I am reviewing the fundamentals of time series modelling and came across the following question regarding the concept of stationarity:

A time series is strictly stationary if its unconditional distribution (all moments) is invariant in time.

Consider the following stationary time series $\{{X_t}\}_{t\in{Z}}$ and let us focus on the first two moments of the distribution, namely:

  1. $E[X_t]= constant$
  2. $var[X_t]= constant$

The time invariant unconditional moments does not imply that the conditional moments ($E[X_t|X_{t-1}]$ and $var[X_t|X_{t-1}]$) are time invariant as well. I understand that the conditional expectation is random, due the random nature of conditioning part of $X_{t-1}$. But, I am not sure about $var[X_t|X_{t-1}]$ .

If we show a typical path of time series generated by an $AR(1)$ model, we observe a time-varying conditional distribution. Notice that the One realized path has a time varying conditional mean.

Simulated example

Given that a time series is strictly stationary, what can we say about the conditional distribution?

How is it possible to determine that a time series is stationary based on a single realization path?


I believe the conceptual issue I am facing relates to conditioning. Moreover, I think that the question may be answered by:

Law of total expectation: $E(x)=E[E(x|y)]$

Law of total variance: $var(x)=E[var(x|y)] + var[E(x|y)]$


1 Answer 1


I think you can think about it as long term versus short term values with the unconditional mean being the long term value and the conditional mean being the short term value. for AR(1) process $$ y_t = \phi y_{t-1} + \epsilon_t, \quad \phi < 1$$ you have the unconditional mean $E(y_t)=0$
and the conditional mean $E(y_t|y_{t-1},\ldots,y_1) = \phi y_{t-1}$

As you see, the unconditional mean is time-invariant. You can think of the multiple-path version of the process as being an approximation to a long term one-path version of the process.

If the conditional mean was 0, you then could use the law of iterated expectations to find the unconditional mean like so: $$E(y_t) = E[E[y_t|y_{t-1}]] = 0$$

  • $\begingroup$ I think I get it since: $E(y_t)=E[E(y_t|y_{t-1})]=E[\phi y_{t-1}]=\phi E[y_{t-1}]=0$. What can we say about the conditional variance $var[y_t|y_{t-1}]$ of the time series? Does the conditional variance vary or remain constant? $\endgroup$ Nov 17, 2019 at 15:56
  • $\begingroup$ @NadiaMerquez In AR(1) process the variance is assumed to be constant (assuming of course that $\phi < 1$), if it's not constant it is probably because $\phi$ is not less than 1. Therefore in a standard AR(1) process $\textrm{Var}(y_t) = \textrm{Var}(y_{t-1})$. Regarding what you wrote with the expectations here, its not entirely correct. I said if the conditional mean was zero then you can infer on the unconditional mean (so you already know that the inner expectation in the first equality is zero). $\endgroup$
    – Corel
    Nov 17, 2019 at 20:26
  • $\begingroup$ It is a rather trival result that if the conditional mean is zero, the unconditional mean is also zero. I just showed that if the conditional mean is not equal to zero, the unconditional mean could be equal to zero. Regarding the conditional variance, I believe it should be: $var(y_t|y_{t-1})=var(\phi y_{t-1}+\epsilon_t|y_{t-1})=0+var(\epsilon_t|y_{t-1})=\sigma^2_\epsilon$, therefore the conditional variance remains time invariant. $\endgroup$ Nov 18, 2019 at 9:09
  • $\begingroup$ @NadiaMerquez - It could but its not necessarily so. $\textrm{Var}(y_t) = \textrm{Var}(\phi y_{t-1} + \epsilon_t) = \phi^2\textrm{Var}(y_{t-1}) + \textrm{Var}(\epsilon_t) = \phi^2\textrm{Var}(y_{t-1}) + \sigma^2$ And now you solve the difference equation: $\textrm{Var}(y_t)(1-\phi^2L) = \sigma^2$ which yields $\textrm{Var}(y_t) = \frac{\sigma^2}{1-\phi^2}$. And so as I said, if $\phi < 1$ the variance is constant with time and thus $\textrm{Var}(y_t) = \textrm{Var}(y_{t-1})$ $\endgroup$
    – Corel
    Nov 18, 2019 at 9:50

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