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