In jolars/TSAsolutions: Does not matter.
Models for stationary time series
First principles
We have the process
[
Y_t = 5 + e_t - \frac{1}{2}e_{t-i} + \frac{1}{4}e_{t-2}
]
and begin by working out its variance
[
\begin{aligned}
\text{Var}(Y_t) & = \text{Var}(5 + e_t - \frac{1}{2}e_{t-i} + \frac{1}{4}e_{t-2})\
& = \text{Var}(e_t) + \frac{1}{4}\text{Var}(e_t) + \frac{1}{16}\text{Var}(e_t)\
& = \frac{21}{16}\sigma_e^2
\end{aligned}
]
and then the autocovariance at lag 1
[
\begin{aligned}
\text{Cov}(Y_t, Y_{t-1}) & = \text{Cov}(5+e_t-\frac{1}{2}e_{t-1}+\frac{1}{4}e_{t-2}, 5+e_{t-1}-\frac{1}{2}e_{t-2}+\frac{1}{4}e_{t-3})\
& = \text{Cov}(-\frac{1}{2}e_{t-1},e_{t-1}) + \text{Cov}(\frac{1}{4}e_{t-2},-\frac{1}{2}e_{t-2}) \
& = -\frac{1}{2}\text{Var}(e_{t-1}) -\frac{1}{8}\text{Var}(e_{t-2})\
& = -\frac{5}{8}\sigma_e^2
\end{aligned}
]
lag 2
[
\begin{aligned}
\text{Cov}(Y_t, Y_{t-2}) & = \text{Cov}(5+e_t-\frac{1}{2}e_{t-1}+\frac{1}{4}e_{t-2}, 5+e_{t-2}-\frac{1}{2}e_{t-3}+\frac{1}{4}e_{t-4})\
& = \frac{1}{4}\text{Var}(e_{t-2}) \
& = \frac{1}{4}\sigma_e^2
\end{aligned}
]
and lag 3
[
\text{Cov}(Y_t, Y_{t-2}) = \text{Cov}(5+e_t-\frac{1}{2}e_{t-1}+\frac{1}{4}e_{t-2}, 5+e_{t-2}-\frac{1}{2}e_{t-3}+\frac{1}{4}e_{t-4}) = 0
]
which results in the autocorrelation
[
\rho_k = \begin{cases}
1 & k = 0\
\frac{-\frac{5}{8}\sigma_e^2}{\frac{21}{16}\sigma_e^2}=-\frac{10}{21} & k = 1\
\frac{\frac{1}{4}\sigma_e^2}{\frac{21}{16}\sigma_e^2}=\frac{4}{21} & k = 2\
0 & k = 3\
\end{cases} \tag*{$\square$}
]
Sketch autocorrelations
a {-}
tacf(ma = list(-0.5, -0.4))
b {-}
tacf(ma = list(-1.2, 0.7))
c {-}
tacf(ma = list(1, 0.6))
Max and min correlations for MA(1)
For
[
\rho_1 = \frac{-\theta}{1+\theta^2}
]
we retrieve extreme values at
[
\frac{\partial}{\partial \theta}\rho_1 = \frac{-1(1+\theta^2)-2\theta(-\theta)}{(1+\theta^2)^2} = \frac{\theta^2 - 1}{(1+\theta^2)^2} = 0
]
when $t = \begin{cases}-1\1\end{cases}$, which gives us
[
\begin{aligned}
\max \rho_1 & = \frac{-1(-1)}{1+(-1)^2} = 0.5\
\min \rho_1 & = \frac{-1}{1+1^2} = -0.5
\end{aligned}
]
which we graph in figure \@ref(fig:minmax)
theta <- seq(-10, 10, by = 0.01)
p1 <- (-theta) / (1 + theta^2)
plot(theta, p1, type = "l")
points(theta[which.max(p1)], max(p1))
points(theta[which.min(p1)], min(p1))
Non-uniqueness of MA(1)
[
\frac{-\frac{1}{\theta}}{1 + \left( \frac{1}{\theta}\right)^2} = \frac{-\frac{1}{\theta}\times\theta^2}{\left( 1 + \frac{1}{\theta^2} \right) \theta^2} = \frac{-\theta}{1+\theta^2} \tag*{$\square$}
]
Sketch more autocorrelations
theta <- c(0.6, -0.6, 0.95, 0.3)
lag <- c(10, 10, 20, 10)
for (i in seq_along(theta)) {
print(tacf(ar = theta[i], lag.max = lag[i]))
}
Difference function for AR(1)
a {-}
[
\begin{aligned}
\text{Cov}(\triangledown Y_t, \triangledown Y_{t-k}) & = \text{Cov}(Y_t-Y_{t-1}, Y{t-k}-Y_{t-k-1})\
& = \text{Cov}(Y_t, Y_{t-k}) - \text{Cov}(Y_{t-1},Y_{t-k}) - \text{Cov}(Y_t, Y_{t-k-1}) + \text{Cov}(Y_{t-1}, Y_{t-k-1})\
& = \frac{\sigma_e^2}{1-\phi^2}(\phi^2 - \phi^{k-1}-\phi^{k+1}+\phi^k) \
& = \frac{\sigma_e^2}{1-\phi^2}\phi^{k-1}(2\phi-\phi2-1)\
& = - \frac{\sigma_e^2}{1-\phi^2}(1-\phi)^2\phi^{k-1}\
& = - \sigma_e^2 \frac{(1-\phi)^2}{(1-\phi)(1+\phi)}\
& = -\sigma_e^2 \frac{1-\phi}{1+\phi}\phi^{k-1}
\end{aligned}
]
as required.
b {-}
[
\begin{aligned}
\text{Var}(W_t) & = \text{Var}(Y_t-Y_{t-1})\
& = \text{Var}(\phi_1Y_{t-1}+e_t-Y_{t-1})\
& = \text{Var}(Y_{t-1}(\phi-1)+\sigma_e^2)\
& = (\phi-1)^2\text{Var}(Y_{t-1}) + \text{Var}(e_t)\
& = \frac{\sigma_e^2}{1-\phi^2}(\phi^2-2\phi+1) + \sigma_e^2\
& = \frac{\sigma_e^2(\phi^2-2\phi+1+1-\phi^2)}{1-\phi^2}\
& = \frac{2\sigma_e^2(1-\phi)}{1-\phi^2} \
& = \frac{2\sigma_e^2}{1+\phi} \tag*{$\square$}
\end{aligned}
]
Characteristics of several models
a {-}
Only correlation at lag 1.
b {-}
Only autocorrelation at lag 1 and 2. Shape of process depends on values of
coefficients.
c {-}
Exponentially decaying correlation from lag 0.
d {-}
Different patterns in ACF that depends on whether roots are complex or real.
e {-}
Exponentially decaying correlations from lag 1.
