Let $X_1,X_2,X_3$ be independent r.v with $X_i \sim \ exp(1)$, evaluate $$\text{Var}[(X_1+X_2)X_3].$$
I tryed to use $$\text{Var}[(X_1+X_2)X_3]= E[((X_1+X_2)X_3)^2]-(E[(X_1+X_2)X_3])^2$$
But i keep finding 8 when the textbook gives the answer $\text{Var}[(X_1+X_2)X_3]=6$. Any hint for this?
One way to do this is to use the law of total variance: $$\begin{align} \operatorname{Var}[(X_1+X_2)X_3] &= \operatorname{Var}[\operatorname{E}[(X_1 + X_2)X_3 \mid X_3]] + \operatorname{E}[\operatorname{Var}[(X_1 + X_2)X_3 \mid X_3]] \\ &= \operatorname{Var}[X_3 \operatorname{E}[X_1 + X_2]] + \operatorname{E}[X_3^2 \operatorname{Var}[X_1 + X_2]] \\ &\overset{\text{ind}}{=} \operatorname{Var}[X_3 (\operatorname{E}[X_1] + \operatorname{E}[X_2])] + \operatorname{E}[X_3^2 (\operatorname{Var}[X_1] + \operatorname{Var}[X_2])] \\ &=\operatorname{Var}[X_3(1 + 1)] + \operatorname{E}[X_3^2 (1 + 1)] \\ &= \operatorname{Var}[2X_3] + \operatorname{E}[2X_3^2] \\ &= 4 \operatorname{Var}[X_3] + 2 (\operatorname{Var}[X_3] + \operatorname{E}[X_3]^2) \\ &= 6 + 2 \\ &= 8. \end{align}$$