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) $\frac{1}{2-p}$
Best Answer
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: $$ \frac{1+\frac{p}{2-p}}{2} = \boxed{\frac{1}{2-p}}\,. $$