AR(2)
First, we have variance
[
\text{Var}(Y_t) = \text{Var}(\phi_2 Y_{t-2} + e_t) = \phi_2^2 \text{Var}Y_{t-2} + \sigma_e^2
]
which, assuming stationarity, is equivalent to
[
\text{Var}(Y_{t-2}) = \phi_2^2\text{Var}(Y_{t-2}) + \sigma_e^2 \iff \
\sigma_e^2 = (1-\phi_2^2) \text{Var}(Y_{t-2}) \iff \
\text{Var}(Y_{t-2}) = \frac{\sigma_e^2}{1-\phi^2_2}
]
which requires that $-1 < \phi_2 < 1$ since $\text{Var}(Y_{t-2}) \geq 0$.
Sketching AR(2) processes
a {-}
[
\begin{split}
\rho_1 & = 0.6\rho_0 + 0.3\rho{-1} = 0.6 + 0.3\rho_1 = 0.8571 \
\rho_2 & = 0.6\rho_1+0.3\rho_0 = 0.81426\
\rho_3 & = 0.6\rho_2 + 0.3\rho_1 = 0.7457
\end{split}
]
The roots to the characteristic equation are given by
[
\frac{\phi_1 \pm \sqrt{\phi_1^2 + 4\phi_2}}{-2\phi_2} = \frac{0.6 \pm \sqrt{0.6 + 4 \times 0.3}}{-2 \times 0.3} = -1 \pm 2.0817 = {1.0817, -3.0817}.
]
Since both of these roots exceed 1 in absolute value, they are real. Next,
we sketch the theoretical autocorrelation function (\@ref(fig:ar2a)).
tacf(ar = c(0.6, 0.3))
b {-}
Next we write a function to do the work for us.
ar2solver <- function(phi1, phi2) {
roots <- polyroot(c(1, -phi1, -phi2))
cat("Roots:\t\t", roots, "\n")
if (any(Im(roots) > sqrt(.Machine$double.eps))) {
damp <- sqrt(-phi2)
freq <- acos(phi1 / (2 * damp))
cat("Dampening:\t", damp, "\n")
cat("Frequency:\t", freq, "\n")
}
tacf(ar = c(phi1, phi2))
}
ar2solver(-0.4, 0.5)
c {-}
ar2solver(1.2, -0.7)
d {-}
ar2solver(-1, -0.6)
e {-}
ar2solver(0.5, -0.9)
f {-}
ar2solver(-0.5, -0.6)
Sketch ARMA(1,1) models
a {-}
tacf(ar = 0.7, ma = -0.4)
b {-}
tacf(ar = 0.7, ma = 0.4)
ARMA(1,2)
a {-}
[
\begin{aligned}
\text{Cov}(Y_t, Y_{t-k}) & = \text{E}[(0.8Y_{t-1}+e_t+0.7e_{t-1}+0.6e_{t-2})Y_{t-k}] - \text{E}(Y_t)\text{E}(Y_{t-k})\
& = \text{E}(0.8Y_{t-1}Y_{t-k} + Y_{t-k}e_t + 0.7e_{t-1}Y_{t-k}+0.6e_{t-2}Y_{t-k}) - 0\
& = 0.8\text{E}(Y_{t-1}Y_{t-k}) + \text{E}(Y_{t-k}e_t) + 0.7\text{E}(e_{t-1}Y_{t-k}) + 0.7\text{E}(e_{t-2}Y_{t-k})\
& = 0.8\text{E}(Y_{t-1}Y_{t-k})\
& = 0.8\text{Cov}(Y_t,Y_{t-k+1})\
& = 0.8\gamma_{k-1} & \square
\end{aligned}
]
b {-}
[
\begin{split}
\text{Cov}(Y_t,Y_{t-2}) & = \text{E}[0.8Y_{t-1}+e_t+0.7e_{t-1}+0.6e_{t-2})Y_{t-2}]\
& = \text{E}[(0.8Y_{t-1}+0.6e_{t-2})Y_{t-2}]\
& = 0.8\text{Cov}(Y_{t-1},Y_{t-2})+0.6\text{E}(e_{t-2}Y_{t-2})\
& = 0.8\gamma_1 + 0.6\text{E}(e_t Y_t)\
& = 0.8\gamma_1 + 0.6\text{E}[e_t(0.8Y_{t-1}+e_t+0.7e_{t-1}+0.6e_{t-2})]\
& = 0.8\gamma_1 + 0.6\sigma_e^2 \iff \
\rho_2 & = 0.8\rho_1+0.6\sigma_e^2/\gamma_0 & \square
\end{split}
]
Two MA(2) processes
a {-}
For $\theta_1 = \theta_2 = 1/6$ we have
[
\rho_k = \frac{-\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}}{1 + \left(\frac{1}{6}\right)^2 + \left(\frac{1}{6}\right)^2} =
\frac{\frac{1}{6}\left(\frac{1}{6}-1\right)}{1 + \frac{2}{36}} = - \frac{5}{38}.
]
For $\theta_1 = -1$ and $\theta_2 = 6$,
[
\rho_k = \frac{1-6}{1+1^2+36} = - \frac{5}{38} \tag*{$\square$}.
]
b {-}
For $\theta_1 = \theta_2 = 1/6$ we have roots given by
[
\frac{\frac{1}{6} \pm \sqrt{\frac{1}{36}+ 4 \times \frac{1}{6}}}{-2\times \frac{1}{6}} =
- \frac{1}{2} \pm \frac{\sqrt{\frac{25}{36}}}{-\frac{1}{3}} = - \frac{1}{2} \pm \frac{\frac{5}{6}}{\frac{1}{3}} = {-3, -2}
]
and for $\theta_1 = -1$ and $\theta_2 = 6$,
[
\frac{-1 \pm \sqrt{1 + 4 \times 6}}{-2\times6} = \frac{-1\pm 5}{-12} = \frac{1}{12} \pm \frac{5}{12} = \left{-\frac{1}{3}, \frac{1}{2}\right}
]
Autocorrelation in MA(1)
[
\begin{aligned}
& \text{Var}(Y_{n+1}+Y_n+Y_{n-1}+ \dots + Y_1) = \left((n+1)+2n\rho_1\right)\gamma_0 = \left(1+n(1+2\rho_1)\right)\gamma_0\
& \text{Var}(Y_{n+1}-Y_n+Y_{n-1}- \dots + Y_1) = \left((n+1)-2n\rho_1 \right)\gamma_0 = \left(1 + n(1-2\rho_1)\right)\gamma_0
\end{aligned}
]
[
\begin{cases}
\left(1+n(1+2\rho-1)\right) \geq 0 \
\left(1+n(1-2\rho-1)\right) \geq 0 \
\end{cases} \iff
\begin{cases}
1+n+2\rho_1n \geq 0 \
1+n-2\rho_1n \geq 0 \
\end{cases} \iff
\begin{cases}
\rho_1 \geq \frac{-(n+1)}{2n}\
\rho_1 \leq \frac{n+1}{2n}
\end{cases} \iff
\begin{cases}
\rho_1 \geq -\frac{1}{2}(1+\frac{1}{n})\
\rho_1 \leq \frac{1}{2}(1+\frac{1}{n})
\end{cases}
]
where $\rho_1 \geq |1/2|$ for all $n$.
