Prove the Variance of an AR(2) Model

Question

Take a stationary AR(2) model, $y_t=\alpha+\phi_2y_{t-2}+\epsilon_t$

We know that $$Var[y_t]=E[y_t-E[y_t]]^2$$

Which is, $$Var[y_t]=E[y_t^2-E^2[y_t]]\\$$

\begin{aligned} Var[Y_t]&=E[y_t(\alpha+\phi_2y_{t-2}+\epsilon_t~)]-E^2[Y_t] \\ &=\alpha E[y_t]+\phi_2E[y_ty_{t-2}]+E[y_t\epsilon_t]-\frac{\alpha^2}{(1-\phi_2^2)^2} \\ &=\alpha E[y_t]+\phi_2cov[y_ty_{t-2}]+E[y_t]E[y_{t-1}]+E[y_t\epsilon_t]-\frac{\alpha^2}{(1-\phi_2^2)^2} \\ &=\frac{\alpha^2}{1-\phi_2^2}+\phi_2\gamma_2+\frac{\alpha^2}{(1-\phi_2^2)^2}+\sigma^2_n \end{aligned}

where $E[y_t]=\frac{\alpha}{1-\phi_2^2}$

I don't know where to go from the last line, provided that it is even correct, could anyone point me in the right direction? Thanks.

Answer 1

Given an AR(2) process

\begin{align} y_t = \alpha + \phi_1y_{t-1} + \phi_2y_{t-2} + \epsilon_t, \ \epsilon_t \sim \text{i.i.d}(0, \sigma^2), \end{align}

the mean is given by

\begin{align} \mathbb{E}\left[y_t\right] &= \mathbb{E}\left[\alpha + \phi_1y_{t-1} + \phi_2y_{t-2} + \epsilon_t\right] \\ &= \alpha + \phi_1\mathbb{E}\left[y_{t-1}\right] + \phi_2\mathbb{E}\left[y_{t-2}\right] + \underbrace{\mathbb{E}\left[\epsilon_t\right]}_{=0}, \end{align}

where we under the assumption of stationarity ($\mathbb{E}\left[y_t\right]=\mathbb{E}\left[y_{t-1}\right]=\mathbb{E}\left[y_{t-2}\right])$ get

\begin{align} \mathbb{E}\left[y_t\right](1-\phi_1-\phi_2) = \alpha \Leftrightarrow \mathbb{E}\left[y_t\right] = \frac{\alpha}{1-\phi_1-\phi_2}. \end{align}

Defining a new process as the deviation from it's mean, $\tilde{y}_t = y_t - \mu$, gives \begin{align} \tilde{y}_t =\phi_1\tilde{y}_{t-1} + \phi_2\tilde{y}_{t-2} + \epsilon_t, \end{align}

where we note that $\mathbb{V}\left[y_t\right] = \mathbb{E}\left[(y_t-\mu)^2\right] =\mathbb{E}\left[\tilde{y}_t^2\right] = \mathbb{V}\left[\tilde{y_t}\right], \ \mathbb{E}\left[\tilde{y}_t\right] = 0$,

such that we can find the variance of the AR(2) process $y_t$ by finding $\mathbb{E}\left[\tilde{y}_t^2\right]$.

Multiplying the equation for $\tilde{y_t}$ by $\tilde{y_t}$ and taking expectations yields

\begin{align} \mathbb{E}\left[\tilde{y}_t^2\right] \equiv \gamma_0 &=\phi_1\mathbb{E}\left[\tilde{y}_{t-1}\tilde{y}_t\right] + \phi_2\mathbb{E}\left[\tilde{y}_{t-2}\tilde{y}_t\right] + \mathbb{E}\left[\epsilon_t\tilde{y}_t\right] \\ &= \phi_1\gamma_1 + \phi_2\gamma_2 + \sigma^2, \end{align} where $\gamma_1$ and $\gamma_2$ are the autocovariances of first and second order and $\mathbb{E}\left[\epsilon_t\tilde{y}_t\right] = \mathbb{E}\left[\epsilon_t(\phi_1\tilde{y}_{t-1} + \phi_2\tilde{y}_{t-2} + \epsilon_t)\right] = \mathbb{E}\left[\epsilon_t^2\right] = \sigma^2$ while $\epsilon_t \sim \text{i.i.d}(0, \sigma^2)$.

