Finding the variance of the time series defined as $x_t=\phi x_{t-1}+w_t$, for $t=2,3,4,...$.

Question

Let $w_t$ be white noise with variance $\sigma_w^2$ and let $|\phi|<1$ be a constant. Consider the process $x_t=w_1$ and $x_t=\phi x_{t-1}+w_t$ for $t=2,3,...$. I need to find the variance.

I know that since $$x_1=w_1$$ then $$x_2=\phi w_1+w_2$$ $$x_3=\phi^2 w_1+\phi w_2+w_3$$...$$x_t=\phi^{t-1}w_1+\phi^{t-2}w_2+..+w_t=\sum_{i=1}^{t-1}\phi^{t-i}w_i=\sum_{i=0}^{t-1}\phi^{(t+1)-i} w_{i+1}$$

And it follows the variance is $$var(x_t)=E((\sum_{i=1}^{t-1}\phi^{t-i}w_i)^2)-0$$ $$=\phi^{2(t-1)}E(w_1^2)+...+\phi^{2*0}E(w_t^2) $$ $$=\sigma_w^2*(\phi^{2(t-1)}+\phi^{2(t-2)}+...+\phi^{2*0})=\sigma_w^2\sum_{i=0}^{t-1}(\phi^2)^i=\sigma_w^2[\dfrac{1-(\phi^2)^t}{1-\phi^2}] $$.(Follows from the geometric sequence formula). Th other terms zero term since the sequence of white noise variables are uncorrelated for lag greater than 0. Would this right? I just don't know if my answer is off on the power of $\phi$. Thanks.

Answer 1

$\phi^{t-1} w_1 + \phi^{t-2} w_2 + \ldots+ w_t = \sum_{i=1}^{t-1} \phi^{t-i} w_i$

This equation is not correct: On the left-hand side there appears a "$w_t$"-term, but on the right-hand side it doesn't. So, instead it should read

$$\phi^{t-1} w_1 + \phi^{t-2} w_2 + \ldots+ w_t = \sum_{i=1}^{\color{red}{t}} \phi^{t-i} w_i$$

Except for that your calculations are correct.