Question 2.1.

A. Since \(\{\epsilon_{n}\}\) is white noise with variance \(\sigma^{2}\), then \[\begin{eqnarray} \gamma_{h}&=& \mathrm{Cov}(X_{n},\ X_{n+h})\\ &=& \mathrm{Cov}(X_{n},\phi X_{n+h-1}+\epsilon_{n+h})\\ &=&\phi\mathrm{\mathrm{Cov}}(X_{n},\ X_{n+h-1})+ \mathrm{Cov}(X_{n},\ \epsilon_{n+h})\\ &=&\phi\gamma_{h-1}. \end{eqnarray}\]

We can get \(\gamma_{0}\) by simple computing,

\[\begin{eqnarray} \gamma_{0}&=& \mathrm{Cov}(X_{n},\ X_{n})\\ \gamma_{0}&=& \mathrm{Cov}(\phi X_{n-1}+\epsilon_{n},\ \phi X_{n-1}+\epsilon_{n})\\ \gamma_{0}&=&\phi^{2} \mathrm{Cov}(X_{n-1},\ X_{n-1})+ \mathrm{Cov}(\epsilon_{n},\ \epsilon_{n})\\ \gamma_{0}&=&\phi^{2}\gamma_{0}+\sigma^{2}\\ (1\ -\phi^{2})\gamma_{0}&=&\sigma^{2}\\ \gamma_{0}&=&\frac{\sigma^{2}}{1-\phi^{2}} \end{eqnarray}\]

Let \(\gamma_{h}=A\lambda^{h}\), then \[\begin{eqnarray} A\lambda^{h}&=&\phi\mathrm{R}\lambda^{h-1}\\ \lambda^{h}&=&\phi\lambda^{h-1}\\ \lambda&=&\phi. \end{eqnarray}\]

Applying \(\gamma_{0}\) as an initial condition, then \[\begin{eqnarray} A\ \lambda^{0}&=&\gamma_{0}\\ &=&\frac{\sigma^{2}}{1-\phi^{2}}. \end{eqnarray}\] Therefore, \[ \gamma_{h}=\frac{\sigma^{2}}{1-\phi^{2}}\phi^{h}. \]

B. By Taylor series expansion, \[ g(x)=g(0)+g^\prime(0)x + \frac{1}{2}g^{(2)}(0)x^{2}+\frac{1}{3!}g^{(3)}(0)x^{3}+... \] Since \[\begin{eqnarray} g^{(n)}(0)&=&\frac{d^{n}}{dt^{n}}\frac{1}{1-\phi x}\\ &=&n!\phi^{n}x^{n}, \end{eqnarray}\] we have \[ g(x)\ =\sum_{n=0}^{\infty}\phi^{n}x^{n} \] The AR(1) model is equivalent to the following MA \((\infty)\) process \[\begin{eqnarray} X_{n}&=&\phi X_{n-1}+\epsilon_{n}\\ &=&\phi BX_{n}+\epsilon_{n}\\ (1-\phi B)X_{n}&=&\epsilon_{n}\\ X_{n}&=&(1-\phi B)^{-1}\epsilon_{n}\\ &=&\epsilon_{n}+\phi B\epsilon_{n}+\phi^{2} B^2\epsilon_{n}+...\\ &=&\epsilon_{n}+\phi\epsilon_{n-1}+\phi^{2}\epsilon_{n-2}+...\\ &=&\sum_{j=0}^{\infty}\phi^{j}\epsilon_{n-j}. \end{eqnarray}\]

Then, apply the general formular for the autocovariance function of the MA \((\infty)\) process with the constraint \(-1<\phi<1\), \[\begin{eqnarray} \gamma_{h}&=&\sum_{j=0}^{\infty}\psi_{j}\psi_{j+h}\sigma^{2}\\ &=&\sum_{j=0}^{\infty}\phi^{2j+h}\sigma^{2}\\ &=&\phi^{h}\sigma^{2}\sum_{j=0}^{\infty}\phi^{2j}\\ &=&\frac{\phi^{h}\sigma^{2}}{1-\phi^{2}}. \end{eqnarray}\]

which is the same as the answer in A.

C. From the above derivation, we have \[\begin{eqnarray} \rho_{h}&=&\frac{\gamma_{h}}{\gamma_{0}}\\ &=&\frac{\frac{\phi^{h}\sigma^{2}}{1-\phi^{2}}}{\frac{\sigma^{2}}{1-\phi^{2}}}\\ &=&\phi^{h} \end{eqnarray}\]

which is the same as R funtion ARMAacf by the following code.

set.seed(12345)
ar_coefs <- 0.6
phi <-ar_coefs
acf <- phi^(0:100)
Racf <- ARMAacf(ar=ar_coefs,lag.max=100)
all(abs(acf-Racf)<1e-6)
## [1] TRUE
plot(acf,type="l", col='red', xlab ="lag")
lines(Racf, lty =2, col='blue')
legend("topright", legend = c("ACF", "RACF"), col = c("red","blue"), lty =c(1,2))

Question 2.2. The solution of stochastic difference equation of the random walk model is: \[ X_{n}=\sum_{k=1}^{n}\epsilon_{k} \] Therefore, \[\begin{eqnarray} \gamma_{mn}&=&\mathrm{Cov}(X_{m},X_{n})\\ &=&\mathrm{Cov}\left(\sum_{i=1}^{m}\epsilon_{i},\sum_{j=1}^{n}\epsilon_{j}\right)\\ &=&\sum_{i=1}^{\min(m,n)}\mathrm{Var}(\epsilon_{i})\\ &=&\min(m,n)\sigma^{2} \end{eqnarray}\]

Putting together, we have, \[{\gamma_{mn}=\min(m,n)\sigma^{2}}\]

Question 2.3.

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