Re: Overlapping probability distributions

On Feb 1, 4:42 pm, sert <je...@xxxxxxxxxxx> wrote:
We have two probability distributions, A and B. We need to
determine the probability that a value randomly picked from A
will be higher than another value randomly picked from B. Is it
possible to do this in a deterministic way or do we need to do
use a Monte-Carlo simulation?

For example, suppose we know that the heights of men and women
conform to the normal distribution, with known mean and
deviation. What is the probability that a random man is taller
than a random woman?

If X is from A and Y is from B, then (given the assumed *independence*
of the two draws) the pair (X,Y) has the bivariate distribution A(x)*B
(y). Now you want the probability P{X > Y}, computed from the
bivariate distribution. I will assume continuous distributions with
_densities_ a(x) and b(y) in the following----if the distributions are
discrete, instead, just replace integrals by sums. Method: P{X > Y} =
integral of a(x)*b(y) dx*dy over the 2D region R = {x > y}. This can
be expressed as integral{y = -infinity ..infinity} integral{x = y ..
infinity} a(x)*b(y) dx dy = integral{y = -inf..inf} b(y)*AA(y) dy,
where AA(y) = P{X > y} = integral{x=y..inf} a(x) dx. So, the problem
now reduces to a 1-dimensional integration. In some cases the
functions b(y) and AA(y) are simple enough to allow an explicit
integration in closed form; if not, you can easily do /numerical
integration/, which, in 1 dimension is well-studied and has effective
methods widely available. Of course, this assumes that AA(y) is easily
computed. If necessary, do a numerical integration also to get AA(y)
for all the y values needed for the main numerical integration.
Alternatively, you can write P{X > Y} = integral{x=-inf..inf} a(x)*B
(x) dx, where B(x) = P{Y < x} = integral{y=-inf..x} b(y) dy; this may
be easier than the other way round.

For the specific case of normal A and B, say with means m (men) and w
(women) and variances vm and vw, resp., the random variable D = X-Y is
normal again, with mean d = m-w and variance vd = vm + vw. Now you
just want the probability P{D > 0} = P{N(0,1) > -d/sqrt(vd)}, which
can be found from normal tables, or by pushing a button on a
scientific calculator, or by using a spread***, or whatever. [Note:
the fact that D has a normal distribution again is a standard property
about linear combinations of independent normal random variables; see,
eg.,
http://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables
or http://mathworld.wolfram.com/NormalSumDistribution.html or
http://www.xycoon.com/nor_properties6.htm . This last reference has an
incorrect formula for the variance; it should be sigma^2 = sum (c_i)^2
* (sigma_i)^2 [not sum c_i * (sigma_i)^2 ].

R.G. Vickson
.