gaussian random matrix transforms vectors into gaussian random vectors

[[concept]]

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Proposition

Let $G\sim \mathcal{N}(0,1{)}^{\otimes d\times m}$ be a random matrix and $y\in {\mathbb{R}}^m$. Then $\text{Law}(Gy)=\mathcal{N}(0,||y|{|}^2{I}_d)$

Proof

$Gy$ is linear in $G$. The entries of $G$ form a gaussian random vector, and so $Gy$ must be Gaussian as well. Thus, it suffices to calculate the mean and covariance to find its law.

Since each entry of $G$, call them $G_{i\alpha }$, is iid $\mathcal{N}(0,1)$, we get

\mathbb{E}(Gy)_{i} &= \mathbb{E}\left[ \sum_{\alpha=1}^m G_{i\alpha} y_{\alpha} \right] \\ &=0 \end{align}$$ And for the variance $$\begin{align} \text{Cov}(\,(Gy)_{i}, (Gy)_{j}\,) &= \mathbb{E}[\,(Gy)_{i}(Gy)_{j}\,] \\ &= \mathbb{E}\left( \left( \sum_{\alpha=1}^m G_{i\alpha} y_{\alpha} \right) \left( \sum_{\beta=1}^m G_{j\beta}y_{\beta} \right) \right) \\ &= \sum_{\alpha,\beta=1}^m y_{\alpha}y_{\beta} \cdot \mathbb{E} [G_{i\alpha} G_{j\beta}] \\ &= \begin{cases} 0 & i \neq j \\ \lvert \lvert y \rvert \rvert ^2 & i=j \end{cases} \\ \implies \text{Cov}(Gy,Gy) &= \lvert \lvert y \rvert \rvert ^2 I_{d} \end{align}$$ $$\tag*{$\blacksquare$}$$ ^proof-0

Above, we found the expectation and covariance in terms of the entry-wise products and sums. We can also do the calculation in terms of matrices.

Proof (matrices)

Expectation If there is only one random matrix involved, we can write $\mathbb{E}[Gy]=\mathbb{E}[G]y(=0{)}^{(\ast )}$ ${\_}^{(\ast )}$ In terms of our same setting above. This is a simple result of linearity of expectation.

Covariance If $g_1,\dots ,g_m\in {\mathbb{R}}^d$ are the columns of $G$ so that each $g_{\alpha }\sim \mathcal{N}(0,I_d)$ iid, then we see that

\text{Cov}(Gy) &= \mathbb{E}\left[ (Gy)(Gy)^{\intercal} \right] \\ &= \mathbb{E}\left[ \left( \sum_{\alpha=1}^m y_{\alpha} g_{\alpha} \right)\left( \sum_{\beta=1}^m y_{\beta} g_{\beta}^{\intercal} \right) \right] \\ &= \sum_{\alpha,\beta=1}^m y_{\alpha} y_{\beta} \cdot \mathbb{E}\left[ g_{\alpha}g_{\beta}^{\intercal} \right] \\ (*) &= \sum_{\alpha=1}^m y_{\alpha}^2 \cdot I_{d} \\ &= \lvert \lvert y \rvert \rvert ^2 I_{d} \end{align}$$ Where again $(*)$ is in terms of the setting above. In this line we use the law of the $g_{\alpha}$. $$\tag*{$\blacksquare$}$$ ^proof-1

References

References

See Also

Mentions

Mentions

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