Ray, Thankyou once again for your help. I am currently working on bringing myself up to speed with the concepts in your reply.
> 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. The original problem was indeed a trivariate normal. Can I take it that covariances will always be zero if the variables are mutually independent? Could the method here be extended to solve for a k-variate with k >3 ? > 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 I am not sure how your representation (ie Integral_-u^Infinity Integral_-v^Infinity f(x,y;w)dydx) translates to the 'usual' text book representation for integrals. (bear in mind I a a novice here) Could these integrals be solved using numerical methods within Microsoft Excel? Thanks again David . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
