# Uniquely defined autoregression

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?

• 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." – whuber May 30 '19 at 13:23
• @whuber, thank you, it wasn't clear indeed. Please let me know if the question is good now. – 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.

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