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### Compound Probabilities

```
Date: 6/28/96 at 21:6:53
From: Johnny Vogler
Subject: Compound probabilities

Dear Dr. Math,

I want to learn how to find the probability of various values of an
expression that is a linear combination of random numbers with uniform
probability density within a certain range, say zero to one.

If the probability density of a is P(x)=1 for x is between 0 and 1
(and the probability of it being between u and v is the integral of
P(x)dx from u to v), and similarly for b, then I believe the
probability density function of a+b is P(x)=x/2 between 0 and 1 and
P(x)=1-x/2 between 1 and 2, zero everywhere else, but I can't prove
it, and I don't know how to go from there.  What is the probability
density function for the sum of three similar variables?  Four?  How

Thanks,
Johnny Vogler
```

```
Date: 6/29/96 at 16:55:0
From: Doctor Anthony
Subject: Re: Compound probabilities

I shall be using the 'Distribution Function' to answer this question
where the distribution function is defined as Pr(X<x) and is the
integral (area under the density curve) up to the ordinate at X = x.
If you differentiate the distribution function you get the density
function.  We use this method because it is often easier to write down
the distribution function, and then differentiate, rather than go
straight to the density function.

We consider now the distribution function for the sum of two
independent variates X and Y. g(x) is density function for x and h(y)
is density function for y.  G(x) and H(y) are corresponding
distribution functions.

Let Z = X+Y and let F(z) be the distribution function for Z.  By
definition, F(z) is the probability that X+Y<z.  If we plot possible
values of X and Y using rectangular coordinates in a plane, the region
corresponding to X+Y<z is that part of the plane lying below and to
the left of the line X+Y = z

For any given y, the probability that X<z-y is G(z-y) so that the
required probability is obtained by multiplying G(z-y) by the
probability for y and integrating over all y.

So F(z) = INT[G(z-y)*h(y)*dy]   over the range of the y variate.

By differentiating with respect to z we obtain the density function:

f(z) = INT[g(z-y)*h(y)*dy]

This is called the 'CONVOLUTION' of the density functions g and h.  It
is the density function for Z

Now we can look at the question you asked, with X and Y each uniformly
distributed over the interval (0,1)

g(x) = 1, 0<x<1    h(y) = 1,  0<y<1

g(z-y) is 1 for values of y between z-1 and z, and is 0 for all other
values. This is because y = z-x and therefore when x=1, y = z-1 and
when x = 0 then y = z.

f(z) = INT(z-1 to z)[h(y)*dy] = INT[dy] since h(y) = 1

In the range 0<z<1 we have f(z) = INT[dy] = [y] between 0 and z

so f(z) = z   for 0<z<1

In the range 1<z<2 we have f(z) = INT[dy] = [y] between z and 2

so f(z) = 2-z    for 1<z<2

This result can be checked graphically by considering the square (0,1)
on the x and y axes.  The area of the sample space with 0<z<1 is given
by area of a right-angled triangle of area (1/2)z^2, and so
F(z) = (1/2)z^2

Then differentiating we get f(z) = z  (0<z<1)

For 1<z<2 the area of the sample space is 1 - (1/2)(2-z)^2

and differentiating we get f(z) = -2*(1/2)(2-z)(-1)

= 2-z  (1<z<2)

I hope you have enough to go on from here.  The general method is that
given first, but sometimes with simple density functions a graphical
solution is quicker.

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
College Probability

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