Similarly, we can multiply the equation of $\tilde{y_t}$ by $\tilde{y}_{t-1}$ and $\tilde{y}_{t-2}$ and taking expectations as

\begin{align} \mathbb{E}\left[\tilde{y}_t \tilde{y}_{t-1}\right] &=\phi_1\mathbb{E}\left[\tilde{y}_{t-1}\tilde{y}_{t-1}\right] + \phi_2\mathbb{E}\left[\tilde{y}_{t-2}\tilde{y}_{t-1}\right] + \underbrace{\mathbb{E}\left[\epsilon_t\tilde{y}_{t-1}\right]}_{=0} \\ \Leftrightarrow \gamma_1 &= \phi_1\gamma_0 + \phi_2\gamma_1, \end{align}

and \begin{align} \mathbb{E}\left[\tilde{y}_t \tilde{y}_{t-2}\right] &=\phi_1\mathbb{E}\left[\tilde{y}_{t-1}\tilde{y}_{t-2}\right] + \phi_2\mathbb{E}\left[\tilde{y}_{t-2}\tilde{y}_{t-2}\right] + \underbrace{\mathbb{E}\left[\epsilon_t\tilde{y}_{t-2}\right]}_{=0} \\ \Leftrightarrow \gamma_2 &= \phi_1\gamma_1 + \phi_2\gamma_0, \end{align}

such that we have 3 equations with 3 unknowns, called the Yule-Walker equations, given by:

\begin{align} \gamma_0 &= \phi_1\gamma_1 + \phi_2\gamma_2 + \sigma^2, \\ \gamma_1 &= \phi_1\gamma_0 + \phi_2\gamma_1, \\ \gamma_2 &= \phi_1\gamma_1 + \phi_2\gamma_0. \end{align}

Rewriting the equation for $\gamma_1$ as $\gamma_1 = \phi_1\gamma_0 + \phi_2\gamma_1 \Leftrightarrow \gamma_1 = \frac{1}{1-\phi_2}\phi_1\gamma_0$ and substituting this together with $\gamma_2$ into the equation for $\gamma_0$ yields

\begin{align} \gamma_0 &= \phi_1\gamma_1 + \phi_2\gamma_2 + \sigma^2 \\ &= \phi_1\frac{1}{1-\phi_2}\phi_1\gamma_0 + \phi_2(\phi_1\gamma_1 + \phi_2\gamma_0) + \sigma^2 \\ &= \phi_1^2\frac{1}{1-\phi_2}\gamma_0 + \phi_2\phi_1\gamma_1 + \phi_2^2\gamma_0 + \sigma^2 \\ &= \phi_1^2\frac{1}{1-\phi_2}\gamma_0 + \phi_2\phi_1\frac{1}{1-\phi_2}\phi_1\gamma_0+ \phi_2^2\gamma_0 + \sigma^2 \\ &= \phi_1^2\frac{1}{1-\phi_2}\gamma_0 + \phi_2\phi_1^2\frac{1}{1-\phi_2}\gamma_0+ \phi_2^2\gamma_0 + \sigma^2 \Leftrightarrow \\ \gamma_0\left(1-\frac{\phi_1^2}{1-\phi_2} - \frac{\phi_2\phi_1^2}{1-\phi_2} - \phi_2^2\right) &= \sigma^2 \Leftrightarrow \\ \gamma_0\left(1-\frac{\phi_1^2(1-\phi_2)}{1-\phi_2} - \phi_2^2\right) &= \sigma^2 \Leftrightarrow \\ \gamma_0 &= \frac{\sigma^2}{1-\phi_1^2 - \phi_2^2}, \end{align} yielding the variance as a function of the parameters.

In your AR(2) process $\phi_1=0$ such that the variance becomes $\gamma_0 = \frac{\sigma^2}{1 - \phi_2^2} = \mathbb{V}\left[y_t\right]$.

Answer 2

I don’t have enough reputation to comment, but I believe the last line from the accepted answer contains a mistake:

$$ (1 - \frac{\phi_1^2}{1-\phi_2}-\frac{\phi_2 \phi_1^2}{1-\phi_2}-\phi_2^2) $$ $$ = (1 - \phi_1^2 (\frac{1}{1-\phi_2}+\frac{\phi_2}{1-\phi_2})-\phi_2^2) $$ $$ = (1 - \phi_1^2 (\frac{1+\phi_2}{1-\phi_2})-\phi_2^2) $$

The final result still holds since $\phi_1=0$.

See also this answer: Variance of a stationary AR(2) model