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Let $X$ be a multivariate normal random variable with mean $\boldsymbol{\mu}$ and variance matrix $\mathrm{\Sigma}$. Next, define

Suppose that $Y = AX$ where $A$ is appropriate matrix. Can we say that the distribution of $Y$ is same as $X$ if and only if $A$ is an orthogonal symmetric matrix?

Thank you Iosif for pointing out the mistake. I modified the question.

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  • $\begingroup$ If different directions have different variances then rotations can alter the distribution. $\endgroup$
    – user44143
    Commented Nov 20, 2022 at 13:34

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$\newcommand\Si\Sigma$First here, there is no such thing as an orthonormal matrix. So, let us assume that you meant an orthogonal matrix instead.

Then we have this question:

Suppose that $X\sim N(\mu,\Si)$ and $Y = AX$, where $A$ is appropriate matrix. Can we say that the distribution of $Y$ is same as $X$ if and only if $A$ is an orthogonal matrix?

The answer to this is that neither the "if" part of this conjecture nor its "only if" holds in general.

Indeed, the distribution of $Y=AX$ is $N(A\mu,A\Si A^\top)$. So, the distribution of $Y$ is same as $X$ if and only if $$A\mu=\mu \tag{1}\label{1}$$ and $$A\Si A^\top=\Si. \tag{2}\label{2}$$

If $\Si$ is nonsingular, then condition \eqref{2} can be rewritten as $A_\Si A_\Si ^\top=I$, where $$A_\Si:=\Si^{-1/2}A\Si^{1/2}.$$ So, in this case, the distribution of $Y$ is same as $X$ if and only if \eqref{1} holds and $A_\Si$ is an orthogonal matrix.

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  • $\begingroup$ The pdf of $Y$ is $$f_{Y}(y) = \frac{1}{2\pi(det(A\Sigma A^T))^{(1/2)}}exp[-(1/2)(AX-A\mu)^T(A\Sigma A^T)]^{-1}(XA-A\mu) = \frac{1}{2\pi(det(\Sigma ))^{(1/2)}}exp[-(1/2)(X-\mu)^T(\Sigma)]^{-1}(X-\mu)$$. $\endgroup$
    – Satya Prakash
    Commented Nov 20, 2022 at 14:35
  • $\begingroup$ While there is no such thing as an orthonormal matrix, a good argument can be made that the usual term “orthogonal matrix” was a mistake and it should have been “orthonormal matrix” all along, because a real $n\times n$ matrix is an orthogonal matrix exactly when its columns are an orthonormal basis of $\mathbf R^n$, not just an orthogonal basis. Serge Lang, in the 3rd edition of his Algebra, bemoaned the terminology “orthogonal matrix” because such matrices are not just ones with orthogonal columns: the columns must be orthogonal unit vectors. He suggested the term “real unitary matrix”. $\endgroup$
    – KConrad
    Commented Nov 20, 2022 at 16:06
  • $\begingroup$ @KConrad : This is a good point; "orthogonal matrix" is indeed an example of bad generally accepted terminology/notation. $\endgroup$
    – Iosif Pinelis
    Commented Nov 20, 2022 at 16:28

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