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Suppose that $X, Y$, and $Z$ are three random variables such that $\operatorname{Var}(X)=1, \operatorname{Var}(Y)=4$, $\operatorname{Var}(Z)=8, \operatorname{Cov}(X, Y)=1, \operatorname{Cov}(X, Z)=-1$, and $\operatorname{Cov}(Y, Z)=2$. Determine $\operatorname{Var}(3 X-Y-2 Z+1)$.

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Variance of a Constant: The variance of an added constant is zero. $\operatorname{Var}(W+c)= \operatorname{Var}(W)$ Therefore, $\operatorname{Var}(3 X-Y-2 Z+1)=\operatorname{Var}(3 X-Y-2 Z)$.

Variance with a Coefficient: A coefficient is squared when pulled out of the variance.

$$
\operatorname{Var}(a X)=a^2 \operatorname{Var}(X)
$$

Variance of a Sum/Difference: The general formula for the variance of a linear combination of three variables $(a X+b Y+c Z)$ is: $\operatorname{Var}(a X+b Y+c Z)=a^2 \operatorname{Var}(X)+ b^2 \operatorname{Var}(Y)+c^2 \operatorname{Var}(Z)+2 a b \operatorname{Cov}(X, Y)+2 a c \operatorname{Cov}(X, Z)+2 b c \operatorname{Cov}(Y, Z)$

$$
\begin{aligned}
& a=3 \\
& b=-1 \\
& c=-2 \\
& a^2 \operatorname{Var}(X)=(3)^2 \times(1)=9 \times 1=\mathbf{9} \\
& b^2 \operatorname{Var}(Y)=(-1)^2 \times(4)=1 \times 4=\mathbf{4} \\
& c^2 \operatorname{Var}(Z)=(-2)^2 \times(8)=4 \times 8=\mathbf{3 2} \\
& 2 a b \operatorname{Cov}(X, Y)=2 \times(3) \times(-1) \times(1)=-6 \times 1=-\mathbf{6} \\
& 2 a c \operatorname{Cov}(X, Z)=2 \times(3) \times(-2) \times(-1)=-12 \times(-1)=\mathbf{1 2} \\
& 2 b c \operatorname{Cov}(Y, Z)=2 \times(-1) \times(-2) \times(2)=4 \times 2=\mathbf{8} \\
& \operatorname{Var}=(9+4+32)+(-6+12+8) \\
& \quad \operatorname{Var}=45+14 \\
& \quad \operatorname{Var}=\mathbf{5 9}
\end{aligned}
$$
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