# Geometric Distribution $P(X\ge Y)$

I need to show that if $X$ and $Y$ are idd and geometrically distributed that the $P(X\ge Y)$ is $1\over{2-p}$. the joint pmf is $f_{xy}(xy)=p^2(1-p)^{x+y}$, and I think the only way to do this is to use a double sum: $\sum_{y=0}^{n}\sum_{x=y}^m p^2(1-p)^{x+y}$, which leads to me getting quite stuck. Any suggestions?

-

It is easier to use symmetry: $$1 = \mathbb{P}\left(X<Y\right) +\mathbb{P}\left(X=Y\right) + \mathbb{P}\left(X>Y\right)$$ The first and the last probability are the same, due to the symmetry, since $X$ and $Y$ are iid. Thus: $$\mathbb{P}\left(X<Y\right) = \frac{1}{2} \left(1 - \mathbb{P}\left(X=Y\right) \right)$$ Thus: $$\mathbb{P}\left(X\geqslant Y\right) = \mathbb{P}\left(X>Y\right) + \mathbb{P}\left(X=Y\right) = \frac{1}{2} \left(1 + \mathbb{P}\left(X=Y\right) \right)$$ The probability of $X=Y$ is easy: $$\mathbb{P}\left(X=Y\right) = \sum_{n=0}^\infty \mathbb{P}\left(X=n\right) \mathbb{P}\left(Y=n\right) = \sum_{n=0}^\infty p^2 (1-p)^{2n} = \frac{p^2}{1-(1-p)^2} = \frac{p}{2-p}$$

-

Hint: By symmetry, we have $\Pr(X\lt Y)=\Pr(X\gt Y)$. So we will know everything if we know $\Pr(X=Y)$.

By the way, your sums are supposed to be not to $m$ and $n$, but to $\infty$.

We can also calculate using your expression. All you need is the formula for the sum of an (infinite) geometric series. You also need that for the solution that exploits symmetry, but a bit less work is involved.

Remark: Your approach (but summing to infinity) would work for the more general problem where $X$ and $Y$ are independent but have different "$p$."

-