Assignment 6 - SOLUTIONS - Faculté des sciences - Faculty ... · MAT3379 (Winter 2016) Assignment...
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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 =(Z 1 ,Z 2 ) is a random vector with the mean vector 0 and the covariance matrix Σ= σ 11 σ 12 σ 21 σ 22 . Let A be a deterministic matrix defined by A = a 11 a 12 a 21 a 22 . 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 AΣA T . 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 AΣA T Solution to Q1: We have X =(X 1 ,X 2 ) T = a 11 Z 1 + a 12 Z 2 a 21 Z 1 + a 22 Z 2 . By the definition, the expected value of the vector X =(X 1 ,X 2 ) T is a vector with two components being E[X 1 ] and E[X 2 ]. We have E[X 1 ] = E[a 11 Z 1 + a 12 Z 2 ]=0 since E[Z 1 ] = E[Z 2 ] = 0. The same for E[X 2 ]. Also, the covariance matrix of X is given by Var[X 1 ] Cov[X 1 ,X 2 ] Cov[X 1 ,X 2 ] Var[X 2 ] . We calculate (note that we must have σ 12 = σ 21 ) Var[X 1 ] = Var[a 11 Z 1 + a 12 Z 2 ]= a 2 11 Var[Z 1 ]+ a 2 12 Var[Z 2 ]+2a 11 a 12 Cov(Z 1 ,Z 2 )= a 2 11 σ 11 + a 2 11 σ 22 +2a 11 a 12 σ 12 Var[X 2 ] = Var[a 21 Z 1 + a 22 Z 2 ]= a 2 21 Var[Z 1 ]+ a 2 22 Var[Z 2 ]+2a 21 a 22 Cov(Z 1 ,Z 2 )= a 2 21 σ 11 + a 2 21 σ 22 +2a 21 a 22 σ 12 Cov[X 1 ,X 2 ] = Cov[a 11 Z 1 + a 12 Z 2 ,a 21 Z 1 + a 22 Z 2 ] = a 11 a 21 σ 11 + a 12 a 22 σ 22 +(a 11 a 22 + a 12 a 21 ) σ 12 . Hence, the covariance matrix of X is given by a 2 11 σ 11 + a 2 11 σ 22 +2a 11 a 12 σ 12 a 11 a 21 σ 11 + a 12 a 22 σ 22 +(a 11 a 22 + a 12 a 21 ) σ 12 a 11 a 21 σ 11 + a 12 a 22 σ 22 +(a 11 a 22 + a 12 a 21 ) σ 12 a 2 21 σ 11 + a 2 21 σ 22 +2a 21 a 22 σ 12 . You obtain the same when computing AΣA T . 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 AΣA T . 1
Transcript of Assignment 6 - SOLUTIONS - Faculté des sciences - Faculty ... · MAT3379 (Winter 2016) Assignment...
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 .
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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.
