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The number of accidents follows a Poisson distribution with mean $12$. Each accident generates $1,2,$ or $3$ claimants with probabilities $\frac{1}{2},\frac{1}{3},\frac{1}{6}$ respectively. Calculate the variance in the total number of claimants.

This is a SOA exam question so I have been provided the answer and they solve it like so:

Treat as three independent Poisson variables, corresponding to $1,2,$ or $3$ claimants

rate$_1 = 6$, rate$_2 = 4$, rate$_3 = 2$

Var$_1 = 6$, Var$_2 = 16$, Var$_3 = 18$

Total Var $= 6+16+18 = 40$

I understand where the numbers are coming from. The rates are the probabilities of the number of claimants times the mean of $12$ and the variances are the rates times the number of claimants squared.

However, what I don't understand is why we using those numbers? Why is variance calculated that way?

Also, could we possibly do this using Wald's Theorem instead and how would one do that?

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  • $\begingroup$ Let $X$ be the number of one claimant accidents, and define $Y$ and $Z$ similarly. Then $X$, $Y$, and $Z$ are indeed Poisson with the rates given. The total number of claimants is $X+2Y+3Z$. If we assume that the random variables $X,Y,Z$ are independent, the variance of $X+2Y+3Z$ is as given. But I do not see that $X$, $Y$, and $Z$ are independent. $\endgroup$ – André Nicolas Feb 4 '16 at 3:33
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Let $$N \sim \operatorname{Poisson}(\lambda = 12)$$ be the frequency distribution of the number of accidents. Let $X_i$ be the discrete probability distribution of the number of claimants in accident number $i$, with mass function $$\Pr[X_i = 1] = \frac{1}{2}, \quad \Pr[X_i = 2] = \frac{1}{3}, \quad \Pr[X_i = 3] = \frac{1}{6}.$$ Let $$S = X_1 + X_2 + \cdots + X_N$$ be the random total number of claimants. The question implies that $X_i$ and $X_j$ are independent whenever $i \ne j$. (If not, then the phrase "each accident generates 1, 2, or 3 claimants with probabilities $\frac{1}{2}$, $\frac{1}{3}$, and $\frac{1}{6}$, respectively" would not be sufficient to specify the probability distribution of $X_i$.) Then we have, by the law of total variance, $$\begin{align*} \operatorname{Var}[S] &= \operatorname{E}[\operatorname{Var}[S \mid N]] + \operatorname{Var}[\operatorname{E}[S \mid N]] \\ &= \operatorname{E}[N \operatorname{Var}[X_i]] + \operatorname{Var}[N \operatorname{E}[X_i]] \\ &= \operatorname{Var}[X_i] \operatorname{E}[N] + \operatorname{E}[X_i]^2 \operatorname{Var}[N]. \end{align*}$$ Since $$\operatorname{E}[N] = \operatorname{Var}[N] = 12,$$ and $$\operatorname{E}[X_i] = \frac{5}{3}, \quad \operatorname{Var}[X_i] = \frac{5}{9},$$ we have $$\operatorname{Var}[S] = 40.$$

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