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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?

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

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1 Answer 1

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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.

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    $\begingroup$ Re "no" restriction on $\rho$ - we do need to rule out unit root processes, no? $\endgroup$ Commented May 31, 2022 at 12:10
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    $\begingroup$ @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
    Commented May 31, 2022 at 13:00
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    $\begingroup$ 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 $\rho=1$. $\endgroup$ Commented May 31, 2022 at 14:19

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