Prove $p(x,y) = q^2p^{y-2}$ is a probability mass function

Question

Question: Prove $p(x,y) = q^2p^{y-2}$ is a pmf

where $q = 1 - p$ $$x = 1,2,3,...,y-1$$ $$y = 2,3,..., \infty $$

My attempt:

I am trying to prove $p(x,y)$ must add up to 1

$$\sum_{y=2}^{\infty} \sum_{x=1}^{y-1} q^2p^{y-2} = \sum_{y=2}^{\infty} q^2p^{y-2} \sum_{x=1}^{y-1} 1 = \sum_{y=2}^{\infty} q^2p^{y-2} (y-1) $$

to me this is kinda similar to a summation of a binomial however there is no way I can get $y-1$ to equal $y \choose 2$ (also, the range of summation is not the same as of binomial's!!!)

I am stuck there and any help would be appreciated. Thanks!

Answer 1

$$\begin{split}\sum_{y=2}^\infty \sum_{x=1}^{y-1}(1-p)^2p^{y-2} &= \sum_{y=2}^\infty \color{red}{(y-1)}(1-p)^2p^{y-2}\\ &=\frac{(1-p)^2}{p}\sum_{y=2}^\infty (y-1)p^{y-1}\\ &=\frac{(1-p)^2}{p}\sum_{y=1}^\infty yp^y\\&=(*)\end{split}$$

Now, looking at this post:

$$\sum_{y=1}^\infty yp^y = \frac p {(1-p)^2}$$

Thus $$(*)=\frac{(1-p)^2}{p} \frac{p}{(1-p)^2}=1$$

$$$$

Answer 2

I don't think your question is well formed but it's almost certain that the problem can be done using the formula for the sum of a geometric series $$ \sum_{y=Y}^\infty p^y=p^Y{1\over 1-p} $$ if $p<1$.