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STA457H1/STA2202H F-LEC5101 Time Series Analysis
发布时间:2021-12-03
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Department of Statistical Sciences
Time Series Analysis
STA457H1/STA2202H F-LEC5101
Fall 2021
Quiz 6
Q1. a. Let x1 , x2 , . . . , x5 be the observations from MA(1) model xt = wt +θwt − 1 , where wt ~ i.i.d. N (0, 2). Find the best linear predictor (BLP) x5(4) in terms of w1 , w2 , and compute its mean squared error. (5 pts)
Solutions:
Let x5(4) be the best linear predictor of x5 based on x1 , x2 , x3 , x4 .
x5(4) = E[x5 |x4 , x3 , x2 , x1]
= E[w5 + θw4 |x4 , x3 , x2 , x1] (1 pt)
Note that
E[wt |xn , . . . , x1] =..0 t > n
.,wt t < n
(1 pt)
So, x5(4) = E[w5 |x4 , x3 , x2 , x1] + E[θw4 |x4 , x3 , x2 , x1] = θw4 (1 pt)
Next, the mean squared error =E[(x5 - x5(4))2]
= E[(w5 + θw4 - θw4 )2]
= E[w5(2)] = Var(w5 ) = 2 (2 pt)
b. Consider the following AR(3) process
xt - 0.40xt − 1 - 0.30xt −2 - 0.20xt −3 = wt ,
where wt ~ i.i.d N (0, 1). We wish to forecast x14 using the following values: x1 = 3.1, x2 = 0.93, x3 = -2.3, x4 = -3.2, x5 = -1.3
x6 = -2.5, x7 = 2.6, x8 = 2.0, x9 = -3.5, x10 = -0.74.
Give a 90% best linear prediction interval for the value x14 . Assume z0.05 = -1.645 and z0.95 = 1.645. Give your answers to four decimal places. (5 pts)
Solutions:
The BLP of xn+1 is of the form
xn(n)+1 = E[xn(n)+1 |x10 , x9 , . . . x1] = 0.40xn + 0.30xn − 1 + 0.20xn −2
x11(n) = 0.40x10(n) + 0.30x9(n) + 0.20x8(n) = 0.40(-0.74) + 0.30(-3.5) + 0.20(2) = -0.946
(0.50 pt)
x12(n) = 0.40x11(n) + 0.30x10(n) + 0.20x9(n) = 0.40(-0.946) + 0.30(-0.74) + 0.20(-3.5) =
-0.8564
(0.50 pt)
x13(n) = 0.40x12(n)+0.30x11(n)+0.20x10(n) = 0.40(-0.8564)+0.30(-0.946)+0.20(-0.74) =
-0.7744
(0.50 pt)
x14(n) = 0.4x13(n) + 0.3x12(n) + 0.2x11(n) = 0.4(-0.7744) + 0.3(-0.8564) + 0.2(-0.946) =
-0.7559.
(0.50 pt)
Now we need to compute Pn(n)+4 mean square prediction error to find the pre- diction interval x14(n) ± 1.645 .Pn(n)+4 .
We know that Pn(n)+m = σw(2)
1 ψj(2) with initial conditions ψk =
j(k)=1 φj ψk −j
for k < p = 3. When k ≥ p, we have ψk = j(3)=1 φj ψk −j , where φ 1 = 0.4, φ2 = 0.3, φ3 = 0.2 are the coefficients to the given autoregressive polynomial.
ψ0 = 1
ψ 1 = φ 1 ψ0 = 0.40 (0.50 pt)
ψ2 = φ 1 ψ 1 + φ2 ψ0 = 0.4(0.4) - 0.30 = -0.14 (0.50 pt)
ψ3 = φ 1 ψ2 + φ2 ψ 1 + φ3 ψ0 = 0.264 (0.50 pt)
Now, we can compute Pn(n)+m = σw(2)
1 ψj(2)
= ψ0(2) + ψ1(2) + ψ2(2) + ψ3(2) = 1.2493 (0.50 pt)
Now the prediction interval is: x14(n) ± 1.645 .Pn(n)+4 = -0.7559 ± 1.645 ′ 1.2493 = (-2.5946, 1.0828) (1 pt)
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