Multivariate Normal

WeeG

TS Contributor
#1
Hello, I have a small question that I can't solve...

Find a vector a, such that:

Y= ( X2 - a'(x1 x3)') and X2, are independent.

The Data:

X~N(mu,sigma) 3 dimensional

mu'=(2,-3,1)

sigma=( 1 1 1
1 3 2
1 2 2 )

I know that X2~N(-3,3) and I found that Y~N(A*mu,A*sigma*A')
Where

A*mu= -2a1-3-a2
A*sigma*A'=2a2^2-a1^2-4a2+3

So I have the 2 distributions, but how do I find a1 & a2, that makes Y and X2 independent ? I think it has something to do with the covariance being zero, but I don't know how to do it, please help.... :confused:
 

vinux

Dark Knight
#2
Hello, I have a small question that I can't solve...

Find a vector a, such that:

Y= ( X2 - a'(x1 x3)') and X2, are independent.

The Data:

X~N(mu,sigma) 3 dimensional

mu'=(2,-3,1)

sigma=( 1 1 1
1 3 2
1 2 2 )

I know that X2~N(-3,3) and I found that Y~N(A*mu,A*sigma*A')
Where

A*mu= -2a1-3-a2
A*sigma*A'=2a2^2-a1^2-4a2+3

So I have the 2 distributions, but how do I find a1 & a2, that makes Y and X2 independent ? I think it has something to do with the covariance being zero, but I don't know how to do it, please help.... :confused:

Consider the matrix B as
[ 0 1 0
-a1 1 -a2 ]
So your B* [ X1 X2 X3 ]' = [ X2 X2 -a1 X1 - a2X3]' = [ x2 Y]'

Covariance of B* [ X1 X2 X3 ]' = B * sigma * B'

=
[ 0 1 0
-a1 1 -a2 ] * [ 1 1 1 ; 1 3 2 ; 1 2 2] * B'


( It is difficult to write matrix here :shakehead )

finally you will get a 2 x 2 matrix.
Now the theory comes...
If X2 and Y are independent then the cov matrix should be diagonal

that is off diagonal is zero

then we get a1 + 2a2 = 3

Here you will get infinite solution. put a2=0 then a1 =3 like this.

[ Since [ X1 X2 X3 ] is MVN( multivariate normal) so B*[ X1 X2 X3 ] also MVN and and for MVN if cov is diagnal then components should be independnt ]


Vinux