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TIME SERIES ANALYSIS
Directions:
MIDTERM EXAM
The exam is closed book and closed notes.
SHOW ALL NECESSARY WORK; JUSTIFY YOUR ANSWER.
SIGN AND DATE THE FOLLOWING DISCLAIMER.
SUBMIT BOTH THE EXAM PAPER AND YOUR BLUE BOOK. FAILING TO DO SO WILL RESULT IN A PENALTY OF 30% DEDUCTION OF YOUR SCORE.
You may use a calculator (cell phones & tablets are NOT allowed) and a formula sheet (both sides of letter-size paper).
The total number of points is 100. Time limit: 75 minutes
1
TIME SERIES ANALYSIS
SAMPLE MIDTERM
1. Consider the time series = + , where ~ 0,1, ~ 0,1, and is independent of .
(A)(10points)Computethemeanandautocovariancefunctionof .
(B) (15 points) Compute the mean and autocovariance function of the process := , and hence show its weak stationarity. Note: If ~0,1, then = 0 and = 2 − 12 − 3 ⋯ 1 for = 1, 2, ⋯.
2. Consider the ARMA process, − 1.3 + 0.42 = − 2, ~0, !.
(A) (10 points) Show that is causal but not invertible.
(B) (10 points) Compute “#, “, and “ in the MA(∞) expression of = ∑’&(# “&&, and give a recursive formula for computing “& , ) ≥ 3.
1, 1,
=+# ++,++cos063++4sin063+, ~0,1,
3. A model for a monthly time series is
where +#, +, +, and +4 are non-zero constants. Define ≔ ∇∇ = 1 − 91 − 9 .
(A) (5 points) Is weakly stationary? Justify.
(B) (5 points) Identify as an ARMA(p, q) model; make sure to specify the order(s). (C) (10 points) Find the ACVF of .
were 0,1, would be strictly stationary? Justify.
(A) (10 points) Let , , 4, and = be observations from the process. Find the best linear
(D) (5 points) If
4. ConsidertheAR(1)process =: +,where|:|<1and ~0,!.
predictor of > and compute the corresponding MSE.
(B) (15 points) Let and 4 be observations from the process. Find the best linear estimator of the missing value (in terms of and 4) and compute the corresponding MSE.
(C) (5 points) Find the PACF of .
2
TIME SERIES ANALYSIS
Solutions:
1. (A)= + =+ =0
?@ABC, = ?@ABC + BCBC, +
= ?@ABC, + ?@ABCBC, = 1 C(# + 1 C(# = 2 ∙ 1 C(#
(B)==+ =+ +2
=+ +0=2
?@ABC, = BC − BC
= E + + 2
BC BC BC BC BC BC
+2 F−4= +⋯−4 BC
=G2∙1C(# +1H+1+0+1+G8∙1C(# +2∙1C(± +1H+0+0 +0+4∙1C(# −4=14∙1C(# +2∙1C(±
is weakly stationary because (1) KLM = 14 < ∞ and (2) neither nor ?@ABC, depends on ,.
2. (A) AR polynomial :N = 1 − 1.3N + 0.42N = 1 − 0.6N1 − 0.7N, and hence the roots of :N = 0 are 1/0.6 and 1/0.7, both of which are outside the unit circle. So, is causal.
MA polynomial QN = 1 − 2N, and hence the root of QN = 0 is 0.5, which is inside the unit circle. So, is not invertible.
(B) The "-coefficients are determined by setting :N"N = QN, where "N = ∑'&(# "&N&. To be specific, we set
1−1.3N+0.42N"# +"N+"N +⋯=1−2N. Equating the coefficients of N& , ) = 0, 1, 2, on both sides, we obtain
"# = 1,
" −1.3"# =−2 ⇒ " =−2+1.3=−0.7,
" −1.3" +0.42"# =0 ⇒ " =1.3−0.7−0.42=−1.33.
Moreover, equating the coefficients of N& for ) ≥ 3, on both sides, we obtain "& − 1.3"& + 0.42"& = 0.
That is, "& = 1.3"& − 0.42"& for ) ≥ 3.
+
3
3.
(A) = +# + +, + + cos TUV + +4 sin TUV >>
TIME SERIES ANALYSIS
Sincedependson,, isnotweaklystationary.
(B) =1−91−9W+# ++,++cosTUV++4sinTUV+X=1−912+ +
−= − − +4. > > So, is an ARMA(0, 13), or MA(13) model.
(C) We know, for an MA(∞) model defined as =∑’&(#”&&, ~0,!, the ACVF is given by Yh = ! ∑’&(# “&”&B|C|.
Now,! =1and”&’sarezeroexcept”# =1,” =−1,” =−1,&”4 =1.So,the
desired ACVF is
1+−1+−1+1=4 h=0 1−1 + −11 = −2 |h| = 1
Yh = \ and Yh = 0 otherwise.
1−1 + −11 = −2 |h| = 12
1 |h| = 11 & 13
(D) would be strictly stationary because is a linear combination of , , ,
and 4.
4. =0,Y0= ]^^,and`h=:|C| foranyh.
_
(A) a = 4. Applying the operator bc to both sides of > = :d + > we obtain
The corresponding MSE is
b=> = :b=d = ::b== = :=.
>=>>===
−b = −: =1+: Y0−2: Y2=1−: Y0
= 1 + :!.
(B) Write the best linear estimator as L + e4. Then, L and e are determined by
or
So, the desired best linear estimator is
SolvingforLande,weobtainL=e= _ .
0`0 `23 TLV = 0`13, `2 `0 e `1
01 :3TLeV=0:3. : 1
B_^
4
TIME SERIES ANALYSIS
: +. The corresponding MSE is given by 1 + : 4
f − : + g =h1+ 2: iY0− 4: Y1+ 2: Y2 1+: 4 1+: 1+: 1+:
= Y0 h1 + 2: −
1+: 1+: 1+:
4: ∙ : + 2: ∙ :i
= ! ∙ 1−:= = ! . 1−: 1+: 1+:
(C) By definition, j0 = 1 and j1 = GY0HY1 = `1 = :. On the other hand, for AR(1) processes we have jh = 0 for h ≥ 2.