Find $P(X=Y)$ if $X$ and $Y$ are independent random variables with same geometric distribution If $X$ and $Y$ are independent random variables with the same geometric distribution with parameter $p$, find:
$P(X = Y)$ and $P(X \geq Y)$
I have done the joint distribution table of both variables and found, with some sums, these results, but I don't find in books some confirmation of my procedures:
1) $\frac{p}{2-p}$ 
2) $\frac{1}{2-p}$
 A: *

*For the first: you have
$$
\mathbb{P}\{X=Y\} = \mathbb{P}\bigcup_{k=1}^\infty \{X=Y=k\} =  \sum_{k=1}^\infty \mathbb{P}\{X=Y=k\} = \sum_{k=1}^\infty \mathbb{P}\{X=k\} \cdot \mathbb{P}\{Y=k\} 
$$
the second equality since the events are disjoint; the last equality by independence of $X,Y$. Using the pmf of the Geometric distribution, we get
$$
\mathbb{P}\{X=Y\} = \sum_{k=1}^\infty p(1-p)^{k-1}\cdot p(1-p)^{k-1} = p^2 \sum_{k=0}^\infty (1-p)^{2k} = \frac{p^2}{1-(1-p)^2} = \boxed{\frac{p}{2-p}}
$$
where to compute the sum we used the expression of a geometric series.

*For the second, we can do something a bit more "fun": note that
$$
\mathbb{P}\{X\geq Y\} = \mathbb{P}\{X = Y\} + \mathbb{P}\{X > Y\} \tag{1}
$$
and that $\mathbb{P}\{X > Y\} = \mathbb{P}\{X < Y\}$ by symmetry (Why?). Since 
$$
1 = \mathbb{P}\{X = Y\} + \mathbb{P}\{X > Y\} + \mathbb{P}\{X < Y\} \tag{2}
$$
(Why?) we thus get
$$
\mathbb{P}\{X > Y\} = \frac{1-\mathbb{P}\{X = Y\}}{2}
$$ 
and therefore by (1)
$$
\mathbb{P}\{X\geq Y\} =  \mathbb{P}\{X = Y\}+\frac{1-\mathbb{P}\{X = Y\}}{2} = \frac{1+\mathbb{P}\{X = Y\}}{2}
$$
which you can compute given the first part:
$$
\frac{1+\frac{p}{2-p}}{2} = \boxed{\frac{1}{2-p}}\,.
$$
A: Write $q=1-p$
a)
$$P(X=Y) = \sum _{n=1}^{\infty}P((X=n)\cap (Y=n)) $$
$$ =\sum _{n=1}^{\infty}P(X=n)P(Y=n) $$
$$ =\sum _{n=1}^{\infty}q^{n-1}p q^{n-1}p  =p^2\sum _{n=0}^{\infty}q^{2n} $$
$$ = p^2{1\over 1-q^2}= {p\over 1+q} = {p\over 2-p}$$

b)
$$P(X\geq Y) = \sum _{n=1}^{\infty}\sum _{k=1}^nP((X=n)\cap (Y=k)) $$
$$ =\sum _{n=1}^{\infty}P(X=n)\sum _{k=1}^nP(Y=k) $$
$$ =\sum _{n=1}^{\infty}q^{n-1}p \sum _{k=1}^nq^{k-1}p  =p^2\sum _{n=0}^{\infty}q^{n-1}\sum _{k=0}^{n-1}q^{k} $$
$$  =p^2\sum _{n=0}^{\infty}q^{n}{1-q^{n}\over 1-q} $$
$$  =p\sum _{n=0}^{\infty}q^{n}(1-q^{n}) $$
$$  =p\Big(\sum _{n=0}^{\infty}q^{n}-\sum _{n=0}^{\infty}q^{2n}\Big) $$
$$ = p({1\over 1-q}-{1\over 1-q^2}) = 1 - {1\over 1+q} = {q\over 1+q}$$
