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Mathematical Statistics, Winter semester 2020/21 Problem sheet 1

1) Suppose that

Y = Xβ + ε

holds for someβ ∈Rkand thatEε = 0n, Cov(ε) = Σ, where Σ is a regular matrix and the matrix X has rank k.

(i) Show that XTΣ−1X is a regular matrix and that

βb = (XTΣ−1X)−1XTΣ−1Y is an unbiased estimator of β and compute Eβh

(βb−β)(βb−β)Ti . (ii) Letβe=LY be any arbitrary unbiased estimator ofβ.

Show that

Eβ

h

(βe−β)(βe−β)T i

− Eβ

h

(βb−β)(βb−β)T i

is non-negative definite.

Hint: A symmetric and positive definite (n × n)-matrix M can be represen- ted as M = Pn

i=1λieieTi , where λ1, . . . , λn are the (positive) eigenvalues and e1, . . . , en are corresponding eigenvectors with eTi ej = 0 for i 6= j. Then M1/2 :=

Pn i=1

√λieieTi and M−1/2 :=Pn

i=1(1/√

λi)eieTi. To prove (ii), use the fact that

1/2 − (XTΣ−1X)−1XTΣ−1/2

1/2 − (XTΣ−1X)−1XTΣ−1/2 T

is non-negative definite.

2) (i) Let

X =

1 v1 v21 · · · v1k 1 v2 v22 · · · v2k ... ... ... . .. ...

1 vn vn2 · · · vnk

 .

Prove thatXTX is regular if the set {v1, . . . , vn} contains at least k+ 1 different values.

Hint: Choose c= (c1, . . . , ck+1)T 6= 0k+1 := (0, . . . ,0)T and compute cTXTXc.

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(ii) Let

X =

v1 v21 · · · v1k v2 v22 · · · v2k ... ... . .. ...

vn vn2 · · · vnk

 .

Prove that XTX is regular if the set {v1, . . . , vn} contains at least k different non-zero values.

Hint: Consider the matrix

Xe =

1 0 0 . . . 0 1 v1 v21 · · · v1k 1 v2 v22 · · · v2k ... ... ... . .. ...

1 vn vn2 · · · vnk

 .

3) Consider the linear regression model Yi1+xiβ2i, i= 1, . . . , n, where ε1, . . . , εn

are i.i.d. with εi ∼ N(0, σ2). Let βbbe the least squares estimator ofβ.

(i) Suppose that xi 6=xj, for some (i, j).

Compute E[(βbi −βi)2], for i= 1,2.

Hint: The inverse of a regular matrix

a b c d

is given by ad−bc1

d −b

−c a

. (ii) Suppose thatx1, . . . , xncan be chosen by an experimenter, wherexi ∈[−1,1] and

n≥2 is even.

Which choice of x1, . . . , xn minimizes E[(βbi − βi)2]? (Take into account that x1, . . . , xn have to be chosen such that xi 6= xj, for some (i, j); otherwise the least squares estimator is not uniquely defined.)

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