Let $\{w_t\}, t \in \mathbb{Z}$ be a random noise. Given a sequence $\{w_t\}$, does the autoregression

$x_t = x_{t-1} - 0.9 x_{t-2} + w_t$, $ t \in \mathbb{Z}$

uniquely define a sequence $\{x_t\}$?

If not, is it enough to impose the value of the mean of $\{x_t\}$ to make this sequence uniquely defined?

  • 1
    $\begingroup$ Could you please explain what you mean by "fixing" the $w_i$? That sounds like you are specifying a sequence of numbers, but in that case could you explain what sense an expectation might make? And if you intend the $w_i$ to be a stochastic process itself, then please tell us what you mean by "uniquely defined." $\endgroup$ – whuber May 30 '19 at 13:23
  • $\begingroup$ @whuber, thank you, it wasn't clear indeed. Please let me know if the question is good now. $\endgroup$ – toliveira May 30 '19 at 20:50

This is a two-term recursion. Given arbitrary values for $x_0$ and $x_1,$ the formula

$$x_t = x_{t-1} -0.9 x_{t-2} + w_t$$

recursively determines $x_2, x_3, \ldots, x_n, \ldots$ and the inverse formula (found by solving this one for $x_{t-2}$ with $s=2-t$)

$$x_{-s} = \frac{-x_{2-s} + x_{1-s} + w_{2-s}}{0.9}$$

recursively determines $x_{-1}, x_{-2}, \ldots, x_{-n}, \ldots.$

Both recursions make sense whether the $x_t$ and $w_t$ are interpreted as sequences of random variables or just sequences of numbers.

Note that since each formula expresses $x_t$ or $x_{-s}$ as a Real linear combination of the preceding $x$'s and $w$'s, it follows by induction that all terms in the sequence are real linear combinations of $x_0, x_1,$ and the $w$'s. If you are thinking of these objects as numbers, with the $w$'s given, this shows that the set of all such sequences is a two-dimensional vector space.

In particular, imposing a single linear constraint on these sequences will yield a one-dimensional vector space. Thus, the analog for random variables--such as fixing the means--still will not determine the sequence uniquely.

The generalization to longer-term recursions (AR$(k)$ models) should be clear.

  • $\begingroup$ thank you, very clear answer. $\endgroup$ – toliveira May 31 '19 at 0:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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