Compound ProbabilitiesDate: 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? 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/ |
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