Zero-mean stationary process
We set $Y_t=e_t−θe_t−1$ and then we have
$$
\begin{split}
e_t & = \sum_{j=0}^\infty \theta^j Y_{t-j}\quad \text{and expanding into} \
& = \sum_{j=1}^\infty \theta^j Y_{t-j} + \theta^0 Y_{t-0} \
& \iff \
Y_t & = e_t - \sum_{j=1}^\infty \theta^j Y_{t-j}
\end{split}
$$
which is equivalent to
[
Y_t = \mu_0 + (1 + \theta B + \theta^2 B^2 + \dots + \theta^n B^n)e_t
]
which is the definition of a MA(1) process where $B$ is the backshift operator
such that $Y_t B^k =Y_{t-k}$.
Stationarity prerequisite of AR(1)
[
\begin{split}
\text{Var}(Y_t) & = \text{Var}(\phi Y_{t-1}+e_t) = \phi^2\text{Var}(Y_{t-1})+\sigma_e^2\
& = \phi^2 \text{Var}(\phi Y_{t-2}+e_t)+\sigma_e^2\
& = \phi^4 \text{Var}(Y_{t-2})+2\sigma_e^2\
& = \phi^{2n}\text{Var}(Y_{t-n})+n\sigma_e^2
\end{split}
]
And $\lim_{n \rightarrow \infty} \text{Var}(Y_t) = \infty$ if $|\phi$=1$, which
is impossible.
Nonstationary AR(1) process
a {-}
[
\begin{split}
Y_t & = \phi Y_{t-1}+e_t \implies \
- \sum_{j=1}^\infty \left(\frac{1}{3}\right)^j e_{t+j} & = 3 \left(-\sum_{j=1}^\infty \left(\frac{1}{3}\right)^j e_{t-1+j}\right) + e_t\
- \sum_{j=1}^\infty \left(\frac{1}{3}\right)^{j+1} e_{t+j} & = -\sum_{j=1}^\infty \left(\frac{1}{3}\right)^j e_{t-1+j} + \frac{1}{3} e_t\
- \sum_{j=1}^\infty \left(\frac{1}{3}\right)^{j+1} e_{t+j} & = -\sum_{j=2}^\infty \left(\frac{1}{3}\right)^j e_{t-1+j}\
- \sum_{j=1}^\infty \left(\frac{1}{3}\right)^{j+1} e_{t+j} & = -\sum_{j+1=2}^\infty \left(\frac{1}{3}\right)^{j+1} e_{t+j} & \square
\end{split}
]
b {-}
[
\text{E}(Y_t) = \text{E}(\sum_{j=1}^\infty \left(\frac{1}{3}\right)^j e_{t+j}) = 0
]
since all terms are uncorrelated white noise.
[
\begin{gathered}
\text{Cov}(Y_t,Y_{t-1}) = \text{Cov}\left( -\sum_{j=1}^\infty \left(\frac{1}{3}\right)^j e_{t+j}, \sum_{j=1}^\infty \left(\frac{1}{3}\right)^j e_{t+j-1} \right) = \
\text{Cov}\left(-\frac{1}{3}e_{t+1} - \left(\frac{1}{3}\right)^2e_{t+2} - \dots - \left(\frac{1}{3}\right)^n e_{t+n},
-\frac{1}{3}e_{t} - \left(\frac{1}{3}\right)^2e_{t+1} - \dots - \left(\frac{1}{3}\right)^n e_{t+n-1} \right) = \
\text{Cov}\left(-\frac{1}{3}e_{t+1},-\frac{1}{3^2}e_{t+1}\right) +
\text{Cov}\left(-\frac{1}{3}e_{t+2},-\frac{1}{3^3}e_{t+2}\right) + \dots +
\text{Cov}\left(-\frac{1}{3}e_{t+n},-\frac{1}{3^{n+1}}e_{t+n}\right) = \
\frac{1}{26}\sigma_e^2\left(1 + \frac{1}{3} + \frac{1}{3^2} + \dots + \frac{1}{3^n} \right)
\end{gathered}
]
which is free of $t$.
c {-}
It is unsatisfactory because Y_t depends on future observations.
Solution to AR(1)
a {-}
[
\frac{1}{2}\left(10\left(\frac{1}{2}\right)^{t-1} + e_{t-1} + \frac{1}{2}e_{t-2} + \left(\frac{1}{2}\right)^2e_{t-3} + \dots +
\left(\frac{1}{2}\right)^{n-1}e_{t-n}\right) + e_{t-1} = \
10 \left(\frac{1}{2}\right)^t + \frac{1}{2}e_{t-1} + \left(\frac{1}{2}\right)^2e_{t-2} + \left(\frac{1}{2}\right)^3 e_{t-3} +
\dots + \left( \frac{1}{2}\right)^n e_{t-n} + e_{t-1} = \
10\left(\frac{1}{2}\right)^{t-1} + e_{t-1} + \frac{1}{2}e_{t-2} + \left(\frac{1}{2}\right)^2e_{t-3} + \dots +
\left(\frac{1}{2}\right)^{n-1}e_{t-n} \qquad \square
]
b {-}
$\text{E}(Y_t) = 10 \left( \frac{1}{2}\right)^t$ varies with $t$ and thus is
not stationary.
Stationary AR(1)
a {-}
[
\text{E}(W_t) = \text{E}(Y_t + c\phi^t) = \text{E}(Y_t) + \text{E}(c\phi^t) = 0 + c\phi^t = c\phi^t \tag*{$\square$}
]
b {-}
[
\phi(Y_{t-1}+c\phi^{t-1})+e_t = \phi Y_{t-1} + c\phi^t + e_t = \phi \left( \frac{Y_t - e_t}{\phi}\right) + c \phi^t + e_t = Y_t + c\phi^t \tag*{$\square$}
]
c {-}
No, \text{E}(W_t) is not free of $t$.
Find a simpler model (1)
This is similar to an AR(1) with $\rho_k = -(-0.5)^k$.
ARMAacf(ar = -0.5, lag.max = 7)
ARMAacf(ma = -c(0.5, -0.25, 0.125, -0.0625, 0.03125, -0.0015625))
Find a simpler model (2)
This is like an ARMA(1,1) with $\phi = -0.5$ and $\theta = 0.5$.
