Jump to content
Sign in to follow this  
alext87

Bivariate normal distribution

Recommended Posts

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!?

Share this post


Link to post
Share on other sites

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.

Share this post


Link to post
Share on other sites

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

Share this post


Link to post
Share on other sites

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?

Share this post


Link to post
Share on other sites

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.

Share this post


Link to post
Share on other sites

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

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
Sign in to follow this  

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.