Let
$ (X_1, X_2)^T $ be MVN $ (\mu, \sum )$, where $\mu = \begin{pmatrix} -1 \\ 1 \\\end{pmatrix}$ and $ \sum = \begin{pmatrix} 2 & -1 \\ -1 & 3 \\\end{pmatrix}$.
Let $ Y_1 = 2X_1 + X_2 $ and $ Y_2 = -X_1 + 2X_2. $
I then need to find the distribution of $ Y=(Y_1, Y_2)^T.$
So, I found that $ X_1 = \sqrt2 Z_1 -1 $ and $ X_2 = \sqrt3 Z_2 +1. $
Where $Z_1$ and $Z_2$ are standard normal.
I rewrote $ Y=(Y_1, Y_2)^T$ in terms of $Z_1$ and $Z_2$ via what I have about Y in terms of $X=(X_1,X_2)^T$.
I obtained:
$ \begin{pmatrix} Y_1 \\ Y_2 \\\end{pmatrix} = \begin{pmatrix} 2\sqrt2 &\sqrt3 \\ -\sqrt2 & 2\sqrt3 \\\end{pmatrix} \begin{pmatrix} Z_1 \\ Z_2 \\\end{pmatrix} + \begin{pmatrix} -1 \\ 3 \\\end{pmatrix}$.
Since $Y_1$ and $Y_2$ is expressed in terms of $Z_1$ and $Z_2$, I directly calculated the joint density of $Y_1$ and $Y_2$.
I obtained that $ \begin{pmatrix} Y_1 \\ Y_2 \\\end{pmatrix} $ ~ $(\begin{pmatrix} -1 \\ 3 \\\end{pmatrix}, \begin{pmatrix} 11 &2 \\ 2 & 14 \\\end{pmatrix})$,
where $\begin{pmatrix} 11 &2 \\ 2 & 14 \\\end{pmatrix}$ = $\begin{pmatrix} 2\sqrt2 &\sqrt3 \\ -\sqrt2 & 2\sqrt3 \\\end{pmatrix} \begin{pmatrix} 2\sqrt2 &\sqrt3 \\ -\sqrt2 & 2\sqrt3 \\\end{pmatrix}^T $.
However, my variance-covariance matrix is different to that of the answer. The answer was
$ \begin{pmatrix} Y_1 \\ Y_2 \\\end{pmatrix} $ ~ $(\begin{pmatrix} -1 \\ 3 \\\end{pmatrix}, \begin{pmatrix} 7 & -1 \\ -1 & 18 \\\end{pmatrix})$.
I don't see where I went wrong either. Any feedback would be appreciated.
Best Answer
Your error is in asserting that $X_1 = \sqrt2 Z_1 -1$ and $X_2 = \sqrt3 Z_2 +1$. If this were true then $X_1$ and $X_2$ would be independent (assuming $Z_1$ and $Z_2$ are independent), but we know they are correlated.
Rather than express $X:=(X_1,X_2)^T$ in terms of $Z_1$ and $Z_2$, express $Y$ directly in terms of $X$. Use the following fact:
Here $B=\begin{pmatrix}2&1\\-1&2\end{pmatrix}$, so plug in $\operatorname{Cov}(X)=\begin{pmatrix}2&-1\\-1&3\end{pmatrix}$ to obtain the mean vector and covariance matrix for $Y:=BX$.