[
\begin{cases}
\psi_1 = \phi - \theta = 1\
\psi_2 = (\phi - \theta)\phi = -0.5
\end{cases}
\implies
\begin{cases}
\theta = 0.5\
\phi = -0.5
\end{cases}
]
ARMAacf(ma = -c(1, -0.5, 0.25, -0.125, 0.0625, -0.03125, 0.015625))
ARMAacf(ar = -0.5, ma = -0.5, lag.max = 8)
ARMA in disguise
a {-}
[
\begin{gathered}
\text{Cov}(Y_t, Y_{t-k}) = \text{Cov}(e_{t-1}-e_{t-2}+0.5e_{t-3}, e_{t-1-k}-e_{t-2-k}+0.5e_{t-3-k}) = \
\gamma_k = \begin{cases}
\sigma_e^2 + \sigma_e^2 + 0.25\sigma_e^2 = 2.25\sigma_e^2 & k = 0\
-\sigma_e^2-0.5\sigma_e^2=-1.5\sigma_e^2 & k = 1\
0.5\sigma_e^2 & k = 2
\end{cases}
\end{gathered}
]
b {-}
This is an ARMA(p,q) in the sense that $p = 0$ and $q = 2$, that is, it is in
fact an MA(2) process $Y_t = e_t - e_{t-1}+0.5e_{t-2}$ with $\theta_1 = 1, \theta2 = -0.5$.
Equivalence of statements
[
1 - \phi_1x - \phi_2x^2 - \dots - \phi_p x^p \implies
x^k \left( \left(\frac{1}{k}\right)^p - \phi_1 \left(\frac{1}{k}\right)^{p-1} \dots - \phi_p \right)
]
Thus if $x_1 = G$ is a root to the above, $\frac{1}{x_1} = \frac{1}{G}$ must be a root to
[
x^p - \phi_1 x^{p-1} -\phi_2 x^{p-2} - \dots -\phi_p
]
Covariance of AR(1)
a {-}
[
\begin{gathered}
\text{Cov}(Y_t - \phi Y_{t+1}, Y_{t-k} - \phi Y_{t-k+1}) = \
\text{Cov}(Y_t, Y_{t-k}) - \phi \text{Cov}(Y_t, Y_{t+1-k}) - \phi\text{Cov}(Y_{t+1},Y_{t-k}) + \phi^2\text{Cov}(Y_{t+1, Y_{t+1-k}}) = \
\frac{\sigma_e^2}{1-\phi^2}\left(\phi^k - \phi \phi^{k-1} - \phi \phi^{k+1} + \phi^2 \phi\right) =\
\frac{\sigma_e^2}{1-\phi^2}\left( \phi^k - \phi^k - \phi^{k+2} + \phi^{k+2} \right) = 0
\end{gathered}
]
b {-}
First,
[
Y_{t+k} = \phi Y_{t+k-1} + e_t + k \implies
]
[
\begin{split}
\text{Cov}(b_t, Y_{t+k}) & = \text{Cov}(Y_t - \phi Y_{t+1, Y_{t+n}})\
& = \text{Cov}(Y_t,Y_{t+k}) - \phi \text{Cov}(Y_{t+1}, Y_{t+k)})\
& = \frac{\sigma_e^2}{1-\phi^2}\phi^k - \phi \frac{\sigma_e^2}{1-\phi^2}\phi^{k-1}\
& = 0 & \square
\end{split}
]
Recursion
a {-}
[
\begin{split}
E(Y_0) & = E[c_1e_0] = c_1E[e_0] = 0\
E(Y_1) & = E(c_2Y_0 + e_1) = c_2 E(Y_0) = 0\
E(Y_2) & = E(\phi_1 Y_1 + \phi_2 Y_0 + e_t) = 0\
& \vdots \
E(Y_t) & = E(\phi_1 Y_{t-1} + \phi_2 Y_{t-2}) = 0
\end{split}
]
b {-}
We have
[
\begin{cases}
\text{Var}(Y_0) = c_1^2\text{Var}(e_0) = c_1^2\sigma_e^2\
\text{Var}(Y_1) = c_2^2\text{Var}(Y_0) + \text{Var}(e_0) = c_2^2c_1^2\sigma_e^2=\sigma_e^2(1+c_1^2c_2^2)
\end{cases} \implies
]
[
\begin{cases}
c_1^2\sigma_e^2 = \sigma_e^2(1+c_1^2c_2^2) & \iff c_1^2(1-c_2^2) = 1\
c_1^2 = \frac{1}{1-c_2^2} & \iff c_1 = \sqrt{\frac{1}{1-c_1^2}} = \frac{1}{\sqrt{1-c_1^2}}
\end{cases}
]
with covariance
[
\begin{gathered}
\text{Cov}(Y_0, Y_1) = \text{Cov}(c_1e_0,c_2 c_1 e_0 + e_1) = \text{Cov}(c_1e_0,c_2 c_1 e_0) + \text{Cov}(c_1e_0, e_1) =\
c_1^2c_2\sigma_e^2 + c_1\text{Cov}(e_0, e_1) = c_1^2c_2\sigma_e^2 + 0
\end{gathered}
]
and autocorrelation
[
\rho_1 = \frac{c_1^2c_2\sigma_e^2}{\sqrt{(c_1^2)^2}} = c_2
]
so we must choose
[
\begin{cases}
c_2 = \frac{\phi_1}{1-\phi_2} \
c_1 = \frac{1}{\sqrt{1-c_2^2}}
\end{cases}.
]
c {-}
The process can be transformed by scaling and standardizing it and then shifting
with any given mean.
[
\frac{Y_t \sqrt{y_0}}{c_1} + \mu
]
Final exercise
a {-}
[
\begin{split}
Y_t & = \phi Y_{t-1} + e_t\
& = \phi(\phi Y_{t-2}+e_{t-1}) + e_t = \phi^2 Y_{t-2} + \phi e_{t-1} + e_t\
& \vdots \
& = \phi^t Y_{t-t} + \phi e_{t-1} + \phi^2 e_{t-2} + \dots + \phi^{t-1}e_1+e_t & \square
\end{split}
]
b {-}
[
E(Y_t) = E(\phi^t Y_0 +\phi e_{t-1} + \phi^2 e_{t-2} + \dots + \phi^{t-1}e_1+e_t) =\phi^t\mu_0
]
c {-}
[
\begin{split}
\text{Var}(Y_t) & = \text{Var}(\phi^t Y_0 + \phi e_{t-1}+\phi^2e_{t-2} + \dots + \phi^{t-1}e_1)\
& = \phi^{2t}\sigma_0^2+\sigma_e^2 \sum_{k=0}^{t-1}(\phi^2)^k\
& = \sigma_e^2 \frac{1-\phi^{2n}}{1-\phi^2} + \phi^{2t}\sigma_0^2 \quad\text{if $\phi \neq 1$ else}\
& = \text{Var}(Y_0) + \sigma_e^2 t = \sigma_0^2 + \sigma_e^2 t
\end{split}
]
d {-}
If $\mu_0 = 0$ then $E(Y_t) = 0$ but for $\text{Var}(Y_t)$ to be free of t,
$\phi$ cannot be 1.
e {-}
[
\text{Var}(Y_t) = \phi^2 \text{Var}(Y_{t-1}) + \sigma_e^2 \implies \phi^2 \text{Var}(Y_t) + \sigma_e^2
]
and
[
\text{Var}(Y_{t-1}) = \text{Var}(Y_t)(1-\phi^2) = \sigma_e^2 \implies \text{Var}(Y_t) = \frac{\sigma_e^2}{1-\phi}
]
and then we must have $|\phi| < 1$.
jolars/TSAsolutions documentation built on Oct. 20, 2022, 3:49 a.m.
