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SPRING 2012 Problem 4

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Solution to Problem 4, SP 2012

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Problem 6, SPRING 2012

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Solution to Problem 6, SPRING 2012

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AUTUMN 2012, Problem 2

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AUTUMN 2012 Problem 4

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SOLUTION to Problem 4, AU2012

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Problem 6, AUTUMN 2012

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Autumn 2011, Problem 5

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Solution to Problem 5 Autumn 2011

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Spring 2010 Problem 4

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Autumn 2010 Problem 4

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Spring 2009 Problem, 1

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Spring 2009 Problem 6

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Autumn 2009 Problem 4

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Solution to Problem 4 Autumn 2009

3. Consider the general linear models,

M1: Y1 = X1β + ε1 ; M2: Y2 = X2β + ε2 ,

where Y1 is an n1 × 1 vector, Y2 is an n2 × 1 vector, X1 and X2 are notnecessarily of full rank, and ε1 and ε2 are uncorrelated random vectors, withmeans 0 and variance-covariance matrices σ2I.

(a) (3 points) Find the best linear unbiased estimators, T1 and T2, respec-tively, of functions `′β that are estimable under both models M1 andM2. In addition, find Var(Ti) for i = 1, 2.

(b) (3 points) Show that the minimum variance unbiased estimator of es-timable `′β among all convex combinations T (α) = αT1 + (1 − α)T2,where 0 ≤ α ≤ 1, is given by T (α∗) where α∗ = ω1/(ω1 + ω2) whereω−1

i = (Var[Ti])−1, i = 1, 2. Also find Var[T (α∗)].

(c) (3 points) Consider the combined linear model,

M3: Y =( Y1

Y2

)=( X1

X2

)β +

( ε1

ε2

).

Obviously, `′β in part (a) is estimable under model M3. Find its bestlinear unbiased estimator T3, and find Var(T3).

(d) (5 points) Explain why the inequality Var(T3) ≤ Var([T (α∗)], holds ingeneral.

(e) (6 points) Show that if either X1 or X2 is of rank one, then T (α∗) = T3.

(f) (5 points) Let A and B be nonnegative definite matrices. Prove that

[a′A−a][a′B−a] ≥ [a′(A+B)−a][a′A−a+ a′B−a] ,

provided that a ∈ µ(A)∩µ(B), where µ(A) and µ(B) denote the columnspaces of A and B respectively, with equality when either A or B is ofrank one.[Hint: Use A = X ′

1X1 and B = X ′2X2 in parts (d) and (e) above. You

can use these parts here, even if you do not prove the earlier results.]

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Solution: Problem 3

1. Using the notation, A = X ′1X1 and B = X ′

2X2, the function `′β isestimable under each of the two models, provided ` ∈ µ(A) ∩ µ(B).Then T1 = `′A−X ′

1Y1, and T2 = `′B=X ′2Y2. The variances are given by

Var(T1) = σ2`′A−`, and Var(T2) = σ2`′B−`.

2. Since T1 and T2 are unbiased estimators of `′β, so is T (α) for allα. Furthermore, since, (ε1, ε2) are uncorrelated random variables, T1

and T2 are also uncorrelated, and the variance of T (α) is given byVar[T (α)] = α2Var(T1) + (1 − α)2Var(T2). On setting its derivativewith respect to α equal to zero, it is easily seen that α∗ = ω1/(ω1 +ω2),with Var[T (α∗)] = 1/(ω1 + ω2).

3. For the model M3, T3 = `′(A+B)−(X1 X2)′Y , and Var(T3) = σ2`′(A+

B)−`.

4. Since T3 is the best linear unbiased estimator based on Y1 and Y2, it hasminimum variance among all linear unbiased estimators. Furthermore,T (α∗) is also a linear unbiased estimator based on Y1 and Y2. Therefore,Var(T3) ≤ Var([T (α∗)] holds in general.

5. Without loss of generality, assume that the rank of X2 is one and thelength of the vector ` is equal to 1. Since `′β is an estimable functionunder both the models M1 and M2, each row of X2 belongs to the spacespanned by the vector `, and ` ∈ µ(A). One can write the matrix A asa linear combination of the matrix ``′ and other that are orthogonal tothis matrix. Then it is easy to show that T3 = `′(A+B)−(X1 X2)

′Y =T (α∗).

6. This part follows from the part (e) after cancelling out σ2, and rear-ranging terms.

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