Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 6 down vote accepted

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: $$ \mathbb{P}\left(X=Y\right) = \sum_{n=0}^\infty \mathbb{P}\left(X=n\right) \mathbb{P}\left(Y=n\right) = \sum_{n=0}^\infty p^2 (1-p)^{2n} = \frac{p^2}{1-(1-p)^2} = \frac{p}{2-p} $$

share|cite|improve this answer

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 "$p$."

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.