Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$y_{MA}$ = $ε_t$ + $ε_{t-1}$ <- stationary

$y_{AR}$ = $ε_t$ + $y_{AR_{t-1}}$

$y_{AR_{t-1}}$ = $ε_t$ + $ε_{t-1}$ + $y_{AR_{t-2}}$

$y_{AR_{t-2}}$ = $ε_t$ + $ε_{t-1}$ + $ε_{t-2}$ + $y_{AR_{t-3}}$ <- non-stationary?


The MA time series is stationary. This makes sense to me because you're summing up normally distributed mean 0 random variables, but then, an AR process, if the $y_{AR_{t-i}}$ terms are all also sums of normally distributed mean 0 random variables, why is it non-stationary? Again here, you're summing normally distributed mean 0 random variables. I'm obviously not understanding something here, where is the flaw in my logic?

share|cite|improve this question
up vote 1 down vote accepted

For an AR process, the variance changes (and, in fact, increases) over time. Thus, for example,

$$\begin{align} Var(y_{AR_t}) &= Var(\epsilon_t) + Var(y_{AR_{t-1}}) \\ &> Var(y_{AR_{t-1}}). \end{align}$$

Since the definition of a stationary process is one whose probability distribution doesn't change under shifts in time, an AR process cannot be stationary.

You're right that the mean of $y_{AR_t}$ and $y_{AR_{t-1}}$ are both $0$, but stationary requires more than just the means being the same.

share|cite|improve this answer
I realized what I was missing this morning. In an MA process, each step consists of the sum of only two normally distributed mean 0 random variables, allowing the variance to stay constant. Thanks for the response, it helped me arrive my improved understanding :) – Soo Mar 2 '12 at 1:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.