Distribution of X-Y when X and Y are independent geometric

Question

$X$ and $Y$ are independent geometric distributions with parameter $p$. Find the distribution for $X-Y$

I am aware that there are many similar questions. My problem with this specific transformation is that I cannot solve the summation I end up with.

This is what I have done:

$$Z = X - Y$$ $$P(Z=z) = P(X-Y=z)$$ $$= \sum_{k=1}^\infty P(Y=k)P(X=z+k)$$ $$= \sum_{k=1}^\infty p(1-p)^{k-1} p(1-p)^{z+k-1}$$

Somehow the answer should be:

$$\frac{p(1-p)^{|z|}}{2-p} $$

How does this sum turn into this final solution? Or have I done a mistake in the previous steps. Any help is appreciated, thanks!

Answer 1

Suppose $z \ge 0$,\begin{align}\sum_{k=1}^\infty p(1-p)^{k-1} p(1-p)^{z+k-1} &= p^2\sum_{k=1}^\infty (1-p)^{z+2k-2} \\ &=p^2(1-p)^z \sum_{k=1}^\infty ((1-p)^2)^{k-1}\\ &=\frac{p^2(1-p)^z}{1-(1-p)^2}\\ &=\frac{p^2(1-p)^z}{(2-p)p}\\ &=\frac{p(1-p)^z}{2-p}\end{align}