MAT3379 (Winter 2016)Assignment 6 - SOLUTIONS

The following questions will be marked: Q1

Total number of points for Assignment 6: 5

Q1. Assume that Z = (Z1, Z2) is a random vector with the mean vector 0 and the covariance matrix

Σ =

[σ11 σ12σ21 σ22

].

Let A be a deterministic matrix defined by

A =

[a11 a12a21 a22

].

Consider the random vector X = AZ.• Show that E[X] = 0. Note: You need to multiply A by Z and compute the expected value of each

coordinate.• Show that the covariance matrix of the random vector X is given by AΣAT . Note: You need to multiply

A by Z. The resulting vector has two components. Calculate the variance of each component and thecovariances between both components. Check that what you obtain is exactly AΣAT

Solution to Q1:

We have

X = (X1, X2)T =

[a11Z1 + a12Z2

a21Z1 + a22Z2

].

By the definition, the expected value of the vector X = (X1, X2)T is a vector with two components being E[X1]and E[X2]. We have

E[X1] = E[a11Z1 + a12Z2] = 0

since E[Z1] = E[Z2] = 0. The same for E[X2].Also, the covariance matrix of X is given by[

Var[X1] Cov[X1, X2]Cov[X1, X2] Var[X2]

].

We calculate (note that we must have σ12 = σ21)

Var[X1] = Var[a11Z1 + a12Z2] = a211Var[Z1] + a212Var[Z2] + 2a11a12Cov(Z1, Z2) = a211σ11 + a211σ22 + 2a11a12σ12

Var[X2] = Var[a21Z1 + a22Z2] = a221Var[Z1] + a222Var[Z2] + 2a21a22Cov(Z1, Z2) = a221σ11 + a221σ22 + 2a21a22σ12

Cov[X1, X2] = Cov[a11Z1 + a12Z2, a21Z1 + a22Z2]

= a11a21σ11 + a12a22σ22 + (a11a22 + a12a21)σ12 .

Hence, the covariance matrix of X is given by[a211σ11 + a211σ22 + 2a11a12σ12 a11a21σ11 + a12a22σ22 + (a11a22 + a12a21)σ12

a11a21σ11 + a12a22σ22 + (a11a22 + a12a21)σ12 a221σ11 + a221σ22 + 2a21a22σ12

].

You obtain the same when computing AΣAT .

Marking scheme for Q1:

1 point for computation of E[X], 2 points for computation of the covariance matrix of X, two points forcomputation of AΣAT .

1

2

Q2. Assume that Zt = (Zt, Zt), where {Zt} is an i.i.d. sequence with mean vector 0 and variance σ2Z (yes, both

components of the vector are the same). Define the bivariate linear process by

Xt = Zt + ΨZt−1 ,

where

Ψ =

[1 01 θ

].

Find Γ(0), Γ(1) and Γ(−1), where Γ(h) is the covariance matrix function of the bivariate time series Xt.

Solution to Q2:

Here, there is the solution forXt = Zt + ΨZt−1 ,

where

Ψ =

[1 01 θ

].

We have

Xt = (Xt1, Xt2)T =

[Zt + Zt−1

Zt + (1 + θ)Zt−1

].

and

Γ(0) =

[Var(Xt1) Cov(Xt1, Xt2)

Cov(Xt2, Xt1) Var(Xt2)

]=

[2σ2

Z (2 + θ)σ2Z

(2 + θ)σ2Z

{(1 + θ)2 + 1

}σ2Z

].

Γ(1) =

[Cov(Xt1, Xt+1,1) Cov(Xt1, Xt+1,2)Cov(Xt2, Xt+1,1) Cov(Xt2, Xt+1,2)

]=

[σ2Z (1 + θ)σ2

Z

σ2Z (1 + θ)σ2

Z

].

Γ(−1) =

[Cov(Xt1, Xt−1,1) Cov(Xt1, Xt−1,2)Cov(Xt2, Xt−1,1) Cov(Xt2, Xt−1,2)

]=

[σ2Z σ2

Z

(1 + θ)σ2Z (1 + θ)σ2

Z

].

As a bonus, below, there is the solution for

Xt = Zt + ΨZt−1 ,

where

Ψ =

[0 00 θ

]which is the example we did in class. We have

Xt = (Xt1, Xt2)T =

[Zt

Zt + θZt−1

].

and

Γ(0) =

[Var(Xt1) Cov(Xt1, Xt2)

Cov(Xt2, Xt1) Var(Xt2)

]=

[σ2Z σ2

Z

σ2Z (1 + θ2)σ2

Z

].

Γ(1) =

[Cov(Xt1, Xt+1,1) Cov(Xt1, Xt+1,2)Cov(Xt2, Xt+1,1) Cov(Xt2, Xt+1,2)

]=

[0 θσ2

Z

0 θσ2Z

].

Γ(−1) =

[Cov(Xt1, Xt−1,1) Cov(Xt1, Xt−1,2)Cov(Xt2, Xt−1,1) Cov(Xt2, Xt−1,2)

]=

[0 0θσ2

Z θσ2Z

].

Marking scheme for Q2:

This part will not be marked.