can someone explain why this is wrong:
$\epsilon_t$ is WN~ $(0,\sigma^2)$
$X_t=\frac{\epsilon_t}{1-\phi}$
$V(X_t)=V(\frac{\epsilon_t}{1-\phi})=\frac{V(\epsilon_t)}{(1-\phi)^2}$
$V(X_t)=\frac{\sigma^2}{(1-\phi)^{2}}$ Why is this wrong? $\phi <1$ is a constant
The solution which is right is
$V(X_t)=\frac{\sigma^2}{1-\phi^2}$
Its the variance of an AR(1) process.
Thank you in advance.
Hi: You are attempting to derive the variance of an AR(1) but you are skipping some steps. The AR(1) is
$X_t = \phi X_{t-1} + \epsilon_t$ where $\epsilon_t \sim N(0,\sigma^2)$.
First, pull the first term over to the left using the lag operator.
$X_{t}(1-\phi L) = \epsilon_t$. Now when you divide by $(1 - \phi L)$, you end up with an infinite series on the right hand side.
$X_t = \frac{\epsilon_t}{1-\phi L}$.
Note that, as long as $\phi < 1$, then the right hand side is an infinite series that converges. So, we end up with:
$X_t = \sum_{t=0}^{\infty} \phi^{t} \epsilon_{t-i}$
Taking the variance of the sum results in: $var(X_t) = \sum_{t=0}^{\infty} \phi^{2t} \sigma^2 = \frac{\sigma^2}{1-\phi^2}$.
So, the issue was that the lag operator was left out which made the expression look like it wasn't an infinite series.
HERE IS AN EDIT TO PROVIDE AN ALGEBRIAC REASON WHY NORMAL VARIANCE RULES CANNOT BE USED ON A INFINITE SUM.
Let's suppose that the lag operator, $L$, is not even involved in order not to confound the issue.
Let: $ 1 + \rho + \rho^2 \ldots + \rho^n + \ldots = A $.
and Let:
$ \frac{1}{1-\rho} = B $.
I'm not 100 percent certain but I think that the reason your squaring approach doesn't work is because $A^2 \neq B^2$. Take the case of $A^2$. When one takes expectations of the individual terms of the square of the sum, all of the cross product terms are zero because of the independence of the $\epsilon_{i}$ and $\epsilon_{j}$ for $i \neq j$. Conversely, when one squares B, there is no knowledge of individual cross product terms reducing to zero.
If someone else can prove $A^2 \ne B^2$ in a less heuristic way, it's appreciated.