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.
Best Answer
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.
$=3cov(X,X)-cov(X,Y)-cov(X,Z)-6cov(X,Y)+2cov(Y,Y)-4cov(Y,Z)$
$+12cov(X,Z)-2cov(Y,Z)-4cov(Z,Z)$