[EMAIL PROTECTED] (David) wrote in message news:<[EMAIL PROTECTED]>... > [...] Can I please ask you to walk me through to the solution? > A = N(mean 0,variance 1), B = N(2,2) and C = N(3,4) > What is the probability that A is greater than B and C
The first step is to restate the original three-variable problem as a two-variable problem. When you spoke of "A beating B and C in a 3 man field", I took that to mean that numeric values of the three random variables A, B, and C are observed once, together, as in a 3-man race; in particular, that you did not mean separate independent 2-man races, such as A vs B, then A vs C. I also assumed that the largest value wins. Then Pr(A beats both B and C) = Pr(( A > B) & ( A > C)) = Pr((A-B > 0) & (A-C > 0)) = Pr(( X > 0) & ( Y > 0)). The second step is to get the joint distribution of the new random variables X and Y. Knowing that A, B, and C are individually normal is not enough. We need to know if their joint distribution is trivariate normal. If so then there is a relatively straightforward solution. If not then there may be no general solution. (But note that if A, B, and C are mutually independent, as well as individually normal, then their joint distribution is trivariate normal.) I assumed trivariate normality, from which it follows that the joint distribution of X and Y is bivariate normal. The parameters of a k-variate normal distribution are the k means, the k variances, and the k(k-1)/2 covariances. You didn't say what the covariances among A, B, and C are, so I will take them all to be 0. Then the parameters of the bivariate (X,Y) distribution are Mean(X) = Mean(A) - Mean(B) = 0 - 2 = -2 Mean(Y) = Mean(A) - Mean(C) = 0 - 3 = -3 Var(X) = Var(A) + Var(B) - 2*Cov(A,B) = 1 + 2 - 2*0 = 3 Var(Y) = Var(A) + Var(C) - 2*Cov(A,C) = 1 + 4 - 2*0 = 5 Cov(X,Y) = Var(A) - Cov(A,B) - Cov(A,C) + Cov(B,C) = 1 - 0 - 0 + 0 = 1 Bivariate normality implies that these five values are all we need to know. (But note that the general expressions for the means, variances, and covariance always hold -- they do not depend on any assumptions.) The third step is to find the probability that both X and Y are positive. There are two ways to do it, both of which require numeric integration. The obvious way is to integrate the volume under the bivariate normal surface. That's a two-dimensional integral. Pearson showed how to transform the problem into a one-dimensional integral, that is much easier to evaluate numerically. Both ways of solving the problem need u, v, and w: u = Mean(X)/SD(X) = -2/sqrt(3) = -1.1547 v = Mean(Y)/SD(Y) = -3/sqrt(5) = -1.3416 w = Cov(X,Y)/(SD(X)SD(Y)) = 1/sqrt(15) = .2582 The bivariate standard normal density function is f(x,y;r) = exp((-1/2)(x^2 + y^2 - 2rxy)/(1 - r^2)) / (2pi*sqrt(1 - r^2)). After standardizing X and Y, the two-dimensional integral is Integral_-u^Infinity Integral_-v^Infinity f(x,y;w)dydx = .0215 Pearson's solution is Integral_0^w f(u,v;r)dr + F(u)*F(v) = .0103 + .1241*.0899 = .0215 Thus, in spite of the fact that the A mean is substantially below both the B and C means, there is still a 2% chance that A will exceed both B and C. Also, note that Pr(A > B) = F(u) = .1241, and Pr(A > C) = F(v) = .0899, so if A vs B and A vs C were two independent comparisons, with possibly different, unrelated values of A, then Pr((A > B) & (A > C)) would be F(u)*F(v) = .0112. . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
