EDIT
If the random variables $X,Y, Z$ have the expected,
$$\text{ means: }\mu_{x}=2 \qquad \qquad \mu_{y}=-3 \qquad \qquad \mu_{z} = 4$$
$$ \text{variances: }\sigma_{x}^{2}=3 \qquad \qquad \sigma_{y}^{2}=2 \qquad \qquad \sigma^{2}_{z}=8$$
$$\text{covariances: }\text{cov}(X,Y) =1 \quad \quad \text{cov}(X,Z) = -2 \quad \quad \text{Cov}(Y,Z) = 3$$
find the variance of $U = X-2Y+4Z$.
The co-variance of $U$ and $V = 3X-Y-Z$
One must use these formulas in order to solve this problem.
From what I have deduced from the formulas above in order to find the variance one must use this formula $v(U)= \text{var}(a_x+b_y+c_z) =a^2\cdot \text{var}(x) +b^2\cdot \text{var}(Y)+c^2 \cdot \text{var}(z) + 2ab \cdot cov(x,y) +2ac\cdot \text{cov}(x,z)+2abc \cdot \text{cov}(Y,Z)$
To find the co-variance one must use this formula
$\text{cov}(u,v) = \text{cov}(a_1+b_1+c_1,a_2+b_2+c_2)=(a_{1})(a_{2})\text{var}(x)+(b_{1})(b_{2})\text{var}(Y)+(c_{1})(c_{2})\text{var}(Z)+\left[ (a_{1})(b_{2})+(b_{1})(a_{2}) \right] \cdot \text{cov}(X,Y)+\left[ (a_{1})(c_{2})+(c_{1})(a_{2}) \right] \cdot \text{cov}(X,Z)+ \left[ (b_{1})(c_{2})+(c_{1})(b_{2}) \right] \cdot \text{cov}(Y,Z)$
Is the formula that I used above a correct interpretation of what is alluded by in the formulas above?
Lastly, I do not want to make duplicates so the questions I have asked above is different from what I asked before in the previous questions, mainly because am asking about the interpretations of the formulas…… not just the answer.I hope this is enough information so that this question can be its own independent entity.
Best Answer
Let
$$Y := c_1 X_1 + c_2 X_2 + c_3 X_3 = \mathrm c^\top \mathrm X$$
Hence,
$$\mathbb E (Y) = \mathrm c^\top \mathbb E (\mathrm X)$$
and
$$\mbox{Var} (Y) = \mathrm c^\top \left( \mathbb E (\mathrm X\mathrm X^\top) - \mathbb E (\mathrm X) \mathbb E (\mathrm X^\top) \right) \mathrm c = \mathrm c^\top \mbox{cov} (\mathrm X) \, \mathrm c$$
Using the data in the question,
$$\mbox{Var} (Y) = \begin{bmatrix} 1\\ -2\\ 4\end{bmatrix}^\top \begin{bmatrix} 3 & 1 & -2\\ 1 & 2 & 3\\ -2 & 3 & 8\end{bmatrix} \begin{bmatrix} 1\\ -2\\ 4\end{bmatrix} = \cdots = 71$$