Let $X$ denote the loss amount sustained by an insurance company’s policyholder in an auto collision. Let $Z$ denote the portion of $X$ that the insurance company will have to pay. An actuary determines that $X$ and $Z$ are independent with respective density and probability functions $$ f(x) = \begin{cases} \frac{1}{8} e^{-x/8},& x>0\\ 0 & \mbox{otherwise} \end{cases} $$
and
$$ P(Z = z) = \begin{cases} 0.45& z = 1, \\ 0.55&\mbox{otherwise} \end{cases} $$ Calculate the variance of the insurance company’s claim payment $ZX$.
My understanding of $P$ is that there are two events: $\{Z=1\}$ and $\{Z \neq 1\} = \{0 \leq Z <1 \}$, with probability 0.45 and 0.55 respectively. My confusion arises as I don't see how not knowing the distribution of $Z$ on $\{Z \neq 1\}$ can have no impact on the variance of $ZX$, and a few Mathematica simulations
a = 0; b = 1;
Variance[Table[ RandomChoice[{0.45, 0.55} -> {1, RandomVariate[UniformDistribution[{a, b}]]}] RandomVariate[ ExponentialDistribution[1/8]], {100000}]]
seem no confirm this suspicion. However, the answer is supposed to be 44.6. What am I not seeing?
Glad the comment helped. Just to clear this question I will put that comment here also: There was a typo in the question and it should have been: \begin{align*} P[Z=1]&=0.45\\ P[Z=0]&=0.55 \end{align*} from that you can indeed show $Var(XZ)=44.64$.