MAT 3379 (Summer 2022) Midterm
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MAT 3379 (Summer 2022)
Midterm
Q1 (5 points) Let E(X) = 2, Var(X) = 9, E(Y) = 0, Var(Y) = 4, and the correlation between X and Y is ρ = 0.25
Calculate :
(a)
Var(X + Y) = Var(X) + Var(Y) + 2Cov(X,Y)
= Var(X) + Var(Y) + 2ρ ^Var(X) ^Var(Y)
= 9 + 4 + 2 × 0.25 × 3 × 2 = 16
(b)
Cov(X,X+Y) = Cov(X,X)+Cov(X,Y) = Var(X)+ρ ^X ^Y = 9+0.25×3×2 = 10.5
Q2 (5 points) Find the mean function and the covariance function for each of the processes below. For each processes, determine if the process is stationary or not stationary.
(a) The time series {xt }, such that xt = θ+tet and {et } ∼ WN(0,σ2 ),
where θ is a constant.
(b) The process {Yt }, such that Yt = Xt −Xt −1 ,and Xt comes from
Solution for (a): The mean function µX (t) = E(Xt ) = E(θ + tet ) = θ + t(0) = θ .
Thew covariance function :
γX (t,t + h) = Cov(Xt ,Xt+h) = Cov(θ + tet , θ + (t + h)et+h)
= t(t + h)Cov(et , et+h)
Therefore
γX (t,t + h) =
Since the covariance function is dependent on t , so it is not stationary process.
For part(b):
Yt = Xt − Xt −1 = θ + tet − (θ + (t − 1)et −1) = tet − (t − 1)et −1
The mean function :
µY (t) = E(Xt − Xt −1) = E(Xt ) − E(Xt −1) = θ − θ = 0
and the covariance function :
γY (t,t + h) = Cov(Yt ,Yt+h) = Cov(Xt − Xt −1, Xt+h − Xt+h−1)
= γX (t,t + h) − γX (t,t + h − 1) − γX (t − 1,t + h) + γX (t − 1,t + h − 1)
Therefore:
(t2 + (t − 1) )σ2 2 h = 0;
‘ (t − −2(1;) .
Since the covariance function is dependent on t , so it is not stationary process
Q3 (5points) Consider a process {Xt } , such that
Xt + 0.4Xt −1 − 0.21Xt −2 = Zt − 0.09Zt −2 .
{Zt } ∼ WN(0,σ2 )
(a) Does this model have redundant factor? if yes, reduce the model.
(b) Is the process stationary, causal, and invertible?
(c) Does this process have an ARMA(p ,q) model? If yes, give the values to p and q .
Solution For (a) : The polynomial function
ϕ(z) = 1 + 0.4z − 0.21z2 = (1 + 0.7z)(1 − 0.3z)
and
θ(z) = 1 − 0.09z2 = (1 + 0.3z)(1 − 0.3z)
Since both function have common factor 1 − 0.3z , which is redundant
(1 + 0.7B)Xt = (1 + 0.3B)Zt
which is following model :
Xt + 0.7Xt −1 = Zt + 0.3Zt −1
For part(b). Solve the equation and find roots: 0 = ϕ(θ) = 1 + 0.7z
z = −10/7. Since |z| 1, so the process is stationary. Moreover,
the process is invertible.
For part)(c)
The process is a model ARMA(p = 1,q = 1).
Q4 (6 points) Consider three random variables X1 ,X2 ,X3 is a sta- tionary process, such that µX = 0, γX (0) = 10, γX (1) = 3, γX (2) = 1
(a) Give the best linear prediction of X3 knowing = (X2 ,X1 ) (b) Give mean square prediction error (MSPE) of (a).
(c) Give the best linear prediction of 2+5X3 knowing = (X2 ,X1 )
Solution
We want
P(X3 |) = a1X2 + a2X1
Γa = Γ = [3(10) 10(3) ] γ = [1(3) ]
So,
a = [ 3(0) 0(3) ][1(3) ] = [1(27) ]
The best linear forecast of X3 given = (X2 ,X1 ) is
MSPE2 (1) = γX (0) − a\ = 10 − (3) − (1) = 9.0989
(c) The best linear prediction of 2 + 5X3 such that = (X2 ,X1 ) is
91 91
Q5(2 points) Consider an AR(1) model
Xt − ϕXt −1 = Zt ,
where |ϕ| < 1 and Zt are i.i.d. random variables with mean 0 and vari- ance σZ(2) . Derive a linear representation for Xt , i.e. find the coefficients ψj in Xt =对 ψj Zt −j .
Solution: See Lecture Notes, Example 3.14.
We have
ϕ(B)Xt = Zt , (1)
where ϕ(z) = 1 − ϕz . Define
χ(z) = 1
Now, the function 1/(1 − ϕz) has the following power series expansion:
∞
χ(z) = = ϕj zj .
This expansion makes sense whenever |ϕ| < 1. Take equation (1) and multiply both sides by χ(B):
χ(B)ϕ(B)Xt = χ(B)Zt ,
Xt = χ(B)Zt ,
since χ(z)ϕ(z) = 1 for all z . That is, the linear representation of AR(1) is
∞ ∞
Xt = χ(B)Zt =工 ϕj Bj Zt =工 ϕj Zt −j .
j=0 j=0
The above formula gives a linear representation for AR(1) with ψj = ϕj . We note that the above computation makes sense whenever |ϕ| < 1.
That is, under this condition there exists the causal linear representa- tion.
Q6 (3 points) Let {Zt } be independent random variables with mean
0 and variance σ Z(2) . Consider the model Xt = Zt − Zt −1 . Evaluate
Note: You can use the formula from Lecture Notes that represents α(2) in terms of correlations. Then plug-in the correlations for MA(1) model.
SolutionWe can almost copy Example 1.17 in Lecture Notes. Recall
that for Xt = Zt − Zt −1 we have
1 h = 0;
ρX (h) = (
Then
Corr(X1 ,X3 ) − Corr(X1 ,X2 )Corr(X2 ,X3 )
^ (1 − Corr2 (X1 ,X2 ))(1 − Corr2 (X2 ,X3 ))
ρX (2) − ρX(2) (1)
^1 − ρX(2)(1) ^1 − ρX(2)(1)
2022-07-20