For $X=(X_1,X_2,X_3)$ with Gaussian distribution, covariance matrix $2I_3$ ($2$ multiplied by the identity matrix), 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 $2I$ (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.

## Best Answer

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.