Find Variance of AR(2) process $X_i = 0.3X_{i-2} + u_i$

Question

Full question:

$X_0,X_1, …., X_n$ are distributed according to the following AR(2) process

$$X_i = 0.3X_{i-2} + u_i$$

for $i=1,...,n$, $X_0=X_1=0$, and $u_i$ are iid $N(0,3^2)$.

Have no idea where to start with this one.

Any help would be much appreciated.

Here is what I know:

$$E(X_i) = E(.3X_{i-2} + u_i) = .3E(X_{i-2})+E(u_i) = 3(0)+0$$

What I have tried: \begin{align*} Var(X_i)&= Var(.3X_{i-2} + u_i)\\ &=.3^2(Var(X_{i-2}))+Var(u_i)\\ &= .3^2(Var(X_{i-2}))+9\\ &= ? \end{align*}

I don't know how to take the variance of $X_{i-2}$ because when you expand it, you get a ton more $u_i$s that I don't know what to do with

Answer 1

Here's a series of hints. I leave the proof to the reader.

Hint 1: For large enough $i$, we have the recursion $\mathrm{Var}(X_i)=:v_i=\alpha+\beta^2v_{i-2}$, for some constants $\alpha$ and $\beta$.

Hint 2: This recursion has the closed form solution $v_i = C_1\beta^i + C_2(-\beta)^i + \alpha/(1-\beta^2)$, where $C_1$ and $C_2$ are constants that may be found using the initial conditions.

Hint 3: It may be helpful to notice that $v_i \to 9/(1-0.3^2)$ as $i\to\infty$.

Answer 2

You can verify that this AR(2) process is stationary, based on the value of the coefficients (0 and .3). One implication of stationarity is that the variance is constant. You know

$$ {\rm var}(X_t) = .3^2 {\rm var}(X_{t-2}) + 9 $$

So the answer is the $x$ that solves

$$ x= .3^2 x + 9 $$

You can take it from there.