Solution : Problem Set 2 - Sung Y. Park

CUHK Dept. of Economics
Fall 2012 ECON 4120 Sung Y. Park
Question 1
Solution : Problem Set 2
(a) For any general ARMA series {xt}, we may express xt = ∞∑
is white noise.
Alternatively, we may also express ϵt = ∞∑
Hence, we have
E(ϵt+h|ϵt, ϵt−1, ϵt−2, ...) = E(ϵt+h|{ϵt, ϵt−1, ϵt−2, ...}, { ∞∑
= E(ϵt+h|{ϵt, ϵt−1, ϵt−2, ...}, {xt, xt−1, xt−2, ...})
= E(ϵt+h|{ ∞∑
αixt−i, ∞∑
αixt−1−i, ...}, {xt, xt−1, xt−2, ...})
i=0
i=0
= E(ϵt+h|xt, xt−1, xt−2, ...)
∵ {ϵt} is white noise
∴ E(ϵt+h|xt, xt−1, xt−2, ...) = E(ϵt+h|ϵt, ϵt−1, ϵt−2, ...) = 0
(b) AR(1) process: xt = ϕxt−1 + ϵt
=⇒ xt = ϕkxt−k + k−1 ∑
ϕ
i=0
iϵt−i =⇒ xt+h = ϕhxt + k−1 ∑
ϕ
i=0
iϵt+h−i =⇒ xt+h,h = E(ϕhxt + k−1 ∑
ϕ
i=0
iϵt+h−i|xt, xt−1, xt−2, ...)
= ϕh xt
xt+h
if
h > 0
h ≤ 0
AR(2) process: xt = a1xt−1 + a2xt−2 + ϵt
It is trivial that xt+h,t = xt+h for any h ≤ 0
xt+1 = a1xt + a2xt−1 + ϵt+1
βiϵt−i where β0 = 1 and {ϵt}
i=0
αixt−i where α0 = 1 and {ϵt} is white noise.
i=0
βiϵt−i,
i=0
∞∑
βiϵt−1−i, ...})
i=0
ϵt+h
if
h > 0
h ≤ 0
=⇒ xt+1,t = E(a1xt + a2xt−1 + ϵt+1 |xt, xt−1, xt−2, ...) = a1xt + a2xt−1
For any h > 1, xt+h,t can be constructed recursively using the following relation
xt+h,t = a1xt+h−1,t + a2xt+h−2,t + ϵt+h,t = a1xt+h−1,t + a2xt+h−2,t
For both process, xt+1 − xt+1,t = ϵt+1. Therefore, The prediction error variance is simply
V ar(ϵt+1) = σ 2
Question 2
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xt = ϵt + 0.2ϵt−1
x101 = ϵ101 + 0.2ϵt100 =⇒ x101,100 = 0.2ϵt100 = 0.002
x102 = ϵ102 + 0.2ϵt101 =⇒ x102,100 = 0
x101 − x101,100 = ϵ101 =⇒ σ1 = σϵ = √ 0.025 = 0.158
x102 − x102,100 = ϵ102 + 0.2ϵt101 =⇒ σ2 = √ (1 + 0.2 2 )σϵ = 0.161
V ar(xt) = (1 + 0.2 2 )σ 2 ϵ
Cov(xt, xt−1) = Cov(ϵt + 0.2ϵt−1, ϵt−1 + 0.2ϵt−2) = 0.2σ 2 ϵ
Cov(xt, xt−2) = Cov(ϵt + 0.2ϵt−1, ϵt−2 + 0.2ϵt−3) = 0
=⇒ ρ(1) = 0.2
(1+0.2 2 )
Question 3
= 0.192 and ρ(2) = 0
xt = 0.01 + 0.2xt−2 + ϵt
E(xt) = 0.01 + 0.2E(xt−2) =⇒ E(xt) = 0.01 = 0.0125
0.8
V ar(xt) = 0.22V ar(xt−2) + V ar(ϵt) =⇒ V ar(xt) = 0.02
0.96
= 0.0208
Cov(xt, xt−1) = 0 and Cov(xt, xt−2) = Cov(0.01 + 0.2xt−2 + ϵt, xt−2) = 0.2V ar(xt)
=⇒ ρ(1) = 0 and ρ(2) = 0.2
x101 = 0.01 + 0.2x99 + ϵ101 =⇒ x101,100 = 0.01 + 0.2x99 = 0.014
x102 = 0.01 + 0.2x100 + ϵ102 =⇒ x102,100 = 0.01 + 0.2x100 = 0.008
x101 − x101,100 = ϵ101 and x102 − x102,100 = ϵ102 =⇒ σ1 = σ2 = σϵ = √ 0.02 = 0.141
Question 4
To generate 10 ARIMA(1,0,1) series, we provide two methods. The first one is to use random
variable generator, and the second one is to directly use command “arima.sim”. See Code in
Appendix for details.
