Why is the conditional variance of Y equal to (1-rou^2)* variance of y?

[tennishaha]

The attached equation is from
http://en.wikipedia.org/wiki/Multivariate_normal_distribution
can anyone show me why the conditional variance is equal to (1-rou^2)* variance of y

thanks

 

[ych22]

You have to first derive the marginal distribution of X. Then the conditional pdf of Y|X is given by f(x,y)/f(x). It will follow from there that the conditional variance is (1-p^2)Var(Y).

I am having problems with this question myself:

Let X1 and X2 be independent standard normal random variables. Show that the joint distribution of

Y1=aX1 + bX2 + c
Y2=dX1 + eX2 + f

is bivariate normal.

 

[tennishaha]

Your one can be shown using the jacobian

J=|a b| =ae-bd
d e

then the joint pdf of y1 and y2= joing pdf of X1 and X2 / (ae-bd)
X1 and X2 are independent, you can easily get their joint pdf,
and you can get joint pdf of y1 and y2, it is also a joint normal

You have to first derive the marginal distribution of X. Then the conditional pdf of Y|X is given by f(x,y)/f(x). It will follow from there that the conditional variance is (1-p^2)Var(Y).

I am having problems with this question myself:

Let X1 and X2 be independent standard normal random variables. Show that the joint distribution of

Y1=aX1 + bX2 + c
Y2=dX1 + eX2 + f

is bivariate normal.