I would like to know how to implement a bivariate normal distribution.

I have the following formula(see picture) and I know that

R=-2.25 rhoR=0.5 and the result I should get is ~0.17%.

I read the instructions on Mathworks for the mvnpdf but I dont know how to use it with my data.

I am thankful for any help and suggestions on how I need to enter my data into the formula.

Answer 1

[Torsten]

rhoR=0.5;
fun=@(x,y)1/(2*pi*sqrt(1-rhoR^2))*exp(-1/(2*(1-rhoR^2))*(x.^2-2*rhoR*x.*y+y.^2));
intfun=@(R)integral2(fun,-Inf,R,-Inf,R)-0.0017;
R0=-2.25;
sol=fzero(intfun,R0)

gives you the R-value for which the integral equals 0.17 %.

Btw: The "pi" in the exponential term seems to be wrong.

Best wishes

Torsten.

[John D'Errico]

Correct. pi does not belong inside the exponential there.

[Charlotte]

Thanks! So I guess they wrote it wrong in the paper but since they got the same result as I got using the mvncdf, the probably calculated it using the correct formula.

Answer 2

[John D'Errico]

Using the stats toolbox, you want to use mvncdf, not mvnpdf anyway. (Ok, you could integrate the pdf. But why would you here?)

R = -2.25;
rhoR=0.5;
mvncdf([R,R],[0 0],[1 rhoR;rhoR 1])
ans =
    0.0017093

Which is roughly 0.17%.

Note that mvncdf expects a covariance matrix. And you have supplied a correlation, not apparently a covariance. But since the presumed variances were 1 by default, the correlation matrix will be the same as the covariance matrix here.

[Charlotte]

Thanks so much! That is how I implemented it and it obviously worked.