# Lecture 04 (Jan 31, 2022)

It is obvious that $\text{Cov}(x_{t-1},\epsilon_t) = 0$ because $\epsilon$ has mean zero. Thus,

$\text{Var}(x_t) = \text{Var} ( \phi_1 x_{t-1} + \epsilon_t) = \phi_1^2 \text{Var}(x_{t-1}) + \text{Var}(\epsilon_t)$

due to the fact that covariance between $x_{t-1}$ and $\epsilon_t$ is zero.

Because the series is stationary, the variance is the same

$\text{Var}(x_t) = \frac{\sigma^2}{1-\phi_1^2}$

Summary: if $x_t$ is weakly stationary $AR(1)$ model then, $|\phi_1| < 1.$ If $| \phi_1 | < 1$ then $\mathbb{E}(X_t) = \mu$. and

$\text{Var}(x_t) = \sum_{i=0}^\infty \phi_1^{2i} \sigma^2$

The covariance

$\gamma_l = \text{Cov}(x_t, x_{t-l}) = \sum_{i=0}^\infty \phi_1^i \phi_1^{i+l} \sigma^2<\infty$

Then $x_t$ is weakly stationary $AR(1)$ model.

Thus, $x_t$ is an $AR(1)$ model, $\text{abs}(\phi_1) <1$ if and only if $x_t$ is weakly stationary.

Alternaltively, the covariance is

\begin{aligned} \text{Cov} (x_t,\epsilon_t) &= \mathbb{E}[(x_t-\mu)\epsilon_t] \\ &= \mathbb{E}[ (\phi_1(x_{t-1}-\mu) + \epsilon_t )\epsilon_t] \\ &= \text{Var}(\epsilon_t) = \sigma^2 \end{aligned}

Thus

\begin{aligned} \gamma_l &= \text{Cov}(x_t, x_{t-l}) \\ &= \mathbb{E}[(x_t-\mu)(x_{t-l}-\mu)] \\ &= \phi_1 \gamma_{l-1} + \sigma^2 \mathbf{1}_{l=0} \end{aligned}

Thus we have a recursion for $\gamma_l$ :

$\gamma_0 = \frac{\sigma^2}{1-\phi^2_1} \quad \gamma_l = \phi_1\gamma_{l-1}$

For ACF, $\rho_0 =1$ and therefore, $\rho_l = \phi_1^l$. and this decays exponentially fast to zero for $AR(1)$ model.

$AR(2)$ model: $x_t = \phi_0 + \phi_1 x_{t-1} + \phi_2 x_{t-2} +\epsilon_t$.

By weakly stationary,

$\mu = \mathbb{E}(x_t) = \phi_0 + \phi_1 \mu + \phi_2\mu \quad \Rightarrow \mu = \frac{\phi_0}{1-\phi_1-\phi_2}$

similarly to the above process for the $AR(1)$ model, we have the recursion

\left\{ \begin{aligned} &\gamma_l = \phi_1 \gamma_{l-1} + \phi_2\gamma_{l-2} \\ &\rho_1 = \phi_1\rho_{l-1} + \phi_2\rho_{l-2} \end{aligned} \right.

when $l=1$, we have

\begin{aligned} &\rho_1 = \phi_1 + \phi_2 \rho_{-1} = \phi_1 + \phi_2 \rho_1 \\ &\Rightarrow \rho_1 = \frac{\phi_1}{1-\rho_2} \end{aligned}

If we define a shift operator $B$ such that $B \rho_l = \rho_{l-1}$, then we can rewrite the recursion

$0= (1-\phi_1 B - \phi_2 B^2 ) \rho_l$

For $x_{1,2}$ as the two solutions of the polynomial $1- \phi_1x -\phi_2 x^2 = 0$. By some alebra, we can rewrite

$(1-\frac{x}{x_1})(1-\frac{x}{x_2}) = 0$

Thus we can factorize as

$(1-\omega_1 B) (1-\omega_2 B) \rho_l = 0$

This shows $AR(2)$ is almost a two $AR(1)$ model for the auto-correlation function.

If we have complex solution, , then $\omega_{1,2}\in\mathbb{C}$.

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