Suppose that $X $ has a multivariate normal distribution $X\sim MVN (\mu, \Sigma) $, How can I transform $X$ into $Z$ so that $Z\sim MVN(\mu, I) $ where $I$ is the identity matrix?

For instance, let $\mu= \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$ and the variance-covariance matrix $\Sigma= \begin{bmatrix} 0.75 & -0.09& 0.33\\ -0.09 & 0.37& 0.10\\ 0.33 & 0.10& 0.29 \\ \end{bmatrix}$

I tried to use the singular value decomposition (SVD) and calculate the eigenvalues $\lambda_{1},\lambda_{2}, \lambda_{3}$ and eigenvectors $e_{1}, e_{2}, e_{3}$, but I do not know how to continue or if my approach is correct.


1 Answer 1


Your approach using SVD is one way to do it. Let $\Sigma$ decomposed by $UDU^{\rm T}$, where $D={\rm diag}(\lambda_1, \ldots, \lambda_p)$ and $U$ is a matrix with its columns the eigenvectors. Then, define $K=UD^{1/2}$, which satisfies $KK^{\rm T} = \Sigma$. Now multiply the inverse matrix of $K$ on the original vector $X$, i.e. $$K^{-1}(X-\mu) \sim N(0, K^{-1}\Sigma (K^{-1})^{\rm T}).$$ By noting that $K^{-1}\Sigma (K^{-1})^{\rm T}=I$, we get the desired result.

  • $\begingroup$ Is $K$ a Cholesky factor of $\Sigma$? $\endgroup$
    – JTH
    Commented Feb 13, 2020 at 18:58
  • $\begingroup$ The Cholesky decomposition is based on a lower triangular matrix, so $K$ is not a Cholesky factor. $\endgroup$
    – inmybrain
    Commented Feb 13, 2020 at 21:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.