Independent Probablilty DistributionsDate: 12/15/2005 at 07:35:57 From: Phil Subject: Independent Probablilty Distributions I have two independent random variables A and B with known continuous density and distribution functions and their pdf's overlap. How do I calculate the probability that variable A will be lower than variable B? I've looked at the intersections between their pdf and cdf and mused about the ratio between the area under the cdf but can't think how to prove this. Many thanks, Phil Date: 12/16/2005 at 15:51:49 From: Doctor George Subject: Re: Independent Probablilty Distributions Hi Phil, Thanks for writing to Doctor Math. Since A and B are independent, their joint density is f(a)f(b) A B To find P(A<B) we need to integrate the joint density over the half plane a<b. Here is one way to do this. oo b / / P(A<B) = | | f(a)f(b) da db / / A B -oo -oo oo / = | F(b)f(b) db / A B -oo If the joint density is non-zero over only some region of the ab plane you can set the limits of integration accordingly. As you look for the easiest way to carry out the integration it can be useful to note that P(A<B) = 1 - P(B<A) (Note that P(A=B) = 0 since A and B are continuous.) Finally, it can be helpful to think of the solution like this P(A<B) = E[F(B)] A where E denotes expected value. Viewed in this way, P(A<b) for some particular value b is averaged over all possible values of b. Does that make sense? Write again if you need more help. - Doctor George, The Math Forum http://mathforum.org/dr.math/ |
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