If the random variables $X,Y$ and $Z$ have the means $\mu_{x}=2,\mu_{y}=-3 \text{ and } \mu_{z} = 4$ the variances $\sigma_{x}^{2}=3,\sigma_{Y}^{2}=2 \text{ and } \sigma^{2}_{z}=8 \text{ and } \\\text{cov}(X,Y) =1,\text{cov}(X,Z) = -2 \text{ and } \text{Cov}(Y,Z) = 3, \text{ find }$
The covariance of $U$ and $V= 3X-Y-Z.$
$U =71$
How does one get the answer$-54?$
I know that one is suppose to use the formula $\sigma_{xy}= \mu^{'}_{1,1}-\mu_{x}\mu_{Y}$
I know one needs to multiply in a certain way in order to derive the answer. However the process of deriving the exact formula eludes my comprehensions. Any tips on deriving the formula would be beneficial.
Hint: $cov(X-2Y+4Z,3X-Y-Z)$
Just multiply each summand of the first random variable by each summand of the second variable. The constants (numbers) can be factored out.
$=3cov(X,X)-cov(X,Y)-cov(X,Z)-6cov(X,Y)+2cov(Y,Y)-4cov(Y,Z)$
$+12cov(X,Z)-2cov(Y,Z)-4cov(Z,Z)$