How can I do bivariate numerical integration in Matlab for computing joint probabilities?

Hi all, I have to compute the probabilities listed below for each row i of the matrices b and c. I assume that (u_1,u_2) are distributed as a bi-variate normal with mean mu and var-cov matrix sigma. Could you help me please?

b= unidrnd(10,100,2);
c= unidrnd(10,100,2);
sigma=[1 0.3;0.3 1]; 
mu=[0 0];

(1) Pr(u_1<-b(i,1); u_2<-b(i,2))

(2) Pr(u_1>=-c(i,1); u_2>=-c(i,2))

(3) Pr(u_1>-b(i,1); u_2<-b(i,2))

(4) Pr(u_1>-c(i,1);-b(:,2)<=u_2<=-c(i,2))

(5) Pr(u_1>-b(i,1); u_2<-c(i,2))

(6) Pr(u_1<-c(i,1); u_2> -c(i,2))

(7) Pr(u_1<-b(i,1);-b(:,2)<=u_2<=-c(i,2)

(8) Pr(u_1<-c(i,1); u_2>-b(i,2))

Answer 1

[Roger Stafford]

These can all be expressed in terms of the 'mvncdf' function. For (1) it would be:

 P1 = mvncdf(-b,mu,sigma);

For (4) it would be: (I assume you meant "Pr(u_1>-c(i,1);-b(i,2)<=u_2<=-c(i,2))" rather than "Pr(u_1>-c(i,1);-b(:,2)<=u_2<=-c(i,2))" .)

 P4 = mvncdf([-c(:,1),-b(:,2)],[repmat(inf,size(b,1),1),-c(:,2)],mu,sigma);

This latter uses the "rectangle" form of the 'mvncdf' call.

I think you should be able to work out the remaining cases for yourself using the concepts demonstrated here.

[Stella]

Thank you, but I have some doubts regarding (2):

b= unidrnd(10,100,2);
c= unidrnd(10,100,2);
sigma=[1 0.3;0.3 1]; 
mu=[0 0];  
%(2) Pr(u_1>=-c(i,1); u_2>=-c(i,2))
P21=mvncdf([-c(:,1), -c(:,2)],[inf*ones(size(c,1),1), inf*ones(size(c,1),1)], mu, sigma); 
% or, using the property of the cdf
P22=1-mvncdf(-c, mu, sigma);
% I expect sum(P21-P22) to be almost zero, but it is not!

Do you know why?

[Roger Stafford]

I agree with your method for P21. However, I see no reason why you should expect that sum(P21-P22) to be almost zero. P22 includes all the cases where u_1>-c(i,1) OR u_2>-c(i,2) which will in general be a lot larger than P21 which contain those where u_1>-c(i,1) AND u_2>-c(i,2). What I am saying is that P21 contains a rectangle, whereas P22 contains everything outside the diagonally opposite rectangle and is therefore much larger. You have received a sizable negative value for sum(P21-P22). Am I right?

[Stella]

Yes, thanks!