Question: If the random variables $X$ and $Y$ are independent and each have the geometric distribution $G(p)$ - that is, $P(X=k)=P(Y=k)=pq^k$ for $k=0,1,2,\ldots$ (where $q=(1-p)$) show that:
(I) $P(X+Y=n) = (n+1)p^2q^n$ for $n=0,1,2,\ldots$
(II) $P(X=k|X+Y=n) =\frac1{n+1}$ for $0\leqslant k\leqslant n$
(III) $P(X=Y) = \frac p{2-p}$
For (I) I have been given the hint {X+Y=n}= ∪(arbitrary unions running from k=o to n) for {X=k, Y=n-k}-disjoint union
I would greatly appreciate it any feedback with this because I do not know how to start even given the hint for section (I)
Here are some hints:
For (I), as you mentioned in your question, $$\{X+Y=n\} = \bigcup_{k=0}^n \{X=k, Y=n-k\}.$$ Since $X$ and $Y$ are independent, and the events $\{X=k, Y=n-k\}$ are disjoint for different values of $k$, we have $$P(X+Y=n)= P\left(\bigcup_{k=0}^n\{X=k, Y=n-k\}\right) = \sum_{k=0}^n P(X=k, Y=n-k)=\sum_{k=0}^n P(X=k)P(Y=n-k). $$
For (II), by the definition of conditional probability we have $$P(X=k|X+Y=n) = \frac{P(X=k, X+Y=n)}{P(X+Y=n)}=\frac{P(X=k,Y=n-k)}{P(X+Y=n)}=\frac{P(X=k)P(Y=n-k)}{P(X+Y=n)}. $$
For (III), $$\mathbb P(X=Y)=\sum_{n=0}^\infty P(X=n,Y=n)=\sum_{n=0}^\infty P(X=n)P(Y=n). $$
From there it should be straightforward to compute the probabilities.