Question: Suppose $X_1$, $X_2$ and $X_3$ are independent random variables such that $X_1 \sim N(0,1)$, $X_2 \sim N(1,4)$ and $X_3 \sim N(-1,2)$. Let $Y_1=X_1+X_3$ and $Y_2=X_1+X_2-2X_3$. Give the joint distribution of $Y_1$ and $Y_2$.
My attempt: So far I have calculated that $Y_1 \sim N(-1, 3)$ and $Y_2 \sim N(3, 13)$ by use of the fact the independence of the $X_i$ implies that any linear combination of them is also a normally distributed random variable. However, I am unsure how to find their joint distribution.
Theory: Considering the transformation $Z=\mu+BX$ where:
- B is some $m \times n$ matrix
- $X$ is a vector of iid standard normal variables
- $Z$ has expectation vector $\mu$ and covariance matrix $\Sigma = BB^T$
Then we have $Z \sim N(\mu, \Sigma)$.
Application of theory: I presume in my case I would let $Y$ denote the column vector $[Y_1, Y_2]^T$. Then, from my above working, the expectation vector would be $\mu = [-1, 3]^T$. However, I am not sure how to go from here to determine what $X$ would be or how to calculate the covariance matrix $\Sigma$.