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Suppose I have an AR(1) process of the form:

$$y_t = \phi y_{t-1} + \epsilon_t$$

where $\epsilon_t$ is a white noise process with mean zero and variance $\sigma^2$.

If $|\phi| < 1 $ , the model is called 'stable' and hence 'stationary' and I can solve it backward (i.e. backward iteration) to get : $y_t = \sum_{j=0}^{\infty} \phi^j \epsilon_{t-j}$.

If $|\phi |> 1 $ the model is not 'stable' and not 'stationary'. That is, the first two moments should not depend on time $t$. However, when I solve the above equation forward I get:

$y_t = - \sum_{j=0}^{\infty} \phi^{-(j+1)} \epsilon_{t+j+1}$

The sum here will converge and I can take the unconditional mean and variance $y_t$ to see that it does not depend on time, since $\epsilon$ is a white noise I can do this calculation quite easily.

So how come the condition that $|\phi |< 1 $ is given as the equivalent condition for weak stationarity. As I did here, I can just solve the equation forward and get time independent mean and covariances - since the sum will converge and the errors are white noise.

I would appreciate any help.

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  • $\begingroup$ Minor thing, stationarity condition should read $|\phi| < 1$. Stationarity implies conditional joint probability does not change with an arbitrary lag. $\endgroup$ Commented Apr 3, 2022 at 5:10
  • $\begingroup$ @msuzen I didn't say the otherwise since it was the 'if' direction. Anyway added the absolute value anyway lol $\endgroup$
    – Kuantew
    Commented Apr 3, 2022 at 15:07
  • $\begingroup$ @Kuantew: I've seen your argument in some textbook that I can't recall the title of so I'm pretty sure it's correct. The issue is that you would be using noise terms in the future in order predict the value of $y_t$ currently. This is clearly not possible. So, I guess for ease of explanation, the text's just explain non-stationarity as $y_t$'s mean increasing without bound but another way to say is that we don't have the observations in the derived expression because they are in the future. $\endgroup$
    – mlofton
    Commented Apr 3, 2022 at 16:56
  • $\begingroup$ Indeed, both solutions are stationary in that as in the answer below the moments do not depend on time. This example serves to point out that stability and stationarity are not the same thing. The backward solution is what is known as the "causal" solution (where causal in this context means "dependent on the past). $\endgroup$ Commented May 27 at 7:54

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A stochastic process is said to be weak (or covariance) stationary if its first and second moment are both constant and finite, and so time-independent, and if the autocovariance is a function of the lags only. This is exactly the case for your AR(1) process since $\phi$ is lower than one. Indeed:

Condition 1. mean constant and finite, meaning that $E[y{_t}]= \mu_y$ $~~$ and $~~$ $\mu_y<\infty$

$E[y{_t}]= \mu_y =\sum_{j=0}^{\infty}\phi^jE[\epsilon_{t-j}] = 0 $ -> since $E[\epsilon_t]=0$, $\:$ $ \forall t $ $~$; Condition 1 satisfied.

Condition 2. variance constant and finite:

$E[y{_t^2}]$ $= \sum_{j=0}^{\infty}\phi^{2j}E[\epsilon^2_{t-j}]$= $\sigma^2_{\epsilon}$ $\sum_{j=0}^{\infty}\phi^{2j}$ = $\frac{\sigma_{\epsilon}^2}{(1-\phi^2)}$; $~~$ Condition 2 satisfied since $\ |\phi|<1$

As you can see the key point for having a finite variance (a necessary condition for weak stationarity) is that $\phi$ must be lower than 1 in module otherwise the infinite summation will not converge to a finite value

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