If the random variables $X,Y, Z$ have the
means $$\mu_{x}=2 \qquad \qquad \mu_{y}=-3 \qquad \qquad \mu_{z} = 4$$
the variances
$$\sigma_{x}^{2}=1 \qquad \qquad \sigma_{y}^{2}=5 \qquad \qquad \sigma^{2}_{z}=2$$
the covariances$$\text{cov}(X,Y) =-2 \quad \quad \text{cov}(X,Z) = -1 \quad \quad \text{Cov}(Y,Z) = 1$$ Find the mean and the variance of $W= 3X-Y+2Z$
In order to solve this problem one must use this theorem.
So the first part in order to find the mean you do this.
$$E(W) =E(3X-Y+2Z) \\ = 3E(X)-E(Y)+2E(Z) \\= 3\cdot 2 -(-3)+2\cdot 4=17$$
Now here is the part where I do not understnad and formula above plays a tremendous role. Now one must find the variance.
$$Var(W) = 9\text{var}(X)=Var(Y)+4\text{var}(Z)-6\text{cov}(X,Y)+12\text{cov}(X,Z)-4\text{cov}(Y,Z)$$
I want to know how this equation was formed.
Despite me rereading the formula above I can only infer that one must use $a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bc+2bd+2cd$ in order to solve this, but I am unsure am I in the right track?
Since you have only 3 variables the formula is
$Var(3X−Y+2Z)=Var(aX+bY+cZ)$
$=a^2\cdot Var(X)+b^2\cdot Var(Y)+c^2\cdot Var(Z)$
$+2\cdot a\cdot b\cdot Cov(X,Y)+2\cdot a\cdot c\cdot Cov(X,Z)+2\cdot b\cdot c\cdot Cov(Y,Z)$
with $a=3, b=-1, c=2, Var(X)=1, Var(Y)=5, Var(Z)=2,$
$Cov(X,Y)=-2, Cov(X,Z)=-1, Cov(Y,Z)=1$
In your case $b=-1$. Thats why $2\cdot b\cdot c\cdot Cov(Y,Z)=2\cdot (-1)\cdot 2\cdot 1=-4$ is negative.
Thus it begins with
$Var(3X−Y+2Z)=3^2\cdot 1+(-1)^2\cdot 5+\ldots $
Can you go on ?