# Finding the covariance matrix of $Y=AX$ where $X$ is a known multivariate gaussian random variable

covariancemultivariate-statistical-analysisnormal distributionprobabilitystatistics

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.

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.