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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?

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    $\begingroup$ 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
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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.

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

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