ISS 2016 Statistics Paper-2 Solution: Question 1 (BLUE Uncorrelated Estimators)

This is the first post in our ongoing question-by-question walkthrough of ISS Statistics Paper-2 previous year papers, starting right from the beginning of the 2016 paper. The goal of this series is simple: take one question at a time and solve it so thoroughly that you could reconstruct the entire method from scratch, without needing to memorise any shortcut blindly.

Quick Summary

• Topic: Best Linear Unbiased Estimators (BLUE), covariance structure of least squares estimators, Gauss-Markov setup

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

• Correct Answer: Option (a) α = -1

The Question

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: Translate the Model into Matrix Form

Before touching any formula, it helps to see exactly what kind of model this is. We are told the expected values of three observations y1, y2, y3, each written as a linear combination of two unknown parameters β1 and β2. Writing each equation as an actual observation plus an error term:

• y1 = 2β1 + β2 + ε1

• y2 = β1 - β2 + ε2

• y3 = β1 + αβ2 + ε3

This is exactly the setup of a linear model y = Xβ + ε, where β = (β1, β2)' is the vector of unknown parameters, and X is the "design matrix" built purely from the known coefficients attached to β1 and β2 in each equation. Reading off the coefficients row by row:

X = rows (2, 1), (1, -1), (1, α) — one row for each observation, one column for each parameter.

Under the usual Gauss-Markov assumptions (errors have mean zero, equal variance σ², and are uncorrelated with each other), the Gauss-Markov theorem tells us that the Best Linear Unbiased Estimator (BLUE) of β is exactly the ordinary least squares estimator, β̂ = (X'X)-1 X'y. So to study the correlation between β̂1 and β̂2, we need to study the matrix (X'X)-1.

Step 2: Compute X'X

X'X is always a small, symmetric matrix with as many rows and columns as there are parameters — here it's 2 by 2. Each entry is a sum of products of the corresponding columns of X. Let's call column 1 of X (the coefficients of β1) as (2, 1, 1), and column 2 of X (the coefficients of β2) as (1, -1, α).

The (1,1) entry of X'X is the sum of squares of column 1:

2² + 1² + 1² = 4 + 1 + 1 = 6

The (2,2) entry of X'X is the sum of squares of column 2:

1² + (-1)² + α² = 1 + 1 + α² = 2 + α²

The (1,2) entry (which equals the (2,1) entry, since X'X is always symmetric) is the sum of the products of corresponding elements of column 1 and column 2:

(2)(1) + (1)(-1) + (1)(α) = 2 - 1 + α = 1 + α

So we get:

X'X = [ [6, 1+α], [1+α, 2+α²] ]

Step 3: Invert X'X to Find the Covariance Matrix of β̂

For a general 2 by 2 matrix M = [ [p, q], [q, s] ], the inverse is M-1 = (1/det) × [ [s, -q], [-q, p] ], where det = ps - q².

Applying this with p = 6, q = 1+α, s = 2+α², the determinant is:

det = 6(2 + α²) - (1 + α)² = 12 + 6α² - (1 + 2α + α²) = 11 - 2α + 5α²

And the inverse matrix is:

(X'X)-1 = (1/det) × [ [2+α², -(1+α)], [-(1+α), 6] ]

Since the covariance matrix of the BLUE is Cov(β̂) = σ² (X'X)-1, the off-diagonal entry of this covariance matrix — the covariance between β̂1 and β̂2 — is proportional to -(1+α)/det.

Pro Tip: the trickiest part of this question isn't the algebra, it's recognising that you never actually need to simplify the full (X'X)-1 matrix or worry about σ² or the determinant at all. "Uncorrelated" only requires the off-diagonal entry of the covariance matrix to be zero, and since σ² and the determinant are both just scalar multipliers (never zero for a valid design), the entire question collapses to one simple equation: set the numerator of the off-diagonal term to zero. Students who train themselves to spot this kind of structural shortcut — instead of grinding through a full matrix inversion every time — end up finishing these objective questions in under a minute instead of five.