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Models for stationary time series
First principles
We have the process [ Y_t = 5 + e_t - \frac{1}{2}e_{t-i} + \frac{1}{4}e_{t-2} ] and begin by working out its variance [ \begin{aligned} \text{Var}(Y_t) & = \text{Var}(5 + e_t - \frac{1}{2}e_{t-i} + \frac{1}{4}e_{t-2})\ & = \text{Var}(e_t) + \frac{1}{4}\text{Var}(e_t) + \frac{1}{16}\text{Var}(e_t)\ & = \frac{21}{16}\sigma_e^2 \end{aligned} ] and then the autocovariance at lag 1 [ \begin{aligned} \text{Cov}(Y_t, Y_{t-1}) & = \text{Cov}(5+e_t-\frac{1}{2}e_{t-1}+\frac{1}{4}e_{t-2}, 5+e_{t-1}-\frac{1}{2}e_{t-2}+\frac{1}{4}e_{t-3})\ & = \text{Cov}(-\frac{1}{2}e_{t-1},e_{t-1}) + \text{Cov}(\frac{1}{4}e_{t-2},-\frac{1}{2}e_{t-2}) \ & = -\frac{1}{2}\text{Var}(e_{t-1}) -\frac{1}{8}\text{Var}(e_{t-2})\ & = -\frac{5}{8}\sigma_e^2 \end{aligned} ] lag 2 [ \begin{aligned} \text{Cov}(Y_t, Y_{t-2}) & = \text{Cov}(5+e_t-\frac{1}{2}e_{t-1}+\frac{1}{4}e_{t-2}, 5+e_{t-2}-\frac{1}{2}e_{t-3}+\frac{1}{4}e_{t-4})\ & = \frac{1}{4}\text{Var}(e_{t-2}) \ & = \frac{1}{4}\sigma_e^2 \end{aligned} ] and lag 3 [ \text{Cov}(Y_t, Y_{t-2}) = \text{Cov}(5+e_t-\frac{1}{2}e_{t-1}+\frac{1}{4}e_{t-2}, 5+e_{t-2}-\frac{1}{2}e_{t-3}+\frac{1}{4}e_{t-4}) = 0 ] which results in the autocorrelation [ \rho_k = \begin{cases} 1 & k = 0\ \frac{-\frac{5}{8}\sigma_e^2}{\frac{21}{16}\sigma_e^2}=-\frac{10}{21} & k = 1\ \frac{\frac{1}{4}\sigma_e^2}{\frac{21}{16}\sigma_e^2}=\frac{4}{21} & k = 2\ 0 & k = 3\ \end{cases} \tag*{$\square$} ]
Sketch autocorrelations
a {-}
tacf(ma = list(-0.5, -0.4))
b {-}
tacf(ma = list(-1.2, 0.7))
c {-}
tacf(ma = list(1, 0.6))
Max and min correlations for MA(1)
For [ \rho_1 = \frac{-\theta}{1+\theta^2} ] we retrieve extreme values at [ \frac{\partial}{\partial \theta}\rho_1 = \frac{-1(1+\theta^2)-2\theta(-\theta)}{(1+\theta^2)^2} = \frac{\theta^2 - 1}{(1+\theta^2)^2} = 0 ] when $t = \begin{cases}-1\1\end{cases}$, which gives us [ \begin{aligned} \max \rho_1 & = \frac{-1(-1)}{1+(-1)^2} = 0.5\ \min \rho_1 & = \frac{-1}{1+1^2} = -0.5 \end{aligned} ] which we graph in figure \@ref(fig:minmax)
theta <- seq(-10, 10, by = 0.01) p1 <- (-theta) / (1 + theta^2) plot(theta, p1, type = "l") points(theta[which.max(p1)], max(p1)) points(theta[which.min(p1)], min(p1))
Non-uniqueness of MA(1)
[ \frac{-\frac{1}{\theta}}{1 + \left( \frac{1}{\theta}\right)^2} = \frac{-\frac{1}{\theta}\times\theta^2}{\left( 1 + \frac{1}{\theta^2} \right) \theta^2} = \frac{-\theta}{1+\theta^2} \tag*{$\square$} ]
Sketch more autocorrelations
theta <- c(0.6, -0.6, 0.95, 0.3) lag <- c(10, 10, 20, 10) for (i in seq_along(theta)) { print(tacf(ar = theta[i], lag.max = lag[i])) }
Difference function for AR(1)
a {-}
[ \begin{aligned} \text{Cov}(\triangledown Y_t, \triangledown Y_{t-k}) & = \text{Cov}(Y_t-Y_{t-1}, Y{t-k}-Y_{t-k-1})\ & = \text{Cov}(Y_t, Y_{t-k}) - \text{Cov}(Y_{t-1},Y_{t-k}) - \text{Cov}(Y_t, Y_{t-k-1}) + \text{Cov}(Y_{t-1}, Y_{t-k-1})\ & = \frac{\sigma_e^2}{1-\phi^2}(\phi^2 - \phi^{k-1}-\phi^{k+1}+\phi^k) \ & = \frac{\sigma_e^2}{1-\phi^2}\phi^{k-1}(2\phi-\phi2-1)\ & = - \frac{\sigma_e^2}{1-\phi^2}(1-\phi)^2\phi^{k-1}\ & = - \sigma_e^2 \frac{(1-\phi)^2}{(1-\phi)(1+\phi)}\ & = -\sigma_e^2 \frac{1-\phi}{1+\phi}\phi^{k-1} \end{aligned} ] as required.
b {-}
[ \begin{aligned} \text{Var}(W_t) & = \text{Var}(Y_t-Y_{t-1})\ & = \text{Var}(\phi_1Y_{t-1}+e_t-Y_{t-1})\ & = \text{Var}(Y_{t-1}(\phi-1)+\sigma_e^2)\ & = (\phi-1)^2\text{Var}(Y_{t-1}) + \text{Var}(e_t)\ & = \frac{\sigma_e^2}{1-\phi^2}(\phi^2-2\phi+1) + \sigma_e^2\ & = \frac{\sigma_e^2(\phi^2-2\phi+1+1-\phi^2)}{1-\phi^2}\ & = \frac{2\sigma_e^2(1-\phi)}{1-\phi^2} \ & = \frac{2\sigma_e^2}{1+\phi} \tag*{$\square$} \end{aligned} ]
Characteristics of several models
a {-}
Only correlation at lag 1.
b {-}
Only autocorrelation at lag 1 and 2. Shape of process depends on values of coefficients.
c {-}
Exponentially decaying correlation from lag 0.
d {-}
Different patterns in ACF that depends on whether roots are complex or real.
e {-}
Exponentially decaying correlations from lag 1.
