Assignment 5 with solutions
MAT3379 (Winter 2015)
Assignment 5
Due date (Assignment 5): April 13, 2015
Q1. Assume that Z = (Z1 , Z2 ) is a random vector with the mean vector 0 and the covariance matrix
σ11 σ12
Σ=
.
σ21 σ22
Let A be a deterministic matrix defined by
A=
a11
a21
a12
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 the
covariances between both components. Check that what you obtain is exactly AΣAT
Solution to Q1:
We have
T
a11 Z1 + a12 Z2
a21 Z1 + a22 Z2
X = (X1 , X2 ) =
.
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[a11 Z1 + a12 Z2 ] = 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[a11 Z1 + a12 Z2 ] = a211 Var[Z1 ] + a212 Var[Z2 ] + 2a11 a12 Cov(Z1 , Z2 ) = a211 σ11 + a211 σ22 + 2a11 a12 σ12
Var[X2 ] = Var[a21 Z1 + a22 Z2 ] = a221 Var[Z1 ] + a222 Var[Z2 ] + 2a21 a22 Cov(Z1 , Z2 ) = a221 σ11 + a221 σ22 + 2a21 a22 σ12
Cov[X1 , X2 ] = Cov[a11 Z1 + a12 Z2 , a21 Z1 + a22 Z2 ]
= a11 a21 σ11 + a12 a22 σ22 + (a11 a22 + a12 a21 ) σ12 .
Hence, the covariance matrix of X is given by
a211 σ11 + a211 σ22 + 2a11 a12 σ12
a11 a21 σ11 + a12 a22 σ22 + (a11 a22 + a12 a21 ) σ12
a11 a21 σ11 + a12 a22 σ22 + (a11 a22 + a12 a21 ) σ12
a221 σ11 + a221 σ22 + 2a21 a22 σ12
.
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 for
computation of AΣAT .
1
2
2
Q2. Assume that Zt = (Zt , Zt ), where {Zt } is an i.i.d. sequence with mean vector 0 and variance σZ
(yes, both
components of the vector are the same). Define the bivariate linear process by
Xt = Zt + ΨZt−1 ,
where
1 0
Ψ=
.
1 θ
Find Γ(0), Γ(1) and Γ(−1), where Γ(h) is the covariance matrix function of the bivariate time series Xt .
Note: this is the same example as I did in class on Wednesday, April 1st.
Solution to Q2:
Here, there is the correct solution for
Xt = Zt + ΨZt−1 ,
where
1
1
Ψ=
We have
Xt = (Xt1 , Xt2 )T =
and
0
θ
.
Zt + Zt−1
Zt + (1 + θ)Zt−1
.
2
2
Var(Xt1 )
Cov(Xt1 , Xt2 )
2σZ
(2 + θ)σZ
Γ(0) =
=
.
2
2
Cov(Xt2 , Xt1 )
Var(Xt2 )
(2 + θ)σZ
(1 + θ)2 + 1 σZ
2
2
Cov(Xt1 , Xt+1,1 ) Cov(Xt1 , Xt+1,2 )
σZ (1 + θ)σZ
Γ(1) =
=
.
2
2
Cov(Xt2 , Xt+1,1 ) Cov(Xt2 , Xt+1,2 )
σZ (1 + θ)σZ
2
2
Cov(Xt1 , Xt−1,1 ) Cov(Xt1 , Xt−1,2 )
σZ
σZ
Γ(−1) =
=
.
2
2
Cov(Xt2 , Xt−1,1 ) Cov(Xt2 , Xt−1,2 )
(1 + θ)σZ
(1 + θ)σZ
Below, there is the solution for
Xt = Zt + ΨZt−1 ,
where
Ψ=
0
0
0
θ
which is the example we did in class. We have
T
Xt = (Xt1 , Xt2 ) =
and
Zt
Zt + θZt−1
.
2
2
Var(Xt1 )
Cov(Xt1 , Xt2 )
σZ
σZ
=
.
2
2
Cov(Xt2 , Xt1 )
Var(Xt2 )
σZ
(1 + θ2 )σZ
2
Cov(Xt1 , Xt+1,1 ) Cov(Xt1 , Xt+1,2 )
0 θσZ
Γ(1) =
=
.
2
Cov(Xt2 , Xt+1,1 ) Cov(Xt2 , Xt+1,2 )
0 θσZ
0
0
Cov(Xt1 , Xt−1,1 ) Cov(Xt1 , Xt−1,2 )
Γ(−1) =
=
.
2
2
Cov(Xt2 , Xt−1,1 ) Cov(Xt2 , Xt−1,2 )
θσZ
θσZ
Γ(0) =
Marking scheme for Q2:
This part will not be marked.
Total number of points for Assignment 5: 5
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