Re: Inconsistency in different ways of calculating CDF of bivariate

*To*: mathgroup at smc.vnet.net*Subject*: [mg106433] Re: Inconsistency in different ways of calculating CDF of bivariate*From*: Ray Koopman <koopman at sfu.ca>*Date*: Tue, 12 Jan 2010 04:50:26 -0500 (EST)*References*: <higdkt$kgl$1@smc.vnet.net>

On Jan 11, 3:53 pm, Stephen <aultman.step... at gmail.com> wrote: > I would like to calculate the CDF of a bivariate normal in some C++ > code. My first thought was to reduce the dimensionality of the > integral by using the conditonal distribution of the bivariate normal > when one variable is known (see link below). When I try to calculate > the CDF this way I get an incorrect result. In the code pasted below > I defined a function, temp1, that uses mathematica's built in > bivaraite normal CDF function. The function temp2 attempts to > calculate the same quantity using a one dimension integral and the > conditional distribution of the bivariate normal. > > It is more than just rounding errors. For some values of rho, the > error is as much as 0.03, which is a lot in probability terms. Mostly > I am curious why this doesn't work from a mathematical statistics > perspective. There are ways to code around this, but I am curious if > there is a bigger math issue at play. Any thoughts? > > mu1 = 0; mu2 = 0; sigma1 = 1; sigma2 = 1; rho = .1; z1 = 0; z2 = 0; > Needs["MultivariateStatistics`"] > temp1[mu1_, mu2_, sigma1_, sigma2_, rho_, z1_, z2_] := > CDF[MultinormalDistribution[{mu1, > mu2}, {{sigma1^2, rho*sigma1*sigma2}, {rho*sigma1*sigma2, > sigma2^2}}], {z1, z2}] > temp2[mu1_, mu2_, sigma1_, sigma2_, rho_, z1_, z2_] := > NIntegrate[ > PDF[NormalDistribution[mu1, sigma1], x]* > CDF[NormalDistribution[ > mu2 + (sigma2/sigma1)*rho (x - mu1), (1 - rho^2)*sigma2^2], > z2], {x, -Infinity, z1}] > > Plot[temp1[mu1, mu2, sigma1, sigma2, rho, z1, z2] - > temp2[mu1, mu2, sigma1, sigma2, rho, z1, z2], {rho, -.99, .99}] > > http://en.wikipedia.org/wiki/Multivariate_normal_distribution#Conditional_distributions In temp2, change (1 - rho^2)*sigma2^2 to Sqrt[1 - rho^2]*sigma2 . (You want the standard deviation, not the variance.) Here are two other routines that you might want to try. They use equations from: Pearson, K. (1901). Mathematical contributions to the theory of evolution - VII: On the correlation of characters not quantitatively measurable. Philosophical Transactions of the Royal Society of London, Series A, 195, 1-47. x = (z1 - mu1)/sigma1, y = (z2 - mu2)/sigma2, r = rho. cdf get the lower left area, < x and < y; ccdf gets the upper right area, > x and > y (Abramowitz & Stegun's "L"). cdf[x_,y_,r_] := .25 Erfc[-Sqrt[.5]*x] Erfc[-Sqrt[.5]*y] + Block[{u = -.5(x^2 + y^2), v = N[x*y]}, Which[u == 0, ArcSin[r], v == 0, NIntegrate[Exp[u/Cos[t]^2],{t,0,ArcSin@r}], True, NIntegrate[Exp[(u + v*Sin[t])/Cos[t]^2],{t,0,ArcSin@r}]]/(2 Pi)] ccdf[x_,y_,r_] := .25 Erfc[Sqrt[.5]*x] Erfc[Sqrt[.5]*y] + Block[{u = -.5(x^2 + y^2), v = N[x*y]}, Which[u == 0, ArcSin[r], v == 0, NIntegrate[Exp[u/Cos[t]^2],{t,0,ArcSin@r}], True, NIntegrate[Exp[(u + v*Sin[t])/Cos[t]^2],{t,0,ArcSin@r}]]/(2 Pi)]