1
$\begingroup$

$var(x) = \frac{1}{25} \left(\begin{array}{rr} \,66 & -12 \\ -12 & \,59 \end{array}\right)$

This is what I have done so far.

Let $var(x) = \sum$

$det(\sum-\lambda I) = \lambda^2 - 5\lambda +6$

Setting $P(\lambda)=0$ and solving the characteristic polynomial, I obtained $(\lambda_1, \lambda_2) = (2,3)$

$\lambda_1, \lambda_2 >0$ hence $\sum >0$ and $\sum$ is positive-definite.

As a result, Matrix $A$ is nonsingular, $A^{-1}$ exists.

Therefore $var(A^{-1}x) = A^{-1} \sum (A^{-1})^{'}=I$ and elements of vector $A^{-1}x$ are uncorrelated and have equal variance.

I'm not sure what to do from this point onward in order to find $A$ such that $var(Ax)=I$. I also need to show whether or not the components of vector $y=Ax$ are correlated.

Any help would be appreciated.

$\endgroup$
1
  • $\begingroup$ You are looking for $A$ so that $A\Sigma A^{T} = I$. $\endgroup$
    – copper.hat
    Commented Oct 22, 2015 at 22:39

1 Answer 1

1
$\begingroup$

$A=\Sigma^{-1/2}$. This operation is called whitening.

To the second part: $Cov(y) = Cov(Ax) = Cov(\Sigma^{-1/2}x)=Cov(\Sigma^{-1/2} \Sigma \Sigma^{-1/2} )= I$, where I is the identity matrix. Since the covariance matrix of $y$ is diagonal its components ore not correlated.

another thing: $\Sigma^{-1/2}$ is computed by means of SVD: $\Sigma=U\Lambda U^H$ then $\Sigma^{-1/2} = U \Lambda^{-1/2} U^H$.

Note that I used Cov and not Var.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .