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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 
about 3a+5b?

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
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Associated Topics:
College Probability

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