joint of gaussian random transform of finite vectors

[[concept]]

Topics

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Proposition

Let $y_1,\dots ,y_n\in {\mathbb{R}}^m$ and $G\sim \mathcal{N}(0,1{)}^{\otimes d\times m}$ . Then

Gy_{1} \\ \vdots \\ Gy_{n} \end{bmatrix}\right) = {\cal N}\left(0,\, \begin{bmatrix} \lvert \lvert y_{1} \rvert \rvert ^2 I_{d} & \langle y_{1},y_{2} \rangle I_{d} & \dots & \langle y_{1},y_{n} \rangle I_{d} \\ \langle y_{2} , y_{1} \rangle I_{d} & \lvert \lvert y_{2} \rvert \rvert ^2 I_{d} & \dots & \langle y_{2}, y_{n} \rangle I_{d} \\ \vdots & \vdots & \ddots & \vdots \\ \langle y_{n}, y_{1} \rangle I_{d} & \langle y_{n} , y_{2} \rangle I_{d} & \dots & \lvert \lvert y_{n} \rvert \rvert ^2 I_{d} \end{bmatrix}\right)$$ Or, if we define $Y$ such that the $i$th column is $y_{i}$, then the the covariance matrix is given by $$\text{Cov}\left(\begin{bmatrix} Gy_{1} \\ \vdots \\ Gy_{n} \end{bmatrix}\right) = Y^{\intercal}Y \otimes I_{d} $$ ^statement

Proof

This follows immediately from gaussian random matrix transforms vectors into gaussian random vectors by applying the calculation to each block of the covariance matrix. (the expectation is identical)

References

References

See Also

Mentions

Mentions

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