Consider the simple linear regression model Yᵢ = βxᵢ + εᵢ, i = 1, …, n, where E(εᵢ) = 0, Cov(εᵢ, εₖ) = 0 for i ≠ k and Var(εᵢ) = xᵢ²σ². The best linear unbiased estimator of β is

C. Heteroscedastic errors: divide by xᵢ to get Yᵢ/xᵢ = β + εᵢ/xᵢ with constant variance σ²; the BLUE is the plain mean of Yᵢ/xᵢ (weighted least squares with weights 1/xᵢ²).

Part BCSIR NET June 2023weighted-least-squares

Weighted least squares

Consider the simple linear regression model $Y_{i} = \beta{}x_{i} + \varepsilon_{i}, i = 1, \dots, n$ , where $E(\varepsilon_{i}) = 0, Cov(\varepsilon_{i}, \varepsilon_{k}) = 0$ for i ≠ k and $Var(\varepsilon_{i}) = x_{i}^{2}\sigma^{2}$ . The best linear unbiased estimator of $\beta$ is

A. $\sum Y_{i}x_{i} / \sum x_{i}^{2}$
B. $\sum Y_{i} / \sum x_{i}$
C. $(1/n) \sum Y_{i}/x_{i}$ ✓ Correct
D. $(1/n) \sum Y_{i}x_{i} / x_{i}^{2}$

The idea, in plain words

The noise here grows with the predictor, so the far-out points are noisier and should count less. Dividing each observation by its predictor equalises the noise, and then the best estimate is simply the plain average of the rescaled values.

Solution

Heteroscedastic errors: divide by $x_{i}$ to get $Y_{i}/x_{i} = \beta + \varepsilon_{i}/x_{i}$ with constant variance $\sigma^{2}$ ; the BLUE is the plain mean of $Y_{i}/x_{i} ($ weighted least squares with weights $1/x_{i}^{2})$ .

Why each option is right or wrong

Checked against 2 symbolic computations

A.WrongHypothesis dropped
False here

though it is the familiar answer.

$\sum{}Y_{i}x_{i}/\sum{}x_{i}^{2}$ is the ordinary least squares estimator

which is BLUE only when the errors have constant variance.

Here $Var(\varepsilon_{i}) = x_{i}^{2}\sigma^{2}$ grows with $x_{i}$

so OLS over-weights the noisiest observations.

It remains unbiased, just not best.
B.Wrong
False.

$\sum{}Y_{i}/\sum{}x_{i}$ is unbiased — its expectation is $\beta\sum{}x_{i}/\sum{}x_{i} = \beta$ — but it is a different linear estimator with larger variance

since it weights every observation equally regardless of its noise level.
C.Correct
True.

Divide the model through by $x_{i}: Y_{i}/x_{i} = \beta + \varepsilon_{i}/x_{i}$ , and SymPy confirms the transformed error variance is $x_{i}^{2}\sigma^{2}/x_{i}^{2} = \sigma^{2}$ , now constant.

The transformed model is a location problem with homoscedastic errors, whose BLUE is the plain arithmetic mean of the transformed responses — that is $(1/n)\sum{}Y_{i}/x_{i}$ .

The trigger: when $Var(\varepsilon_{i})$ ∝ $x_{i}^{2}$

dividing by $x_{i}$ stabilises the variance and turns weighted least squares into an ordinary average.
D.Wrong
False as intended

though the printed expression is ambiguous.

Read term by term, $(1/n)\sum(Y_{i}x_{i}/x_{i}^{2})$ is literally the same as (C) and so would not be a distinct option; the intended reading is the aggregated form $(1/n)\cdot\sum{}Y_{i}x_{i}/\sum{}x_{i}^{2}$

which is the OLS estimator divided by n.

SymPy confirms that shrinks the expectation to $\beta/n$

so it is biased by a factor of n.

1 of these 3 wrong options correspond to a named reasoning error — the same errors recur across subjects.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: OLS is the BLUE in every linear model

Weighted least squares

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