See Table 1 for required statistics, which is the output of the codes.You should notice that
the average estimates are quite close to the true value as the differences is smaller than 2 times
of estimated standard error.
Question 5
(a) See Figure 1 and Figure 2 for scatter diagrams. Just as the question mentioned, the scatter
plots shows that the particulate level seems to have a linear relation with mortality while
temperature is likely to be either linearly or quadratically related to mortality. To better show
the relationships, we also imposed fitted lines to the scatter plots (See Code for details).
(b) See Table 2 for regression result.
(c) See Figure 3 and Figure 4 for residuals plot and correlogram, respectively.
The residual plot clearly exhibits persistence and reveals no evidence against stationarity.
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The correlogram shows that the sample ACFs are significant at the first 5 lags and the
PACF are significant at the first two lag.
We also computed Q-Statistics, which rejects the null hypothesis at 1%. See Table 3.
All these evidence indicate serial correlation in the residuals.
As suggested by the sample correlogram, we fit the residuals using ARMA(2,q) where
q=0,1,...,4 and perform model selection using AIC.
Since ARMA(2,1) yields the smallest AIC among all the candidates, we proceed with
ARMA(2,1). To check if it is adequate to capture the dynamics in the regression residual,
we need to inspect the correlogram and Q-Statistics.
See Figure 4 and Table 4.
Fortunately, the correlogram and the Box-test indicates the absence of serial correlation.
Therefore, we suggests that regression residuals follow ARIMA(2,0,1).
(d) Since we need to transform the entire dataset, we proceed with ARIMA(2,0,0) for simplicity.
To illustrate how we transform all the input variables, we provide the following deduction.
Let ϕ(L) = 1 − θ1L − θ2L 2
if εt=θ1εt−1 − θ2εt−2 + vt where vt is white noise and
Mt = β00 + β0t + β1(Tt − T ) + β2(Tt − T ) 2 + β3Pt + εt
Then we have
ϕ(L) Mt = ϕ(L)β00 + β0ϕ(L)t + β1ϕ(L)(Tt − T ) + β2ϕ(L)(Tt − T ) 2 + β3ϕ(L)Pt + vt
Therefore, we apply the transformation ϕ(L) = 1 − θ1L − θ2L 2 to all the input variables.
See Code for data transformation.
See Table 5 for regression output with the transfromed data.
Please be noted that regression on transformed variables has a smaller residual standard
error than 6.385 obtained by part (b).
Question 6
We first plot the two input series.From Figure 6, we can see that both series exhibit stochastic
trend, which is further comfimed by correlogram in Figure 7 and Figure 8. Hence, we apply 1st
difference to the input variables.Then we run the first differenced log(future) on first differenced
log(spot).
From Figure 9, the correlogram suggests that serial correlation problem. To tackle this
problem, we include ARMA dynamics to the model. As we can see from the correlogram, both
sample ACFs and PACFs exhibit exponential decline after lag 4. Therefore, we fit the model
with ARMA(p,q) where (p,q)∈ [0, 4] × [0, 4] and perform model selection using AIC.
Table 6 suggests MA(4) as it yields lowest AIC value. We proceed with MA(4) first. Again,
we plot the correlogram to see whether serial correlation problem is eradicated this time. How-
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ever, Figure 10 clearly indicates the inadequacy of MA(4) as the sample ACFs and PACFs
remain significant after the first few lags.
Hence, we try ARMA(2,1), the one of second lowest AIC. Its correlogram seems nice as all
the sample ACF’s and PACF’s are individually insignificant. Let’s perform Box-Pierce test to
seek stronger support.
From Table 8, The Q-statistic confirms that the residuals are white noise. Moreover, the
histogram and density plot suggest that the residuals are normally distributed. Therefore, we
select this as our proposed model.
The estimation results are in Table 9.
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