is worth noticing why the question already calls T = λT1 + (1 − λ)T2 "unbiased" for every value of λ. Since E(T1) = θ and E(T2) = θ (both are given as unbiased estimators of θ), we get:

E(T) = λE(T1) + (1 − λ)E(T2) = λθ + (1 − λ)θ = θ

This holds for any λ, which is exactly why the coefficients are written as λ and (1 − λ) instead of two independent constants — it forces unbiasedness automatically, leaving variance as the only thing left to optimise. This is the whole point of the question: among the infinitely many unbiased combinations of T1 and T2, we must pick the one with the smallest variance.

Step 2: Write Down the Variance of T

Since T1 and T2 are independent, the variance of a linear combination is simply the sum of the variances of the individual terms (there is no covariance term to worry about):

V(T) = V(λT1) + V((1 − λ)T2) = λ2V(T1) + (1 − λ)2V(T2)

Now substitute the given values V(T1) = σ12 and V(T2) = 2σ12:

V(T) = λ2σ12 + (1 − λ)2(2σ12) = σ12[λ2 + 2(1 − λ)2]

Since σ12 is just a positive constant multiplying the whole expression, minimising V(T) over λ is the same as minimising the bracketed function g(λ) = λ2 + 2(1 − λ)2.

Step 3: Differentiate and Solve for λ

To find the minimum, differentiate g(λ) with respect to λ and set the derivative to zero. Expand (1 − λ)2 = 1 − 2λ + λ2 first so the differentiation is straightforward:

g(λ) = λ2 + 2(1 − 2λ + λ2) = λ2 + 2 − 4λ + 2λ2 = 3λ2 − 4λ + 2

Now differentiate term by term:

g'(λ) = 6λ − 4

Setting g'(λ) = 0:

6λ − 4 = 0 ⟹ λ = 4/6 = 2/3

Pro Tip: Many students get the algebra right up to g'(λ) = 0 and then second-guess themselves on whether λ = 2/3 or λ = 1/3 is correct, because both numbers "look similar" to the fractions in the problem. The way this is usually taught to avoid that slip is to always run a quick second-derivative or boundary sanity check: g''(λ) = 6, which is positive, confirming λ = 2/3 is indeed a minimum, not a maximum. As a rough intuition check, T1 has the smaller variance (σ12) compared to T2 (2σ12), so it makes sense that the more precise estimator T1 should get the larger weight — and 2/3 > 1/2, so T1 is indeed weighted more heavily than T2, which matches intuition.

Step 4: Confirm It Is a Minimum, Not a Maximum

Differentiating g'(λ) = 6λ − 4 once more gives g''(λ) = 6, which is strictly positive for all λ. A positive second derivative confirms that λ = 2/3 is a point of minimum variance, exactly as the question asks for — not a maximum or a saddle point.

Final Answer

The unbiased estimator T = λT1 + (1 − λ)T2 has minimum variance when λ = 2/3, which is option (a). This matches the official ISS 2016 Statistics Paper-2 answer key, which also lists (a) as the correct response for Question 28.

Why This Question Matters

This question is a compact test of three ideas working together: linearity of expectation (to see why unbiasedness is automatic here), the additivity of variance under independence, and basic calculus-based optimisation. ISS Paper-2 repeatedly rewards candidates who can set up a variance expression correctly and then differentiate cleanly under exam pressure, since a single sign error in expanding (1 − λ)2 is enough to flip the final answer. Students who drill this exact pattern — express variance as a function of the weighting parameter, differentiate, and sanity-check with the second derivative — until it becomes automatic tend to solve this entire family of "optimal weighting" questions in under two minutes, which matters a great deal in a two-hour, 80-question paper.

Frequently Asked Questions

Why is there no covariance term in the variance formula for T?

The question explicitly states that T1 and T2 are independent. For independent random variables, Cov(T1, T2) = 0, so the variance of any linear combination reduces to the sum of the individually weighted variances. If T1 and T2 were merely uncorrelated but not independent, the same simplification would still hold, but here independence is given directly, so there is no ambiguity.

What would change if V(T2) were σ12 instead of 2σ12?

If both estimators had equal variance, the symmetry of the problem would force the optimal weight to λ = 1/2, splitting the weight equally between T1 and T2. The general result for this type of problem is that the optimal weight on T1 equals V(T2) divided by [V(T1) + V(T2)], which you can verify gives 2σ12 / (σ12 + 2σ12) = 2/3 for this exact question.

Is this related to the concept of a "pooled estimator" in statistics?

Yes, this is precisely the logic behind inverse-variance weighting, which appears again later in topics like combining estimates from stratified samples or meta-analysis. The estimator with smaller variance (more precision) is always given a larger weight, and this question is essentially a two-estimator special case of that general principle.

Could I have solved this without calculus, just using the general formula?

Yes. If you memorise the general result that the optimal weight for T1 in a two-estimator unbiased combination is V(T2)/[V(T1) + V(T2)], you can substitute directly and get 2σ12/(3σ12) = 2/3 in a single line. However, working through the differentiation at least once, as done above, is important so you understand where that formula comes from and can rederive it if you forget it under exam pressure.

Does this estimator T achieve the Cramer-Rao lower bound?

Not necessarily. This question only asks for the minimum variance among the linear combinations of T1 and T2 of this specific form; it says nothing about whether that minimum variance touches the Cramer-Rao lower bound for θ in general. Achieving the CRLB is a separate, stronger property that depends on the underlying distribution and the specific estimators T1 and T2 chosen.

Why does the answer key mark this as option (a) and not (c) 1/3?

Option (c), λ = 1/3, is a common distractor for students who correctly differentiate but then either drop a sign or solve 6λ − 4 = 0 incorrectly as λ = 4/12. Carefully redoing the last algebraic step, 6λ = 4 so λ = 4/6 = 2/3, confirms that (a) is correct and (c) is simply the reciprocal-looking wrong answer designed to catch a rushed calculation.

If any step above felt unclear, drop a comment below and we will walk through it in more detail — and if you know a fellow ISS aspirant working through the same paper, feel free to share this post with them.