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

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 $N(0,3^2)$.

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

• I assume this questions asks what is ${\rm var}(X_i)$. To answer that, think about how to calculate the variance of a sum of random variables... – linksys Feb 10 '16 at 17:39
• Hint: Try taking the variance of both sides of the equation. – Brent Kerby Feb 10 '16 at 17:40
• Please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. – gung - Reinstate Monica Feb 10 '16 at 17:51
• I updated the question. Please Advise – Keith Feb 10 '16 at 18:03
• @Keith, start by determining ${\rm var}(X_2)$ and try to find a recursive relationship between ${\rm var}(X_k)$ and ${\rm var}(X_{k-2})$ – linksys Feb 10 '16 at 18:06

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 $i\to\infty$.

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.

• No, the process is not stationary, and it is not true that the variance is constant. Indeed, both $X_0$ and $X_1$ have $0$ variance, whereas $X_2$ has variance $0.3^2\times\operatorname{var}(X_0) + 9 = 9$, $X_4$ has variance $0.3^2\times\operatorname{var}(X_2) + 9 = 0.3^2\times 9+9 = 9.81$, etc. – Dilip Sarwate Feb 10 '16 at 20:07
• Obviously I was talking about steady state, not during its arbitrary initial conditions. Wherever you start from, the variance quickly converges to the answer I said above, but I know you're more concerned with mathematical pedantry than the spirit of the answer, so maybe you should take me to school by posting the "correct" answer. Thanks for playing – linksys Feb 10 '16 at 20:43
• @Student001, like the other guy, you can be as pedantic as you want about stationarity (it is extremely well approximated by a stationary process by about $t=4$) but I think post answers the question the OP probably meant to ask, so I won't be changing this answer. If you think downvotes or the condescending comments here affect me at all, you're pumping a dry well. I suggest you post the "right" answer, if you're so interested. – linksys Feb 10 '16 at 23:15
• It's good that your latest comment points out that you answer a different question than the one asked -- that you are using an approximation. I think it would be nice to mention that in the answer, but I can see this discussion won't lead to that, or anywhere else for that matter. – ekvall Feb 11 '16 at 1:45
• The stated question -- "Find Variance of AR(2) process $X_i=...$" -- is open to interpretation. I guess you interpreted it as asking the answerer enumerate the infinitely long list of variances ${\rm var}(X_0), {\rm var}(X_1), {\rm var}(X_2), ....$, which is certainly your prerogative. As I've made clear, I interpreted this as asking for the steady state variance of the process ($\approx 9.89$). – linksys Feb 11 '16 at 16:06