AR(2)
First, we have variance [ \text{Var}(Y_t) = \text{Var}(\phi_2 Y_{t-2} + e_t) = \phi_2^2 \text{Var}Y_{t-2} + \sigma_e^2 ] which, assuming stationarity, is equivalent to [ \text{Var}(Y_{t-2}) = \phi_2^2\text{Var}(Y_{t-2}) + \sigma_e^2 \iff \ \sigma_e^2 = (1-\phi_2^2) \text{Var}(Y_{t-2}) \iff \ \text{Var}(Y_{t-2}) = \frac{\sigma_e^2}{1-\phi^2_2} ] which requires that $-1 < \phi_2 < 1$ since $\text{Var}(Y_{t-2}) \geq 0$.
Sketching AR(2) processes
a {-}
[ \begin{split} \rho_1 & = 0.6\rho_0 + 0.3\rho{-1} = 0.6 + 0.3\rho_1 = 0.8571 \ \rho_2 & = 0.6\rho_1+0.3\rho_0 = 0.81426\ \rho_3 & = 0.6\rho_2 + 0.3\rho_1 = 0.7457 \end{split} ]
The roots to the characteristic equation are given by
[ \frac{\phi_1 \pm \sqrt{\phi_1^2 + 4\phi_2}}{-2\phi_2} = \frac{0.6 \pm \sqrt{0.6 + 4 \times 0.3}}{-2 \times 0.3} = -1 \pm 2.0817 = {1.0817, -3.0817}. ]
Since both of these roots exceed 1 in absolute value, they are real. Next, we sketch the theoretical autocorrelation function (\@ref(fig:ar2a)).
tacf(ar = c(0.6, 0.3))
b {-}
Next we write a function to do the work for us.
ar2solver <- function(phi1, phi2) { roots <- polyroot(c(1, -phi1, -phi2)) cat("Roots:\t\t", roots, "\n") if (any(Im(roots) > sqrt(.Machine$double.eps))) { damp <- sqrt(-phi2) freq <- acos(phi1 / (2 * damp)) cat("Dampening:\t", damp, "\n") cat("Frequency:\t", freq, "\n") } tacf(ar = c(phi1, phi2)) }
ar2solver(-0.4, 0.5)
c {-}
ar2solver(1.2, -0.7)
d {-}
ar2solver(-1, -0.6)
e {-}
ar2solver(0.5, -0.9)
f {-}
ar2solver(-0.5, -0.6)
Sketch ARMA(1,1) models
a {-}
tacf(ar = 0.7, ma = -0.4)
b {-}
tacf(ar = 0.7, ma = 0.4)
ARMA(1,2)
a {-}
[ \begin{aligned} \text{Cov}(Y_t, Y_{t-k}) & = \text{E}[(0.8Y_{t-1}+e_t+0.7e_{t-1}+0.6e_{t-2})Y_{t-k}] - \text{E}(Y_t)\text{E}(Y_{t-k})\ & = \text{E}(0.8Y_{t-1}Y_{t-k} + Y_{t-k}e_t + 0.7e_{t-1}Y_{t-k}+0.6e_{t-2}Y_{t-k}) - 0\ & = 0.8\text{E}(Y_{t-1}Y_{t-k}) + \text{E}(Y_{t-k}e_t) + 0.7\text{E}(e_{t-1}Y_{t-k}) + 0.7\text{E}(e_{t-2}Y_{t-k})\ & = 0.8\text{E}(Y_{t-1}Y_{t-k})\ & = 0.8\text{Cov}(Y_t,Y_{t-k+1})\ & = 0.8\gamma_{k-1} & \square \end{aligned} ]
b {-}
[ \begin{split} \text{Cov}(Y_t,Y_{t-2}) & = \text{E}[0.8Y_{t-1}+e_t+0.7e_{t-1}+0.6e_{t-2})Y_{t-2}]\ & = \text{E}[(0.8Y_{t-1}+0.6e_{t-2})Y_{t-2}]\ & = 0.8\text{Cov}(Y_{t-1},Y_{t-2})+0.6\text{E}(e_{t-2}Y_{t-2})\ & = 0.8\gamma_1 + 0.6\text{E}(e_t Y_t)\ & = 0.8\gamma_1 + 0.6\text{E}[e_t(0.8Y_{t-1}+e_t+0.7e_{t-1}+0.6e_{t-2})]\ & = 0.8\gamma_1 + 0.6\sigma_e^2 \iff \ \rho_2 & = 0.8\rho_1+0.6\sigma_e^2/\gamma_0 & \square \end{split} ]
Two MA(2) processes
a {-}
For $\theta_1 = \theta_2 = 1/6$ we have [ \rho_k = \frac{-\frac{1}{6}+\frac{1}{6}\times\frac{1}{6}}{1 + \left(\frac{1}{6}\right)^2 + \left(\frac{1}{6}\right)^2} = \frac{\frac{1}{6}\left(\frac{1}{6}-1\right)}{1 + \frac{2}{36}} = - \frac{5}{38}. ]
For $\theta_1 = -1$ and $\theta_2 = 6$, [ \rho_k = \frac{1-6}{1+1^2+36} = - \frac{5}{38} \tag*{$\square$}. ]
b {-}
For $\theta_1 = \theta_2 = 1/6$ we have roots given by [ \frac{\frac{1}{6} \pm \sqrt{\frac{1}{36}+ 4 \times \frac{1}{6}}}{-2\times \frac{1}{6}} = - \frac{1}{2} \pm \frac{\sqrt{\frac{25}{36}}}{-\frac{1}{3}} = - \frac{1}{2} \pm \frac{\frac{5}{6}}{\frac{1}{3}} = {-3, -2} ]
and for $\theta_1 = -1$ and $\theta_2 = 6$, [ \frac{-1 \pm \sqrt{1 + 4 \times 6}}{-2\times6} = \frac{-1\pm 5}{-12} = \frac{1}{12} \pm \frac{5}{12} = \left{-\frac{1}{3}, \frac{1}{2}\right} ]
Autocorrelation in MA(1)
[ \begin{aligned} & \text{Var}(Y_{n+1}+Y_n+Y_{n-1}+ \dots + Y_1) = \left((n+1)+2n\rho_1\right)\gamma_0 = \left(1+n(1+2\rho_1)\right)\gamma_0\ & \text{Var}(Y_{n+1}-Y_n+Y_{n-1}- \dots + Y_1) = \left((n+1)-2n\rho_1 \right)\gamma_0 = \left(1 + n(1-2\rho_1)\right)\gamma_0 \end{aligned} ]
[ \begin{cases} \left(1+n(1+2\rho-1)\right) \geq 0 \ \left(1+n(1-2\rho-1)\right) \geq 0 \ \end{cases} \iff
\begin{cases} 1+n+2\rho_1n \geq 0 \ 1+n-2\rho_1n \geq 0 \ \end{cases} \iff
\begin{cases} \rho_1 \geq \frac{-(n+1)}{2n}\ \rho_1 \leq \frac{n+1}{2n} \end{cases} \iff
\begin{cases} \rho_1 \geq -\frac{1}{2}(1+\frac{1}{n})\ \rho_1 \leq \frac{1}{2}(1+\frac{1}{n}) \end{cases} ] where $\rho_1 \geq |1/2|$ for all $n$.
