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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.

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1 Answer 1

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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.

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  • $\begingroup$ 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

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