Exchanging and dividing the order of double summations

Question

I want to divide the following equation into two independent parts.

$$ S_n=\sum_{i=0}^{n-1}\sum_{j=0}^{i-1}\alpha^{i-j-1}\omega_{j}, \qquad S_0=0 $$

Here, $\omega_{j}$ is a Gaussian process with zero mean and unit variance.
I want to derive the variance of $S_n$.

First, I tried to divide these two summations like,

$$ \frac{1}{\alpha}\sum_{i=0}^{n-1}\alpha^{i}\sum_{j=0}^{i-1}\alpha^{-j}\omega_{j} $$

I'm wondering if I can simply just divide like above or not.

Also, I tried to put specific $i$ and $j$ from $0$. Is is possible to calculate $$ \sum_{j=0}^{-1}\alpha^{-j} $$ which is the situation that $i=0$ in double summation.

I appreciate any help that you can provide.

Answer 1

First, I tried to divide these two summations like,

$$ \frac{1}{\alpha}\sum_{i=0}^{n-1}\alpha^{i}\sum_{j=0}^{i-1}\alpha^{-j}\omega_{j} $$

I'm wondering if I can simply just divide like above or not.

The terminology is "distribute", and yes, it is possible to distribute the factors of this nested series like so.

Also, I tried to put specific $i$ and $j$ from $0$. Is is possible to calculate $$ \sum_{j=0}^{-1}\alpha^{-j} $$ which is the situation that $i=0$ in double summation.

By definition the series of an empty sequence equals zero, and that is what this is, so indeed you actually have:

$$ S_n~{=\sum_{i=0}^{n-1}\sum_{j=0}^{i-1}\alpha^{i-j-1}\omega_{j}\\= \begin{cases}\displaystyle 0+\frac{1}{\alpha}\sum_{i=1}^{n-1}\alpha^{i}\sum_{j=0}^{i-1}\alpha^{-j}\omega_{j}&:& n>1\\ 0 &:& \text{otherwise}\end{cases} }$$

Answer 2

That last sum is zero by definition.