Zero-mean stationary process
We set $Y_t=e_t−θe_t−1$ and then we have
$$ \begin{split} e_t & = \sum_{j=0}^\infty \theta^j Y_{t-j}\quad \text{and expanding into} \ & = \sum_{j=1}^\infty \theta^j Y_{t-j} + \theta^0 Y_{t-0} \ & \iff \ Y_t & = e_t - \sum_{j=1}^\infty \theta^j Y_{t-j} \end{split} $$ which is equivalent to [ Y_t = \mu_0 + (1 + \theta B + \theta^2 B^2 + \dots + \theta^n B^n)e_t ] which is the definition of a MA(1) process where $B$ is the backshift operator such that $Y_t B^k =Y_{t-k}$.
Stationarity prerequisite of AR(1)
[ \begin{split} \text{Var}(Y_t) & = \text{Var}(\phi Y_{t-1}+e_t) = \phi^2\text{Var}(Y_{t-1})+\sigma_e^2\ & = \phi^2 \text{Var}(\phi Y_{t-2}+e_t)+\sigma_e^2\ & = \phi^4 \text{Var}(Y_{t-2})+2\sigma_e^2\ & = \phi^{2n}\text{Var}(Y_{t-n})+n\sigma_e^2 \end{split} ] And $\lim_{n \rightarrow \infty} \text{Var}(Y_t) = \infty$ if $|\phi$=1$, which is impossible.
Nonstationary AR(1) process
a {-}
[ \begin{split} Y_t & = \phi Y_{t-1}+e_t \implies \ - \sum_{j=1}^\infty \left(\frac{1}{3}\right)^j e_{t+j} & = 3 \left(-\sum_{j=1}^\infty \left(\frac{1}{3}\right)^j e_{t-1+j}\right) + e_t\ - \sum_{j=1}^\infty \left(\frac{1}{3}\right)^{j+1} e_{t+j} & = -\sum_{j=1}^\infty \left(\frac{1}{3}\right)^j e_{t-1+j} + \frac{1}{3} e_t\ - \sum_{j=1}^\infty \left(\frac{1}{3}\right)^{j+1} e_{t+j} & = -\sum_{j=2}^\infty \left(\frac{1}{3}\right)^j e_{t-1+j}\ - \sum_{j=1}^\infty \left(\frac{1}{3}\right)^{j+1} e_{t+j} & = -\sum_{j+1=2}^\infty \left(\frac{1}{3}\right)^{j+1} e_{t+j} & \square \end{split} ]
b {-}
[ \text{E}(Y_t) = \text{E}(\sum_{j=1}^\infty \left(\frac{1}{3}\right)^j e_{t+j}) = 0 ] since all terms are uncorrelated white noise.
[ \begin{gathered} \text{Cov}(Y_t,Y_{t-1}) = \text{Cov}\left( -\sum_{j=1}^\infty \left(\frac{1}{3}\right)^j e_{t+j}, \sum_{j=1}^\infty \left(\frac{1}{3}\right)^j e_{t+j-1} \right) = \ \text{Cov}\left(-\frac{1}{3}e_{t+1} - \left(\frac{1}{3}\right)^2e_{t+2} - \dots - \left(\frac{1}{3}\right)^n e_{t+n}, -\frac{1}{3}e_{t} - \left(\frac{1}{3}\right)^2e_{t+1} - \dots - \left(\frac{1}{3}\right)^n e_{t+n-1} \right) = \ \text{Cov}\left(-\frac{1}{3}e_{t+1},-\frac{1}{3^2}e_{t+1}\right) + \text{Cov}\left(-\frac{1}{3}e_{t+2},-\frac{1}{3^3}e_{t+2}\right) + \dots + \text{Cov}\left(-\frac{1}{3}e_{t+n},-\frac{1}{3^{n+1}}e_{t+n}\right) = \ \frac{1}{26}\sigma_e^2\left(1 + \frac{1}{3} + \frac{1}{3^2} + \dots + \frac{1}{3^n} \right) \end{gathered} ] which is free of $t$.
c {-}
It is unsatisfactory because Y_t depends on future observations.
Solution to AR(1)
a {-}
[ \frac{1}{2}\left(10\left(\frac{1}{2}\right)^{t-1} + e_{t-1} + \frac{1}{2}e_{t-2} + \left(\frac{1}{2}\right)^2e_{t-3} + \dots + \left(\frac{1}{2}\right)^{n-1}e_{t-n}\right) + e_{t-1} = \ 10 \left(\frac{1}{2}\right)^t + \frac{1}{2}e_{t-1} + \left(\frac{1}{2}\right)^2e_{t-2} + \left(\frac{1}{2}\right)^3 e_{t-3} + \dots + \left( \frac{1}{2}\right)^n e_{t-n} + e_{t-1} = \ 10\left(\frac{1}{2}\right)^{t-1} + e_{t-1} + \frac{1}{2}e_{t-2} + \left(\frac{1}{2}\right)^2e_{t-3} + \dots + \left(\frac{1}{2}\right)^{n-1}e_{t-n} \qquad \square ]
b {-}
$\text{E}(Y_t) = 10 \left( \frac{1}{2}\right)^t$ varies with $t$ and thus is not stationary.
Stationary AR(1)
a {-}
[ \text{E}(W_t) = \text{E}(Y_t + c\phi^t) = \text{E}(Y_t) + \text{E}(c\phi^t) = 0 + c\phi^t = c\phi^t \tag*{$\square$} ]
b {-}
[ \phi(Y_{t-1}+c\phi^{t-1})+e_t = \phi Y_{t-1} + c\phi^t + e_t = \phi \left( \frac{Y_t - e_t}{\phi}\right) + c \phi^t + e_t = Y_t + c\phi^t \tag*{$\square$} ]
c {-}
No, \text{E}(W_t) is not free of $t$.
Find a simpler model (1)
This is similar to an AR(1) with $\rho_k = -(-0.5)^k$.
ARMAacf(ar = -0.5, lag.max = 7) ARMAacf(ma = -c(0.5, -0.25, 0.125, -0.0625, 0.03125, -0.0015625))
Find a simpler model (2)
This is like an ARMA(1,1) with $\phi = -0.5$ and $\theta = 0.5$.
