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Probability Using Geometric Random Variables

Date: 01/15/2004 at 00:47:40
From: Jamie
Subject: statistics/probability

Let X be a geometric random variable with parameter p1 and let Y be a 
geometric random variable with parameter p2. Find the probability that
X is less than or equal to Y.

I'm not too sure where to even begin with this problem or what to do.
 I know what the geometric distribution is, and I thought of possibly 
conditioning on X to solve this problem.



Date: 01/15/2004 at 09:30:06
From: Doctor Mitteldorf
Subject: Re: statistics/probability

Hi Jamie,

Here's how you might start:  Think of the geometric random variable as
the probability of requiring X trials to turn up your first success,
if the probability for success on each trial is p.  So let's say you
are flipping a biased coin so the probability of heads is p1 and I am
flipping another biased coin with a probability of heads equal to p2.
 What is the probability that you will flip your first heads either
before my first heads or at the same time?

Answer:  

Let's use q1 to mean 1 - p1 and q2 = 1 - p2.  Your very first flip,
you might get heads with probability p1, and then you win.  Otherwise
you might get heads on your second try (q1*p1), and you win if I get
tails on my first try (q2).  So far we have
  
         p1 + q1*p1*q2

Continuing, the probability of your winning on the third flip is
q1*q1*p1*q2*q2.

Write down the fourth and fifth terms yourself, and keep going until
you see a pattern.  Once you have the pattern, see if it leads you to
an idea for finding the infinite sum of all the terms.

Please write back if a further hint would be useful, or let me know
when you've got an answer!

- Doctor Mitteldorf, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
College Probability

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