Bivariate normal distribution

[alext87]

Let (X,Y) have a bivariate normal pdf with correlation coefficient p and variance 1 for both X and Y.

 

So the joint density is:

f(x,y) = (1/((2*PI)SQRT(1-p^2)))*exp(-0.5*(1-p^2)^-1 * (x^2-2pxy+y^2))

 

Show that X and Z=(Y-pX)/SQRT(1-p^2) are independent N(0,1) variables, and deduce that

 

P(X>0,Y>0) = 1/4 +arcsin(p)/((2*PI))

 

??

 

Does anybody know how to do this!?

[Bignose]

alex,

 

as this is sure looks homework-ish, it is against the forum rules to tell you directly the answer. But, I will do my best to help.

 

Firstly, what is the definition of independent variables? What about that property can you use to determine if your two variables are independent?

 

Can you re-write the distribution in terms of x & z?

 

For the second question, what does it mean to find the probability of all cases x>0 and y>0? What procedure do you have that performs such a task? Do you remember how to do it with one variable? It is a natural extension to 2.

[thedarkshade]

it's very hard to follow Alex. Could you please LaTeXise it? If you don't know, browse the forum for a LaTeX tutorial. It's easy to use it and will make things much clearer!

 

I hope you understand!

 

Cheers

[alext87]

Okay I can show that they are independent... That they are N(0,1) comes from the marginal distributions. I need to integrate over the first quadrant but the integration is what I am struggling on...

 

One method I tried was to change variables to x and z and then separate as they are independent. However, I can't get the answer. Do you what the limits are of z? Do you know a substitution I can make in the integrand for z?

[Bignose]

Why are you integrating over z? The P you are looking for is written in terms of x and y, right? I don't think that you need to do that in z variables.

 

Though, even then, if you still did it in z, the limits of x and y can be used to find what the limit of z is. x is always positive and y is always positive, so you should be able to figure out how z is limited. Look at the definition of z again. Think about how x and y are allowed to vary, and what that does to z according to its definition.

[alext87]

I have worked out how to do it now. There was no need to work out the integral (in fact it cannot be done analytically). You need to note that the bivariate normal distribution is centrally symmetric about the origin if X and Z are independent and N(0,1).

 

Hence the angle was proportion to the angle the area made with the origin.

 

Thanks for your help.

 

AT