MAT 3379, Summer 2016
Solution to Assignment 3
2.1 .
We solve this problem for a general time series firs (stationarity is not required). To minimize
g(a, b) = E(Xn+h − aXn − b)2
solve for a and b
−2E(Xn+h − aXn − b) = 0
−2E(Xn (Xn+h − aXn − b) = 0.
We get
E(Xn+h ) = aE(Xn ) + b
E(Xn Xn+h ) = aE(Xn2 ) = bE(Xn ).
Solve for a and b to get
a=
E(Xn Xn+h ) − E(Xn )E(Xn+h )
= ρ(h).
E(Xn2 − (E(Xn ))2
Also we get
b = E(Xn+h ) − ρ(h)E(Xn ).
For the stationary case we have E(Xn+h = E(Xn ) = µ and result follows easily.
2.12. Note that γ(o) = 0 and γ(1) = θ = −0.6 and γ(h) = 0 for |h| > 1. The confidence interval for
µ is
√
x̄ ± 1.96 ν
where
1
1
(γ(0) − 2(0.99)γ(1)) =
(1 + 0.36 + 2(0.99)(−0.6)) = 0.00172
100
100
√
(See the formula in page 58, line -4 of your textbook). Therefore ν = 0.04147288. The C.I. is
0.157 ± (0.04147288)(1.96) which gives (0.07571316, 0.2383) which does not contains 0 and we reject
that µ = 0 at 5% level.
ν=
2.14, a. We know both the autocovariance and autocorrelation function is ρ(h) = cos(ωh). As we
discussed in class for stationary processes
P1 X2 = φ11 X1
where φ11 = ρ(1) = cos(ω). Therefore
P1 X2 = cos(ω)X1 .
Also the mean square error is
ν1 = E(X2 − cos(ω)X1 )2 = γ(0) − ρ(1)γ(1) = 1 − cos2 (ω).
1
2.14, b. As we discussed in class
P2 X3 = φ21 X1 + φ22 X2
where we proved
ρ(2) − φ211
cos(2ω) − cos2 (ω)
=
= −1
2
1 − cos2 (ω)
1 − φ11
φ22 =
and
φ21 = φ11 − φ22 φ11 = cos(ω) + cos(ω) = 2 cos(ω).
The mean squared error is
ν2 = ν1 (1 − φ222 ) = 0.
(c) Similarly
P̃n Xn+1 = 2 cos(ω)Xn − Xn−1
and
E[(Xn+1 − 2 cos(ω)Xn − Xn−1 )2 ] = 0.
2.15. We need to minimize
g(a0 , . . . , an ) = E[(Xn+1 − a0 − a1 Xn − · · · − an X1 )2 ].
Differentiating with respct to a0 and letting
E(Xn+1 − a0 − a1 Xn − · · · − an X1 ) = 0
gives
E(Xn+1 ) = a0 + a1 E(Xn ) + · · · + an E(Xn )
which concludes a0 = 0. Therefore all we need to show is that
X̂n+1 = φ1 Xn + · · · + φp Xn+1−p
satisfies
E[(Xn+1 − X̂n+1 )Xn+1−i ] = 0
for i = 1, 2, . . . , n which is obvious.
Pn Xn+1 = φ1 Xn + · · · + φp Xn+1−p .
We also get
2
ν = E[(Xn+1 − φ1 Xn − · · · − φp Xn+1−p )2 ] = E(Zn+1
) = σ2.
2.22(a). We need to minimize
g(a, b) = E(X3 − aX1 − bX2 )2 .
Noting that
ρ(h) = φh
2
and differentiating with respect to bot a and b and putting
E(X1 X3 ) = aE(X12 ) + bE(X1 X2 )
and
E(X2 X3 ) = aE(X1 X2 ) + bE(X22 )
gives
ρ(2) = a − bρ(1), ρ(1) = aρ(1) − b
and solving for a and b gives a = 0 and b = φ. This result is not surprising as we know from the
class notes that P (X3 |X1 , X2 ) = φX2
(b) Minimize
g(a, b) = E[(X3 − aX4 − bX5 )2 .
Similar to part (a) conclude that a = φ and b = 0. Therefore
P (X3 |X4 , X5 ) = φX4 .
(c) Minimize
g(a, b, c, d) = E[(X3 − aX1 − bX2 − cX4 − dX5 )2 ].
In this part you need to solve a system of equations with 4 values to calculate. We will get with
some tedious calculation that in fact a = d = 0 and
b=c=
φ(1 − φ2 )
φ
=
.
1 − φ4
1 + φ2
Therefore
P (X3 |X1 , X2 , X4 , X5 ) =
φ
(X2 + X4 ).
1 + φ2
(d) Let the M.S.E. for parts (a), (b) and (c) are denoted by νa , νb and νc . We have
νa = E(X3 − φX2 )2 = E(Z32 ) = σ 2 ,
νb = E(X3 − φX4 )2 = E(X42 ) + φ2 E(X42 ) − 2φE(X3 X4 ) =
σ2
(1 + φ2 − 2φ) = σ 2 .
1 − φ2
and
2
φ
νc = E X3 −
(X2 + X4 )
1 + φ2
Therefore
σ2
νc =
1 − φ2
=
1+
E(X32 )+
φ
1 + φ2
3.4. We have
Xt =
2
φ
1 + φ2
2
φ
V ar(X2 +X4 )−2
Cov(X3 , X2 +X4 ).
1 + φ2
∞
X
1
Z
=
(0.8)i Zt−2i .
t
1 − 0.8B 2
i=0
3
=
σ2
.
1 + φ2
!
φ
(2 + 2φ ) − 4
φ
1 + φ2
2
This gives
V ar(Xt ) = σ 2
∞
X
0.82i = σ 2
i=0
∞
X
(0.64i ) =
i=0
1
= (25/9)σ 2 .
1 − 0.64
Notice that also for k ≥ 1,
γ(k) = 0.8γ(k − 2).
(Multiply both sides by Xt−k and take expectation). This gives
γ(2k) = (25/9)σ 2 (0.8)k
and γ(2k + 1) = 0. Similarly
ρ(0) = 1, ρ(2k + 1) = 0, ρ(2k) = (0.8)|k| .
To find PACF Note that
P (Xh+1 |Xh ) = X̂h+1 = φ11 Xh = 0
and we get pacf (1) = α1 = 0 and for pacf (2) = α2 = φ22 we need to write
P (Xh+2 |Xh , Xh+1 ) = X̂h+2 = φ21 Xh+1 + φ22 Xh = 0.8Xh .
Therefore pacf (2) = α2 = 0.8. Also we get pacf (h) = 0 for h > 2.
3.6. We have in general for a stationary process M A(1)
Ut = Zt + βZt−1
such that Z1 , Z2 , . . . ∼ W N (0, a2 ), we have
V ar(Ut ) = a2 (1 + β 2 ),
γU (h) = 0
for |h| > 1,
γU (0) = a2 (1 + β 2 )
and finally
γU (h) = a2 β, |h| = 1.
In the first model let
a = σ2, β = θ
and in the second model let
a = σ 2 θ2 , β = 1/θ
to see the results will be the same in both cases.
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