Step 4: Solve for α

Setting the off-diagonal entry of (X'X)-1 to zero means setting its numerator to zero (the denominator, det, is just a positive scaling constant here and can never make the fraction zero on its own):

-(1 + α) = 0

1 + α = 0

α = -1

Step 5: Confirm the Final Answer

So the value of α for which β̂1 and β̂2 are uncorrelated is α = -1, which is option (a). This matches the official answer key for this question, confirming the derivation above is correct.

As a sanity check, you can plug α = -1 back into X'X: the off-diagonal entry becomes 1 + (-1) = 0, meaning X'X itself becomes a diagonal matrix. A diagonal X'X always inverts to another diagonal matrix, which is a clean, quick way to double-check your answer without redoing the full inversion.

Why This Question Matters

This question tests a concept that shows up repeatedly across ISS Paper-2 linear models and regression questions: the link between the structure of the design matrix X and the statistical properties (variance, covariance) of the resulting parameter estimates. Many candidates know the formula β̂ = (X'X)-1X'y by heart but freeze the moment a question asks about a property of the estimator — like independence or uncorrelatedness — rather than the estimator itself. Being comfortable translating "uncorrelated" into "off-diagonal entry of the covariance matrix is zero" is a transferable skill that pays off in ANOVA, regression, and design-of-experiments questions as well.

Frequently Asked Questions

What exactly does BLUE mean, and why is it the same as the OLS estimator here?

BLUE stands for Best Linear Unbiased Estimator — among all estimators that are linear functions of the data and unbiased for the true parameter, it is the one with the smallest variance. The Gauss-Markov theorem guarantees that under the standard assumptions (zero mean errors, constant variance, uncorrelated errors), the ordinary least squares estimator achieves this minimum variance, so BLUE and OLS refer to the same estimator in this setting.

Why is the covariance matrix of β̂ equal to σ²(X'X)⁻¹?

Since β̂ = (X'X)⁻¹X'y is a linear function of y, and y has covariance matrix σ²I (constant variance, uncorrelated observations), standard rules for the variance of a linear transformation give Cov(β̂) = (X'X)⁻¹X' [σ²I] X(X'X)⁻¹, which simplifies directly to σ²(X'X)⁻¹ because X'X cancels neatly with itself.

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

Two random variables are uncorrelated if their covariance is zero, meaning there is no linear association between how one varies and how the other varies. For the vector estimator β̂, this corresponds exactly to the off-diagonal entries of its covariance matrix being zero — which is precisely the condition this question asks you to enforce.

How do I build the X matrix when a model is given only in terms of expected values like this?

Write each E(yi) equation as an observation equal to its expression in β1, β2, etc. The coefficient attached to each parameter in each equation becomes one entry of X, with each equation forming one row and each parameter forming one column. No error terms or data values are needed to build X itself — only the coefficients.

Does this off-diagonal-equals-zero shortcut work for models with more than two parameters?

The same logic extends, but with more parameters you must examine the specific pair of off-diagonal entries corresponding to the two estimators you want uncorrelated, since (X'X)⁻¹ will have more entries overall. The principle — zero covariance means the relevant off-diagonal entry of σ²(X'X)⁻¹ is zero — still holds exactly the same way.

Why doesn't the actual value of σ² or the determinant matter in the final answer?

Because σ² and the determinant of X'X are both scalar multipliers applied to the entire off-diagonal entry, not the entry itself. As long as they are finite and non-zero (which they are for any valid design), they can never force a non-zero numerator to become zero, so the condition for uncorrelatedness depends purely on the numerator, -(1+α), being zero.

If any step above still feels unclear, drop a comment with exactly where you got stuck — working through the confusion in the comments usually helps the next reader too. And if you know a fellow ISS aspirant preparing for Paper-2, consider sharing this with them; working through these derivations together tends to make the ideas stick far better than reading them alone.