[ \begin{cases} \psi_1 = \phi - \theta = 1\ \psi_2 = (\phi - \theta)\phi = -0.5 \end{cases}
\implies
\begin{cases} \theta = 0.5\ \phi = -0.5 \end{cases} ]
ARMAacf(ma = -c(1, -0.5, 0.25, -0.125, 0.0625, -0.03125, 0.015625)) ARMAacf(ar = -0.5, ma = -0.5, lag.max = 8)
ARMA in disguise
a {-}
[ \begin{gathered} \text{Cov}(Y_t, Y_{t-k}) = \text{Cov}(e_{t-1}-e_{t-2}+0.5e_{t-3}, e_{t-1-k}-e_{t-2-k}+0.5e_{t-3-k}) = \ \gamma_k = \begin{cases} \sigma_e^2 + \sigma_e^2 + 0.25\sigma_e^2 = 2.25\sigma_e^2 & k = 0\ -\sigma_e^2-0.5\sigma_e^2=-1.5\sigma_e^2 & k = 1\ 0.5\sigma_e^2 & k = 2 \end{cases} \end{gathered} ]
b {-}
This is an ARMA(p,q) in the sense that $p = 0$ and $q = 2$, that is, it is in fact an MA(2) process $Y_t = e_t - e_{t-1}+0.5e_{t-2}$ with $\theta_1 = 1, \theta2 = -0.5$.
Equivalence of statements
[ 1 - \phi_1x - \phi_2x^2 - \dots - \phi_p x^p \implies x^k \left( \left(\frac{1}{k}\right)^p - \phi_1 \left(\frac{1}{k}\right)^{p-1} \dots - \phi_p \right) ]
Thus if $x_1 = G$ is a root to the above, $\frac{1}{x_1} = \frac{1}{G}$ must be a root to [ x^p - \phi_1 x^{p-1} -\phi_2 x^{p-2} - \dots -\phi_p ]
Covariance of AR(1)
a {-}
[ \begin{gathered} \text{Cov}(Y_t - \phi Y_{t+1}, Y_{t-k} - \phi Y_{t-k+1}) = \ \text{Cov}(Y_t, Y_{t-k}) - \phi \text{Cov}(Y_t, Y_{t+1-k}) - \phi\text{Cov}(Y_{t+1},Y_{t-k}) + \phi^2\text{Cov}(Y_{t+1, Y_{t+1-k}}) = \ \frac{\sigma_e^2}{1-\phi^2}\left(\phi^k - \phi \phi^{k-1} - \phi \phi^{k+1} + \phi^2 \phi\right) =\ \frac{\sigma_e^2}{1-\phi^2}\left( \phi^k - \phi^k - \phi^{k+2} + \phi^{k+2} \right) = 0 \end{gathered} ]
b {-}
First,
[ Y_{t+k} = \phi Y_{t+k-1} + e_t + k \implies ]
[ \begin{split} \text{Cov}(b_t, Y_{t+k}) & = \text{Cov}(Y_t - \phi Y_{t+1, Y_{t+n}})\ & = \text{Cov}(Y_t,Y_{t+k}) - \phi \text{Cov}(Y_{t+1}, Y_{t+k)})\ & = \frac{\sigma_e^2}{1-\phi^2}\phi^k - \phi \frac{\sigma_e^2}{1-\phi^2}\phi^{k-1}\ & = 0 & \square \end{split} ]
Recursion
a {-}
[ \begin{split} E(Y_0) & = E[c_1e_0] = c_1E[e_0] = 0\ E(Y_1) & = E(c_2Y_0 + e_1) = c_2 E(Y_0) = 0\ E(Y_2) & = E(\phi_1 Y_1 + \phi_2 Y_0 + e_t) = 0\ & \vdots \ E(Y_t) & = E(\phi_1 Y_{t-1} + \phi_2 Y_{t-2}) = 0 \end{split} ]
b {-}
We have [ \begin{cases} \text{Var}(Y_0) = c_1^2\text{Var}(e_0) = c_1^2\sigma_e^2\ \text{Var}(Y_1) = c_2^2\text{Var}(Y_0) + \text{Var}(e_0) = c_2^2c_1^2\sigma_e^2=\sigma_e^2(1+c_1^2c_2^2) \end{cases} \implies ]
[ \begin{cases} c_1^2\sigma_e^2 = \sigma_e^2(1+c_1^2c_2^2) & \iff c_1^2(1-c_2^2) = 1\ c_1^2 = \frac{1}{1-c_2^2} & \iff c_1 = \sqrt{\frac{1}{1-c_1^2}} = \frac{1}{\sqrt{1-c_1^2}} \end{cases} ] with covariance [ \begin{gathered} \text{Cov}(Y_0, Y_1) = \text{Cov}(c_1e_0,c_2 c_1 e_0 + e_1) = \text{Cov}(c_1e_0,c_2 c_1 e_0) + \text{Cov}(c_1e_0, e_1) =\ c_1^2c_2\sigma_e^2 + c_1\text{Cov}(e_0, e_1) = c_1^2c_2\sigma_e^2 + 0 \end{gathered} ] and autocorrelation [ \rho_1 = \frac{c_1^2c_2\sigma_e^2}{\sqrt{(c_1^2)^2}} = c_2 ] so we must choose [ \begin{cases} c_2 = \frac{\phi_1}{1-\phi_2} \ c_1 = \frac{1}{\sqrt{1-c_2^2}} \end{cases}. ]
c {-}
The process can be transformed by scaling and standardizing it and then shifting with any given mean. [ \frac{Y_t \sqrt{y_0}}{c_1} + \mu ]
Final exercise
a {-}
[ \begin{split} Y_t & = \phi Y_{t-1} + e_t\ & = \phi(\phi Y_{t-2}+e_{t-1}) + e_t = \phi^2 Y_{t-2} + \phi e_{t-1} + e_t\ & \vdots \ & = \phi^t Y_{t-t} + \phi e_{t-1} + \phi^2 e_{t-2} + \dots + \phi^{t-1}e_1+e_t & \square \end{split} ]
b {-}
[ E(Y_t) = E(\phi^t Y_0 +\phi e_{t-1} + \phi^2 e_{t-2} + \dots + \phi^{t-1}e_1+e_t) =\phi^t\mu_0 ]
c {-}
[ \begin{split} \text{Var}(Y_t) & = \text{Var}(\phi^t Y_0 + \phi e_{t-1}+\phi^2e_{t-2} + \dots + \phi^{t-1}e_1)\ & = \phi^{2t}\sigma_0^2+\sigma_e^2 \sum_{k=0}^{t-1}(\phi^2)^k\ & = \sigma_e^2 \frac{1-\phi^{2n}}{1-\phi^2} + \phi^{2t}\sigma_0^2 \quad\text{if $\phi \neq 1$ else}\ & = \text{Var}(Y_0) + \sigma_e^2 t = \sigma_0^2 + \sigma_e^2 t \end{split} ]
d {-}
If $\mu_0 = 0$ then $E(Y_t) = 0$ but for $\text{Var}(Y_t)$ to be free of t, $\phi$ cannot be 1.
e {-}
[ \text{Var}(Y_t) = \phi^2 \text{Var}(Y_{t-1}) + \sigma_e^2 \implies \phi^2 \text{Var}(Y_t) + \sigma_e^2 ] and [ \text{Var}(Y_{t-1}) = \text{Var}(Y_t)(1-\phi^2) = \sigma_e^2 \implies \text{Var}(Y_t) = \frac{\sigma_e^2}{1-\phi} ]
and then we must have $|\phi| < 1$.
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