variance of autoregressive process

60 Views Asked by At

Let $X_t=\alpha X_{t-1}+\varepsilon_t, t\in\Bbb N$ be a process where $\epsilon_t$ are iid and follow some distribution $F$ with mean $0$ and variance $\sigma^2$, and $X_0\sim G(0,\sigma^2)$ follows some distribution $G$ with mean $0$ and variance $\sigma^2$, and $|\alpha|<1$.

I don't know how to get Var$(X_t)=\alpha^2\sum\limits_{h=1}^t\alpha^{2h}$

1

There are 1 best solutions below

0
On BEST ANSWER

You can do it with recurrence, by putting $X_{t-1}=\alpha X_{t-2}+\epsilon_{t-1}$ into your equation and so on with $X_{t-2}$,... you obtain: $$X_t=\epsilon_t+\alpha\epsilon_{t-1}+\alpha^2\epsilon_{t-2}+\dots$$ $$Var(X_t)=\sigma^2\sum_{i=0}^{\infty} \alpha^{2i}=\frac{\sigma^2}{1-\alpha^2}$$ The last equality holds because $|\alpha|<1$