Causal and Noncausal Stationary Solutions of an AR(1) Model
Summary
The document explains why the lagged value in a stationary AR(1) model is not always independent of, or uncorrelated with, the current white-noise innovation. Under the causal condition, the stationary process can be represented using past innovations, which are uncorrelated with the current one; this makes the cross term vanish when calculating variance. Under the noncausal condition, the stationary representation instead uses future innovations, including the current innovation, so the cross term is nonzero.
The answer gives the corresponding cross-term result in the noncausal case and connects it to the variance recursion. It also notes that noncausal stationary solutions depend on future noise and are often recast as causal models with different innovations. These conclusions rely on the stationary-solution setup and white-noise assumptions described in the document; the discussion is about dependence structure, not a general claim that uncorrelated variables are independent.
Key ideas
- A causal stationary AR(1) solution depends on past white-noise innovations.
- In the causal case, the lagged process value is uncorrelated with the current innovation.
- A noncausal stationary solution depends on future innovations, so the current innovation can enter the lagged value.
- The variance recursion must retain the cross term in the noncausal case.
- Uncorrelatedness alone does not establish independence.
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Full text
# Why is $Z_t$ uncorrelated with $X_{t-1}$ in $X_t=\theta X_{t-1}+Z_t$?
# Why is $Z_t$ uncorrelated with $X_{t-1}$ in $X_t=\theta X_{t-1}+Z_t$?
In a solution to the problem below, the teaching assistant solves it by calculating $\mathbb{E}[X_t^2]$ and ends up with also having to calculate $\mathbb{E}[X_{t-1}Z_t]$ after expanding the square. To do this, he states that "$\mathbb{E}[X_{t-1}Z_t]=0$ since $X_{t-1}$ is independent of $Z_t$ and $X_{t-1}$ is uncorrelated with $Z_t$".
Questions:
- Why is $X_{t-1}$ independent of $Z_t$? I don't see that we assume that the time series is causal.
- Why is $X_{t-1}$ uncorrelated with $Z_t$?
Problem:
> Let a timeseries model $X:=(X_t, t\in\mathbb{Z})$ be given by $$X_t=\phi X_{t-1}+Z_t, \quad \text{where} \quad Z_t\sim \text{WN}(0,\sigma^2)\quad \text{and} \quad |\phi|\neq 1.$$ Assume that the stochastic process satisfying this model is stationary. Compute the variance of $X.$
Note: Yes, I know that one simply can calculate $\text{Var}[X_t]$ directly in one line, but I'm trying to understand the motivations behind the instructors steps.
## Answer by Jose Avilez (score 4, accepted)
https://quant.stackexchange.com/a/66514
$E(X_{t-1}Z_t) = 0$ in the causal case $|\phi | < 1$, but not in the non-causal case $|\phi | >1$.
Causal case $(|\phi| < 1)$
In this case, the unique stationary solution to the AR(1) equation is given by $$X_t = \sum_{j=1}^\infty \phi^j Z_{t-j}$$ Thus, $$E(X_{t-1} Z_t) = E \left( Z_t \sum_{j=1}^\infty \phi^j Z_{t-1-j} \right) = \sum_{j=1}^\infty \phi^j E (Z_t Z_{t-1-j}) = 0$$ where the last equality follows from $Z_i$ and $Z_j$ being uncorrelated for $i \neq j$, and all the exchanges of integration are guaranteed by Fubini's theorem.
Non-causal case $(|\phi| > 1)$
The unique stationary solution to the AR(1) equation is $$X_t = - \sum_{j=1}^\infty \phi^{-j} Z_{t+j}$$ This equation can be arrived at by performing the recursion forward, rather than backward. Thus,
$$E(X_{t-1}Z_t) = -\sum_{j=1}^\infty \phi^{-j} E(Z_{t-1+j} Z_t) = - \frac{\sigma_Z^2}{\phi}$$
This finding is confirmed if you use the AR equations to get $$Var(X_t) = \phi^2 Var(X_{t-1}) + Var(Z_t) + 2 \phi E(X_{t-1} Z_t)$$
Using $Var(X_{t-1}) = \frac{\sigma_Z^2}{\phi^2 - 1}$ you get $E(X_{t-1} Z_t) = - \frac{\sigma_Z^2}{\phi}$.
Aside: most people restrict the study of ARMA processes to the first case for two reasons: (i) the non-causal future-dependent case is strange, and (ii) in the non-causal stationary case you may always find a white noise process $\tilde{Z}_t$ such that $X_t$ is the causal solution to $X_t = \phi^{-1} X_{t-1} + \tilde{Z}_t$. This may explain why your TA simply assumed $E(X_{t-1}Z_t) = 0$ always.
Edit: There seems to be some confusion in the comments about what the solutions to a time series equation are. I find most textbooks skim over this technical detail.
A solution to a stochastic equation $$g(X_t, Z_t) = 0$$ is a pair $(X_t, Z_t)$ such that $(Z_t)_{t \in \mathbb{Z}}$ is a white noise and for each $t \in \mathbb{Z}$, $X_t$ is a (measurable) function of the entire white noise sequence (i.e. $X_t = h((\epsilon_{k})_{k\in\mathbb{Z}})$) and $X_t$ satisfies the stochastic equation in some sense (e.g. almost surely). Note that $X_t$ can depend on the "past" noise, on the "future" noise, or the entire sequence restrictions. The solution $(X_t, Z_t)$ is said to be stationary if $X_t$ is a stationary process.
Notice that for the AR(1) equation $X_t = \phi X_{t-1} + Z_t$ we can find unique stationary solutions when $|\phi| \neq 1$; you can do this by recursion into the past if $|\phi| < 1$ or by recursion into the future (time-reversal) if $|\phi| >1$. See Example 3.1.2 and Theorems 3.1.1-3.1.3 of Brockwell and Davis for more details.
In the case $|\phi| > 1$, we may also fix a stochastic process $(X_t)_{t \in \mathbb{N}}$ such that $X_0 = x$ and for $t \geq 1$, $X_t = X_0 + \sum_{j=1}^t Z_j$. While this is indeed a solution to the AR(1) equation, it is clearly not stationary. Thus, we can remove this case, as the OP asked us to consider the solution to the AR(1) equation when $X_t$ is stationary.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.