ISS 2016 Statistics Paper-2 Solution: Question 1 (Condition for Uncorrelated BLUEs)

Welcome to a new question-by-question run through the ISS (Indian Statistical Service) Statistics Paper-2 previous year papers. This series works strictly in order, one MCQ at a time, and today we are opening the very first question of the 2016 paper — a neat little result from linear estimation theory that trips up a lot of first-time solvers because it looks harder than it actually is.

Quick Summary

• Topic: BLUE (Best Linear Unbiased Estimator) and the Gauss-Markov linear model

• Question Reference: ISS 2016, Statistics Paper-2, Question 1

• Correct Answer: Option (a), α = −1

Here is the question exactly as it appears in the test booklet:

Question 1.For what value of α, the BLUE of β1 and β2 are uncorrelated for the model

E(y1) = 2β1 + β2, E(y2) = β1 − β2, E(y3) = β1 + αβ2 ?

• (a) −1

• (b) 0

• (c) 1

• (d) 2

Step 1: Write the Model in Matrix Form

Before touching any formula, the first job is always to translate the three expectation equations into a standard linear model y = Xβ + ε. We have three observations y1, y2, y3, and two unknown parameters β1 and β2. Reading off the coefficients of β1 and β2 in each equation gives us the design matrix X:

Row 1 (from E(y1) = 2β1 + β2): coefficients (2, 1) Row 2 (from E(y2) = β1 − β2): coefficients (1, −1) Row 3 (from E(y3) = β1 + αβ2): coefficients (1, α)

So the design matrix is

X = [[2, 1], [1, −1], [1, α]]

with β = (β1, β2)′. As is standard for this kind of problem (and as the Gauss-Markov theorem requires), we assume the errors ε1, ε2, ε3 have mean zero, common variance σ², and are uncorrelated with each other.

Step 2: Recall What "BLUE" Actually Means Here

Under exactly these Gauss-Markov assumptions, the Gauss-Markov theorem tells us that the Best Linear Unbiased Estimator (BLUE)of β is simply the ordinary least squares estimator:

β̂ = (X′X)−1 X′y

and its variance-covariance matrix is

Var(β̂) = σ² (X′X)−1

This single fact is really the whole engine of the problem: since β̂1 and β̂2 are the two components of β̂, whether they are correlated or not is completely decided by the off-diagonal entry of Var(β̂) — equivalently, by the off-diagonal entry of (X′X)−1, since σ² is just a positive scalar multiplier that cannot change whether an entry is zero.

Step 3: Compute X′X

Now we actually build the 2×2 matrix X′X from the design matrix above. Writing the two columns of X as column 1 = (2, 1, 1)′ and column 2 = (1, −1, α)′, the entries of X′X are just the dot products of these columns with themselves and with each other:

(X′X)11 = 2² + 1² + 1² = 4 + 1 + 1 = 6

(X′X)22 = 1² + (−1)² + α² = 1 + 1 + α² = 2 + α²

(X′X)12 = (X′X)21 = (2)(1) + (1)(−1) + (1)(α) = 2 − 1 + α = 1 + α

So X′X = [[6, 1 + α], [1 + α, 2 + α²]].

Step 4: Invert the 2×2 Matrix (Conceptually)

For any invertible 2×2 symmetric matrix M = [[a, b], [b, d]], the standard inverse formula is

M−1 = (1 / (ad − b²)) × [[d, −b], [−b, a]]

Look carefully at the off-diagonal entry of M−1: it is −b divided by the determinant. The determinant (ad − b²) is just a nonzero number when the matrix is invertible — it can never make the off-diagonal entry equal to zero by itself. The only way for the off-diagonal entry of M−1 to vanish is for the numerator, −b, to be zero. In other words:

(X′X)−1 has a zero off-diagonal entry  ⇔  X′X has a zero off-diagonal entry (i.e., b = 0)

This is the shortcut that saves you from actually writing out the full inverse matrix on an MCQ — you only need to set the off-diagonal entry of X′X itself to zero.

