How to solve this double sum?

Question

I need help with a double sum. I would like to express the following as one sum only: $$\sum\limits_{T=t}^{\infty} \beta^{T-t} \sum\limits_{S=t}^{T-1}(x_S-y_{S+1})$$.

Thanks a lot for your help!

Answer 1

To make things easier, let's consider $$ \sum_{T=t}^\infty \beta^{T-t}\sum_{S=t}^{T-1}x_S. $$ Now, let's consider reversing the order of summation, $$ \sum_S x_S\sum_T\beta^{T-t}. $$ Now, as $T$ gets large in the original sum, $S$ can also get large, so the indices on the outer summation is $$ \sum_{S=t}^\infty x_S\sum_T\beta^{T-t}. $$ Now, we have to figure out which terms in the original sum have the term $x_S$. This only happens when $S\leq T-1$, or $T\geq S+1$. Therefore, the second sum is $$ \sum_{S=t}^\infty x_S\sum_{T=S+1}^\infty \beta^{T-t}. $$ Now, the inner sum is a geometric sum, so you can find an explicit formula for it.

Can you take it from here?