Let $W_1$ and $W_2$ be independent geometric random variables with parameters $p_2$ and $p_2$ , respectively.
Find $P (W_1 = W_ 2 )$.
To me this question seems unsolvable. I think we can assume $W_1=p_1$ and $W_2=P_2$. Also, I think we can assume that $W_1+W_2=1$. However it mentions that they are geometric random variables, which we have not learned in the class. So, I am guessing that it is very closely related to Poisson distribution.
For context, the section this comes from was about Poisson distribution.
$$\binom{n}{k}p^k(1-p)^{(n-k)}\rightarrow \frac{e^{-\mu}\mu^{k}}{k!}$$
$$\text{as }n\rightarrow \infty \text{ and } p\rightarrow 0 \text{ with } np=\mu$$
Best Answer
A geometric random variable with parameter $p$ has probability mass function $$ p(k) = (1-p)^{k-1}p,\quad k=1,2,\ldots. $$ So if $W_1\sim\mathrm{Geo}(p_1)$ and $W_2\sim\mathrm{Geo}(p_2)$ are independent, we have \begin{align} \mathbb P(W_1=W_2) &= \mathbb P\left(\bigcup_{k=1}^\infty\{W_1=W_2, W_1=k\} \right)\\ &= \sum_{k=1}^\infty \mathbb P(W_1=W_2,W_1=k)\\ &= \sum_{k=1}^\infty \mathbb P(W_1=k,W_2=k)\\ &= \sum_{k=1}^\infty \mathbb P(W_1=k)\mathbb P(W_2=k)\\ &= \sum_{k=1}^\infty (1-p_1)^{k-1}p_1(1-p_2)^{k-1}p_2\\ &= p_1p_2\sum_{k=0}^\infty ((1-p_1)(1-p_2))^k\\ &= \frac{p_1p_2}{1-(1-p_1)(1-p_2)}. \end{align}