Consider a stochastic process $\{X_t, t = 1, 2, \ldots\}$ following the model $$X_t = \alpha X_{t-1} + e_t,$$ where $e_t \thicksim f$.

Can I say that the distribution of the initial point, $X_1$, is the same as $f$?

Can I say that the stationary marginal density, if it exists, of $\{X_t\}$ is same as $X_2 (\stackrel{D}{=}\alpha X_1 + e_2)$?

I think that the stationary marginal density, if it exists, of $\{X_t\}$ is the same as $X_2$, but not necessarily same as $X_1$.


1 Answer 1


Time-series processes defined only by a recursive equation are not fully specified, since they depend on specification of a "starting distribution". Unless there is some additional restriction you are required to meet, you can use any starting distribution you like, and you will still have an AR model that conforms with the specified recursive equation. However, having said this, it is usually the case that we want to specify a stationary AR model, which imposes an additional restriction beyond the recursive equation. If you would like your AR model to be stationary, then you require $|\alpha|<1$ and you also need to choose the marginal distribution of the initial value to be equal to the asymptotic distribution of the process.

To obtain the marginal distribution required for a stationary model, you set the variance for the marginal distribution by setting its variance to $\sigma_X^2 = \mathbb{V}(X_t) = \mathbb{V}(X_{t-1})$. From the recursive equation defining your AR process you have:

$$\begin{equation} \begin{aligned} \sigma_X^2 = \mathbb{V}(X_t) &= \mathbb{V}(\alpha X_{t-1} + e_t) \\[6pt] &= \alpha^2 \mathbb{V}(X_{t-1}) + \mathbb{V}(e_t) \\[6pt] &= \alpha^2 \sigma_X^2 + \sigma^2. \\[6pt] \end{aligned} \end{equation}$$

Solving for $\sigma_X$ yields:

$$\sigma_X^2 = \frac{\sigma^2}{1-\alpha^2}.$$

Hence, in order to obtain a (strongly) stationary AR model (with zero mean), you would use the starting distribution:

$$X_i \sim \text{N} \Big( 0, \frac{\sigma^2}{1-\alpha^2} \Big).$$

Using this starting distribution ensures that the time-series values all have that same marginal distribution, which gives you a stationary process. You will notice from this result that the marginal variance of the series is larger if the absolute value of the auto-correlation parameter is closer to one. That is because such processes have high auto-correlation, which leads to large swings in the process, leading to higher (marginal) variance.

One more thing to note here is that you do not have a mean term in your AR model, so it has an asymptotic mean of zero, and so we have used a mean of zero in the starting distribution. You could generalise your model to have a mean parameter if you wanted to, but that would change the recursive equation slightly. I have discussed this issue for the more general AR model in another answer to a similar question here, and I recommend you read that answer to supplement this one.

  • $\begingroup$ Hi @Ben, thank you very much for the answer! I was wondering why you have taken the asymptotic distribution of the starting point to be normal. Can it be any stable distribution? $\endgroup$
    – Joy
    Jul 18, 2018 at 20:40
  • $\begingroup$ The stable distribution in this model is the asymptotic distribution, but yes, you are just looking for stationarity, so any stable distribution will do it. $\endgroup$
    – Ben
    Jul 18, 2018 at 23:03

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