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For $X=(X_1,X_2,X_3)$ with Gaussian distribution, covariance matrix $2I_3$, and mean vector $\mu$ = $(3,3,3)^T$, I want to find the covariance matrix of $Y=(Y_1,Y_2,Y_3)^T$ where

$$Y=\begin{pmatrix}1/\sqrt 2 & 0 & -1/\sqrt 2 \\1/\sqrt 3 & 1/\sqrt 3 & 1/\sqrt 3\\ 1/\sqrt 6 & -2/\sqrt 6 & 1/\sqrt 6\\ \end{pmatrix}X$$

I know that when I use the definition of the covariance matrix in terms of expectations, my answer is $2I_3$ (which is the correct answer).

However when I substitute in $$Y=AX=A(2Z+\mu)=2AZ+A\mu$$ the covariance matrix of $Y$ is $(2A)(2A)^T$ which simplifies to $4I_3$.

However, this contradicts the correct answer $2I_3$.

I hope someone could clarify this for me.

FD_bfa
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2 Answers2

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Careful, your substitution for $X$ (and thus your substitution for $Y$) is not correct. For $X$ to have covariance $2I$, then you must have $$Y = A \underbrace{(\sqrt{2} Z + \mu)}_{=^D X}$$ with the equality being in distribution.

After you retrace your steps, you'll get the desired covariance.

Jose Avilez
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  • Thanks. Just realised I used Variance instead of standard deviation in my substitution. Thanks. – FD_bfa Jan 03 '22 at 19:41
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In the case of the linear transformation, $\mathbf{y=Ax}$, the covariance of $\mathbf{y}$ is given by $\mathbf{\Sigma}_y=\mathbf{A}\mathbf{\Sigma}_x\mathbf{A}^T$ which in your case simplifies into $\mathbf{\Sigma}_y=2\mathbf{A}\mathbf{A}^T=2 \mathbf{I}_3$

Steph
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