Need help simplyfying this Geometric Series

W(1) and W(2) are independent geometric series distributions, where W(1) has parameter p(1), and W(2) has parameter p(2).

I'm trying find P(W(1) < W(2)), and have gotten as far as this...

***(Summation from w=0 to infinity - sorry, not sure how to put summation and other math signs here)Pr(W(1) < w) * Prob(W(2) = w)

We know that Prob(W(2)) = (1 - p(2))^(w-1) * p(2), where w = 1, 2, 3, ..., w

However, where I'm lost is to find Pr(W(1) < w) - is it (1 - p(1))^(w-2) * p(1) ? The reason why put (w-2) in the exponential part is because we're looking for W(1) < w, so I'm assuming that w = 1, 2, 3, ..., w-1, because W(1) < w. Is this part correct?

Now - after everything is all said and done, need to find *** above...how do I do this? My confusion is how to put this all in a geometric series. If W(1) = w, then it would be easier because both have exponents (w-1), however, the (w-2) one makes it a bit more tricker.

Thanks!

What happened to w = 0.

This seems to be what you had in mind:

\(\displaystyle \sum_{m=0}^{\infty}\left[(1-p_{2})^{m}\cdot p_{2}\cdot\sum_{n=0}^{m-1}\left[(1-p_{1})^{n}\cdot p_{1}\right]\right]\)

The inner summation should be relatively easy. I get

\(\displaystyle 1 - (1-p_{1})^{m}\)

Where does that leave us?

P.S. It woudl be a valuable learning experience for you to learn a little TeX. There are various conversion programs. I usually jsut type the whole thing in by hand. That does get a little tedious.