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Let $X$ be an $n$-dimensional Gaussian random vector with mean $0$ and covariance $AA^T$. Clearly, if $Z$ is an $n$-dimensional Gaussian random vector with mean $0$ and covariance $I$ (the identity matrix), then $X$ and $AZ$ have the same distribution.

However, can I say that there exists $Z$ with mean $0$ and covariance $I$ such that $X = AZ$? If $A$ were invertible, I can let $Z = A^{-1}X$. What about for the general case where $A$ is an aribtrary matrix.

secondrate
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  • This is a comment rather than an answer. Because $AA^T$ is the covariance matrix for $X$, then $A$ is constrained such that $AA^T$ must be positive definite. In general, for an arbitrary matrix $A$, it is not guaranteed that this is true of $AA^T$, as you can see here: https://math.stackexchange.com/questions/730421/is-aat-a-positive-definite-symmetric-matrix – Sam Mar 17 '20 at 15:15

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