Answers to Selected Exam Questions (2018/19)
1. 22
2. ωj = 2pij/n, where j/n, the frequency of the harmonic wave, is the number of cycles per
observation; PERIOD = wave length = n/j, the inverse of the frequency.
3. A = 2pi(1/44) = 0.143148, B = 44, C = 2.
4. a0 = 133.218, intercept estimate = a0/2 = 66.61.
5. D = n2 (a
2
0 + b
2
0) = 22(133.218
2) = 390434.78.
6. H0 : µ = 0. Note that nY¯
2 = D/2 = 195217.39. s2 = TSS/(n− 1) = 127.23636/43 = 2.959.
Hence F = 195217.39/2.959 = 65974.1. F(0.05;1,43) = 4.05. ∴ Reject H0. (See answer to
Question 7 of 2017/18 exam for more explanations).
7. The two tests yield F statistics of 35.21 and 10.72 respectively. Reject the null hypothesis in
both cases.
8. SINE coefficient estimate = 0; COSINE coefficient estimate = 0.155/2 = 0.0775.
9. The time period 2017 Q-1 corresponds to t = 49 (note that t = 0, 1, 2, · · · , 43). The rest is
straightforward.
10. E=127.23636; F=124.23238.
11. Let β1, β2 and β3 be the coefficients corresponding to the three seasonal dummies.
H0 : β1 = β2 = β3 = 0. F =
(3.57084−3.00398)/3
3.00398/39 = 2.453. F(0.05;3,39) = 2.84. ∴, cannot reject
H0.
12. No, the harmonic wave corresponding to the seasonal cycle, i.e, PERIOD=4, has a rather
small line spectrum (0.06). It is unlikely to be significant.
13. R2 = 124.23238/127.23636 = 0.976391.
14. R2 = 80.41+16.61+10.67+5.1127.23636 = 0.88646.
16. Could consider including lagged dependent variables in the set of regressors; or use a correction
procedure (e.g., Cochrane-Orcutt)
17. Note that t = 31 for 2012 Q-3. As t = 0, 1, · · · , 43, initial values are indexed by the subscript
-1 (i.e., initial period corresponds to period t = −1). From Table 3, l−1 = 69.44758 and
b−1 = −0.13202. Hence Ŷ31 = 69.44758 + (−0.13202)(31) = 65.35496. Now e31 = Y31− Ŷ31 =
0.01260. Hence Y31 = 65.36756. ∴ G = Y31/Ŷ31 = 65.36756/65.35496 = 1.000193.
1
19. Y0 and Y1 can be retrieved from information in Table 2 and values of Ŷ0 and Ŷ1 as in Ques-
tion 17. Note that t = 1 corresponds to the second observation of the series. The rest is
straightforward.
21. When α = 1, lt = Yt/snt−L. ∴ snt = δsnt−L + (1 − δ)snt−L = snt−L, i.e., snt of a given
season remains unchanged from the initial value.
22. The observation corresponds to t = 32. When α = 0, bt = γ(lt−1 +bt−1− lt−1)+(1−γ)bt−1 =
bt−1, i.e., bt remains unchanged from the initial value b−1, i.e., b−1 = b0 = b1 = b2 = b3 · · · .
Now, l0 = l−1 + b−1 = 69.44758 − 0.13202 = 69.31556, l1 = l0 + b0 = 69.31556 − 0.13202 =
69.18354 = l−1 + b−1(2),· · · , l32 = l−1 + b−1(33) = 69.44758 − 0.13202(33) = 65.09092.
23. No, all seasonal estimates are near 1, meaning that seasonal variations are minimal.
24. rj = cov(Yt, Yt−j)/var(Yt). One can compute the sample variance of Yt using either the
unbiased estimator TSS/(n − 1) or the maximum likelihood estimator TSS/n. The rest is
straightforward.
29. Let g = be the number of coefficients in the ARMA(p,q) model, and n′ = n− 1 = 43 be the
number of observations after first differencing. Now,
AIC1 = −2lnL1 + 2g
AIC2 = −2lnL2 + 2(g − 1)
BIC1 = −2lnL1 + gln(n′) = AIC1 − 2g + gln(n′)
BIC2 = AIC2 − 2(g − 1) + (g − 1)ln(n′)
Hence,
BIC2 −BIC1
=AIC2 −AIC1 + 2 − ln(n′)
=0.73 + 2 − ln(43)
= − 1.0312
So while the AIC favours the ARMA(p,q) model, the BIC favours the ARMA(p,q-1) model.
30. One can consider other diagnosis like the t and Q tests, or combining the models.
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