rishi wrote:

> Given two standard normal variables - e1 and e2 - both IID.
> Given also x and y, with correlation p between them.
> 
> Let x = e1
> 
> Then y = p*e1 + ((1-p)^.5)*e2  (I think)
> 
> Here is my question: How do I extend this to the three variable case?
> 
> In this case, there are three variables, x, y, and z with correlations
> Pxy, Pxz, and Pyz. You can have as many gaussian e disturbances as you
> want, but I think you'd only need e1, e2, and e3.
> 
> I must be stupid, but I can't generalize from the two to the three
> variable case. Help!

First: Although your formula for y does cause x and y to have 
correlation p, it's wrong if you want y to be standard-normal, as you 
should be able to prove easily.  In order to correct your expression in 
that case, write an expression for the variance of y = p*e1 + d*e2 and 
then solve for the value of d you'd need in order to make the variance 
of y equal unity.

Second: The generalization that you're looking for can be worked out as 
follows.  Suppose that you've generated jointly-normal r.v. x and y by 
the method you described and that you know the correlation (Pxy) and 
variances of x and y.  Moreover, suppose that a third r.v. z is 
generated from z = a*x + b*y +c*e3, where e3 is standard-normal and 
uncorrelated with both x and y, which will be the case if e1, e2 and e3 
are standard-normal and IID, for example.  You should be able to write 
expressions for the variance of z, the correlation Pxz and the 
correlation Pyz in terms of a, b, c, Pxy, and the variances of x and y. 
  Solving these expressions for a, b and c in terms of the other 
parameters indicates what values of a, b and c you'll need in order to 
get the set of variances and correlations you want.  The way in which 
this approach can be extended from three to four r.v., from four to 
five, etc., should now be obvious.

   Charles Metz



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