Variance of combination of random variables (not independent) I am doing this question.
You have random variables X, Y and Z which have means 1,2,3 respectively and variances 2,4,6 respectively. Also, Cov(X,Y) = Cov(Y,Z) = 1 and Cov(X,Z) = 0.
I have to find the mean and variance of U = X-2Y+Z.
Now, I have done the mean of U and found it to be 0, which I hope is correct. However, I am stuck on the variance part - the covariances are messing me up and I don't know how to proceed. Any help? Thanks.
 A: The mean follows straight from additivity : $$E[U] = E[X - 2Y + Z] = E[X]  - 2E[Y] + E[Z] = 1 - 4 + 3 = 0$$.
The variance, on the other hand, is not so simple, but still doable:
$$
var(X - 2Y+Z) = E[(X-2Y+Z)^2] - E[X-2Y+Z]^2 \\
= E[X^2  + 4Y^2 + Z^2 - 4XY - 4YZ + 2XZ] - E[X - 2Y + Z]^2 \\
= \color{red}{E[X^2]}  \color{green}{ + 4E[Y^2]} \color{orange}{+ E[Z^2]} \color{blue}{- 4E[XY]} \color{brown}{- 4E[YZ]} \color{gray}{+ 2E[XZ]} \\
\color{red}{- E[X]^2} \color{green}{- E[2Y]^2}  \color{orange}{- E[Z]^2} \color{blue}{+ 4E[X]E[Y]} \color{brown}{+ 4E[Y]E[Z]} \color{gray}{- 2E[X]E[Z]} \\   = \color{red}{var(X)} + \color{green}{4var(Y)} + \color{orange}{var(Z)} \color{blue}{- 4Cov(X,Y)} \color{brown}{- 4Cov(Y,Z)} \color{gray}{+ 2Cov(X,Z)} 
$$
where the parts with similar color combine to give the respective part on the next line. Hence, the answer would be $2 + 16 +6-4-4 = 16$. Please check and tell me if this answer is correct.
A: Covariance is bilinear and symmetric, so that:
$$\begin{aligned}\mathsf{Var}U & =\mathsf{Cov}\left(U,U\right)\\
 & =\mathsf{Cov}\left(X-2Y+Z,X-2Y+Z\right)\\
 & =\mathsf{Cov}\left(X,X\right)+4\mathsf{Cov}\left(Y,Y\right)+\mathsf{Cov}\left(Z,Z\right)-4\mathsf{Cov}\left(X,Y\right)+2\mathsf{Cov}\left(X,Z\right)-4\mathsf{Cov}\left(Y,Z\right)\\
 & =\mathsf{Var}X+4\mathsf{Var}Y+\mathsf{Var}Z-4\mathsf{Cov}\left(X,Y\right)+2\mathsf{Cov}\left(X,Z\right)-4\mathsf{Cov}\left(Y,Z\right)\\
 & =2+4\cdot4+6-4\cdot1+2\cdot0-4\cdot1\\
 & =16
\end{aligned}
$$
