Question 4.1.

Part A.

Given the AR2 model \[X_{n}=1.5X_{n-1}-0.8X_{n-2}+\epsilon_{n},\] we have \[\epsilon_{n}=X_{n}-1.5X_{n-1}+0.8X_{n-2}.\] We are going to use the fact that the autocovariance of the white noise process is such that \(\gamma_\epsilon(h) = \sigma^2\) for \(h=0\) and \(\gamma_\epsilon(h) = 0\) otherwise. Thus the spectrum of the white noise is \[\lambda_\epsilon(\omega) = \sum_{h=-\infty}^\infty \gamma_\epsilon(h) e^{-i\omega h} = \sigma^2\] for all \(\omega\).

Now we observe that the ACF for \(\left\{\epsilon_{n}\right\}\) equals \[\begin{split} \gamma_{\epsilon}\left(h\right)&=&\mathrm{Cov}\left(\epsilon_{n+h},\epsilon_{n}\right)\\ &=&\mathrm{Cov}\left(X_{n+h}-1.5X_{n+h-1}+0.8X_{n+h-2},X_{n}-1.5X_{n-1}+0.8X_{n-2}\right)\\ &=&(1+1.5^2+0.8^2)\gamma_{X}\left(h\right)+(-1.5-1.5\times0.8)\left[\gamma_{X}\left(h+1\right)+\gamma_{X}\left(h-1\right)\right]+0.8\left[\gamma_{X}\left(h+2\right)+\gamma_{X}\left(h-2\right)\right]. \end{split}\]

Thus we see that \[\begin{split} \lambda_\epsilon(\omega) &= \sum_{h=-\infty}^\infty \gamma_\epsilon(h) e^{-i\omega h}\\ &= \sum_{h=-\infty}^\infty \left\{ 3.89 \gamma_X(h) - 2.7 [\gamma_X(h+1)+\gamma_X(h-1)] + 0.8[\gamma_X(h+2)+\gamma_X(h-2)] \right\} e^{-i\omega h}. \end{split}\] Note that \[\begin{split} \sum_{h=-\infty}^\infty \gamma_X(h+1)e^{-i\omega h} &= \sum_{h=\infty}^\infty \gamma_X(h+1) e^{-i\omega (h+1)} e^{i\omega} \\ &= e^{i\omega} \sum_{h'=-\infty}^\infty \gamma_X(h') e^{-i\omega h'}\\ &= e^{i\omega} \lambda_X(\omega) \end{split}\] where we used the change of variable \(h' = h+1\). Similarly, we have \[ \sum_{h=-\infty}^\infty \gamma_X(h-1)e^{-i\omega h} = e^{-i\omega} \lambda_X(\omega), \qquad \sum_{h=-\infty}^\infty \gamma_X(h+2)e^{-i\omega h} = e^{2i\omega} \lambda_X(\omega), \qquad \sum_{h=-\infty}^\infty \gamma_X(h-2)e^{-i\omega h} = e^{-2i\omega} \lambda_X(\omega). \]

It follows that \[ \sigma^2 = \lambda_\epsilon(\omega) = 3.89 \lambda_X(\omega) - 2.7 (e^{i\omega} + e^{-i\omega}) \lambda_X(\omega) + 0.8 (e^{2i\omega} + e^{-2i\omega})\lambda_X(\omega). \] Therefore, \[ \lambda_{X}(\omega)=\frac{\sigma^{2}}{3.89-5.4\cos(\omega)+1.6\cos(2\omega)}. \]

The plot of the spectral density and autocovariance are given as follows. Note that \(\omega = 2\pi f\), where \(f\) denotes the frequency.

library('TSA')
ARMAspec(model = list(ar = c(1.5, -0.8)))

model_AR = arima.sim(model = list( ar = c(1.5, -0.8)), n=1000)
acf(model_AR, type="covariance")

Part B.

Given the MA2 model

\[ X_{n}=\epsilon_{n-2}+\epsilon_{n-1}+\epsilon_{n} \] with \(\mathrm{Var}\left(\epsilon_{n}\right)=\sigma^{2}\), we start by writing the ACF \[\begin{eqnarray} \gamma\left(0\right)&=&\mathrm{Cov}\left(X_{n},X_{n}\right)=3\sigma^{2}\\ \gamma\left(1\right)&=&\gamma\left(-1\right)=\mathrm{Cov}\left(X_{n},X_{n-1}\right)=2\sigma^{2}\\ \gamma\left(2\right)&=&\gamma\left(-2\right)=\mathrm{Cov}\left(X_{n},X_{n-2}\right)=\sigma^{2}\\ \gamma\left(h\right)&=&0\;\forall|h|\geq3. \end{eqnarray}\] So we have \[\begin{split} \lambda(\omega)&=\sum_{h=-\infty}^\infty \gamma\left(h\right)e^{- i\omega h}\\ &=\sum_{h=-2}^{2}\gamma(h)e^{-i\omega h}\\ &=\gamma(0)+2 \gamma(1)\cos(\omega)+2\gamma(2)\cos(2\omega)\\ &=\sigma^{2}\left[3+4\cos(\omega)+2\cos(2\omega)\right] \end{split}\] The plot of the spectral density and autocovariance are given as follows:

ARMAspec(model = list(ma = c(1, 1)))

model_MA = arima.sim(model = list( ma = c(1, 1)), n=1000)
acf(model_MA, type="covariance", plot= TRUE)

Part C.

For part A, if we look closely at the spectrum density plot, we can see a peak at frequency around 0.1 (actually slightly less than 0.1), which is the dominant frequency. This indicates that the dominant period is around 10 (precisely larger than 10). If we look at the ACF, we can see there exists an oscillatory behavior characteristic of period = 11. We see that these two are matched with each other.

For part B, there is no appearant peak on the spectrum density plot, and there is no periodic behavior on the ACF plot either. Again, these two correspond to each other.

Question 4.2.
We first read in the data from the source. We seek to find out the relationship between time and number. We now make a time plot of the data to explore.

mydata = read.table(file="http://ionides.github.io/531w16/hw/sunspots.txt", header = TRUE)
year = mydata$Time
number = mydata$Number
plot(year, number, type= "l")

From the plot, we see that there is a periodic behavior, with regularly spaced peaks. The time interval between these peaks are about 11 years. This is as expected from reading

http://solarscience.msfc.nasa.gov/SunspotCycle.shtml.

Now we take a look at spectrum density.

spectrum(number, main = "Unsmoothed periodogram")

We first use R to see an unparametric method result of the data.

smoothed_r = spectrum(number, spans=c(30,30), main = "Smoothed periodogram")

We now determine the dominant frequency.

smoothed_r$freq[which.max(smoothed_r$spec)]
## [1] 0.007407407

We see that the dominant frequency is 0.007407407, which corresponds to a period of 11.25 years.

Now we use parametric method to estimate the spectral density.

estimated = spectrum(number, method = "ar", main = "Spectrum estimated via AR model picked by AIC")
abline(v=estimated$freq[which.max(estimated$spec)], lty="dotted")

The dominant frequecy is given by

estimated$freq[which.max(estimated$spec)]
## [1] 0.008016032

We see that the dominant frequency is 0.008016032, which corresponds to a period of 10.4 years.

the results from two estimations are very close to each other and they are close to the 11 year period. Therefore we can conclude that both of them are reliable and can be used.

Question 4.3.

All statements of sources were given full credit as long as they were consistent with the solutions presented.

Acknowledgements

Parts of this solution are adapted from a previous homework submission by Xiang Gao