Pro Tip: The step most students fumble here is assuming they must fully invert X′X, compute the determinant, and then set the resulting off-diagonal fraction to zero. That is unnecessary extra algebra under exam time pressure. Because the determinant of an invertible matrix is just some nonzero real number, it can never force a ratio to zero — only the numerator can. The way this shortcut is usually taught is: for any invertible 2×2 (or even larger, diagonal-block) covariance-type matrix, "uncorrelated estimators" is equivalent to "zero off-diagonal entry in X′X," full stop. Once this observation is internalised, this question becomes a 20-second calculation instead of a two-minute one — and it is exactly the kind of pattern-recognition that repays itself many times over across a Paper-2 sitting.

Step 5: Solve for α

Applying the condition from Step 4 to our matrix:

(X′X)12 = 0

1 + α = 0

α = −1

Step 6: Final Answer and Confirmation Against the Official Key

The value of α for which β̂1 and β̂2 — the BLUEs of β1 and β2 — are uncorrelated is α = −1, which is option (a). This matches the official ISS 2016 Statistics Paper-2 answer key exactly, which also lists option (a) as correct for Question 1. So our from-scratch derivation and the official key agree — no discrepancy to flag on this one.

Why This Question Matters

This question is a compact test of three things at once: whether you can translate a word-and-symbol description of a linear model into a design matrix, whether you remember that the Gauss-Markov theorem hands you the BLUE as the OLS estimator with variance σ²(X′X)−1, and whether you can read a correlation condition off a 2×2 matrix without unnecessary algebra. Questions built exactly this way — a short setup that quietly tests two or three linked concepts — show up again and again across ISS Paper-2 linear models and design of experiments sections, so getting comfortable with this exact chain of reasoning (model → X′X → off-diagonal condition) pays off well beyond this single question.

Frequently Asked Questions

What does it mean for two estimators to be "uncorrelated" in this context?

It means the covariance between β̂1 and β̂2, the estimated values of β1 and β2, is zero. Since Var(β̂) = σ²(X′X)−1, this covariance is exactly σ² times the off-diagonal entry of (X′X)−1, so uncorrelated estimators correspond to a zero off-diagonal entry in that inverse matrix.

Why is the OLS estimator the same as the BLUE here?

The Gauss-Markov theorem states that under the assumptions of zero-mean, equal-variance, uncorrelated errors, the ordinary least squares estimator β̂ = (X′X)−1X′y is the Best Linear Unbiased Estimator — meaning it has the smallest variance among all linear unbiased estimators of β. No question stem needs to explicitly say "assume Gauss-Markov conditions"; it is the standard backdrop whenever "BLUE" is mentioned for a linear model like this one.

Do I always need to compute the full inverse of X′X for these questions?

No, and this is the main time-saver in this question. Since the determinant of an invertible matrix is a nonzero scalar, it can never make an off-diagonal ratio equal to zero on its own — only a zero numerator can. So for a 2×2 X′X matrix, you only need to set the off-diagonal entry of X′X itself equal to zero, skipping the full inverse computation entirely.

What if X′X were not invertible?

If X′X is singular, ordinary least squares does not have a unique solution, and the BLUE in the usual sense does not exist without further restrictions (such as using a generalized inverse or reparametrising the model to remove non-identifiable combinations of parameters). This question implicitly assumes X′X is invertible for every value of α except any value that makes it singular, and α = −1 does not cause that problem here.

Is this result specific to a 3-observation, 2-parameter model, or does it generalise?

The underlying idea generalises completely: for any linear model y = Xβ + ε satisfying Gauss-Markov assumptions, two components β̂i and β̂j of the BLUE are uncorrelated exactly when the (i, j) entry of (X′X)−1 is zero. The specific shortcut used here — reducing that to a condition on X′X directly — works cleanly whenever you are dealing with a 2×2 block, and with more parameters you would need the corresponding cofactor of X′X to vanish instead.

How should I practise this type of question further?

Try rebuilding the design matrix from a few other linear model statements you find in Paper-2 papers, computing X′X by hand each time, and asking what condition on the model coefficients would zero out a chosen off-diagonal entry. Doing this repeatedly until the pattern feels automatic — rather than re-deriving the logic from first principles every time — is exactly how this class of BLUE and Gauss-Markov questions is meant to be drilled.

If any step above felt unclear, or you worked through it differently, drop a comment below — questions and alternate approaches are always welcome. And if you know a fellow ISS aspirant working through the same Paper-2 papers, feel free to share this post with them; solving these one at a time, together, tends to make the whole syllabus far less intimidating.