5
$\begingroup$

I have a question about deriving the asymptotic distribution of an explosive AR(1) process:

$$X_t=\rho X_{t-1}+\epsilon_t; X_0=0; \rho\gt1 $$

In particular, I have been given the following identity, but I do not understand how to derive it. Can anyone lend me a helping hand?

$$\sum_{t=1}^{n} X_{t-1}^2=\frac{1}{\rho^2-1} \left\{X_n^2-\sum_{t=1}^{n}\epsilon^2_t-2\rho\sum_{t=1}^{n}X_{t-1}\epsilon_t \right\} $$

Improve this question
$\endgroup$

1 Answer 1

Reset to default
9
$\begingroup$

This is ordinary algebra: the fact that the $X_t$ and $\epsilon_t$ are random variables is immaterial. You only need $X_0=0$ (and no restriction on $\rho$) to carry out these simple steps based on the recursive definition of $X_t:$

$$\begin{aligned} (\rho^2-1)\sum_{t=1}^n X_{t-1}^2 &= \rho^2 \sum_{t=1}^n X_{t-1}^2 -\sum_{t=1}^n X_{t-1}^2\\&= \rho^2 \sum_{t=1}^n X_{t-1}^2 + \left(X_n^2 -\sum_{t=1}^n X_t^2\right)\\ &= \rho^2 \sum_{t=1}^n X_{t-1}^2 + X_n^2 - \sum_{t=1}^n \left(\rho X_{t-1} + \epsilon_t\right)^2 \\ &= \rho^2 \sum_{t=1}^n X_{t-1}^2+ X_n^2 - \sum_{t=1}^n \left(\rho^2 X^2_{t-1} + \epsilon_t^2 + 2\rho X_{t-1}\epsilon_t\right)\\ &= X_n^2 - \sum_{t=1}^n \epsilon_t^2 -2\rho \sum_{t=1}^n X_{t-1}\epsilon_t. \end{aligned}$$

Because $\rho^2 - 1\ne 0$ you can divide both sides by it, QED.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Re "no" restriction on $\rho$ - we do need to rule out unit root processes, no? $\endgroup$
    – Christoph Hanck
    May 31, 2022 at 12:10
  • 1
    $\begingroup$ @Christoph Doesn't the condition $\rho^2-1\ne 0$ do that? And even when $\rho^2=1,$ we obtain the potentially useful relation $X_n^2 = \sum^n\epsilon_t^2+2\rho\sum^n X_{t-1}\epsilon_t.$ Please bear in mind that this derivation doesn't assume anything about the process, or even that this is a process at all: it's solely a property of a sequence of numbers $(X_t), t=0,1,2,\ldots,$ defined recursively from another sequence of numbers $\epsilon_t.$ $\endgroup$
    – whuber ♦
    May 31, 2022 at 13:00
  • 3
    $\begingroup$ Yes, indeed, the condition does that - I was referring to what you wrote in the first paragraph in brackets. Clearly, your derivation "works" for any $\rho$, but I wanted to express that the desired result of the OP seems problematic at $\rho=1$. $\endgroup$
    – Christoph Hanck
    May 31, 2022 at 14:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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