Actuarial Science - Part 1

$$ Let $\text{N}$ be the random variable that denote the number of claims. Let $\text{G}$ and $\text{B}$ be random variable that denote our sample come from good drivers and bad drivers, respectively. Hence $$ $\begin{aligned} &\text{Pr}[\text{G}]=0.75\\ &\text{Pr}[\text{B}]=0.25\\ &\text{Pr}[\text{N}=0|\text{G}]=0.7\\ &\text{Pr}[\text{N}=1|\text{G}]=0.2\\ &\text{Pr}[\text{N}=2|\text{G}]=0.1\\ &\text{Pr}[\text{N}=0|\text{B}]=0.5\\ &\text{Pr}[\text{N}=1|\text{B}]=0.3\\ &\text{Pr}[\text{N}=2|\text{B}]=0.2\\ \end{aligned}$ $$ The marginal probability for $\text{N}_1=0$ and $\text{N}_2=1$ is $$ $\begin{aligned} \text{Pr}[\text{N}_1=0\cap\text{N}_2=1]&=\text{Pr}[\text{N}=0|\text{G}]\cdot\text{Pr}[\text{N}=1|\text{G}]\cdot\text{Pr}[\text{G}]+\text{Pr}[\text{N}=0|\text{B}]\cdot\text{Pr}[\text{B}=1|\text{B}]\cdot\text{Pr}[\text{B}]\\ &=0.7\cdot0.2\cdot0.75+0.5\cdot0.3\cdot0.25\\ &=0.1425. \end{aligned}$ $$ The actuary want to observe that the third observation $\text{N}_3\ge1$ given that $\text{N}_1=0$ and $\text{N}_2=1$ . Therefore, the joint probability of all posibilities is $$ $\begin{aligned} \text{Pr}[\text{N}_1=0\cap\text{N}_2=1\cap\text{N}_3=1]&=\text{Pr}[\text{N}=0|\text{G}]\cdot\text{Pr}[\text{N}=1|\text{G}]\cdot\text{Pr}[\text{N}=1|\text{G}]\cdot\text{Pr}[\text{G}]+\text{Pr}[\text{N}=0|\text{B}]\cdot\text{Pr}[\text{B}=1|\text{B}]\cdot\text{Pr}[\text{N}=1|\text{B}]\cdot\text{Pr}[\text{B}]\\ &=0.7\cdot0.2\cdot0.2\cdot0.75+0.5\cdot0.3\cdot0.3\cdot0.25\\ &=0.03225,\\ \text{Pr}[\text{N}_1=0\cap\text{N}_2=1\cap\text{N}_3=2]&=\text{Pr}[\text{N}=0|\text{G}]\cdot\text{Pr}[\text{N}=1|\text{G}]\cdot\text{Pr}[\text{N}=2|\text{G}]\cdot\text{Pr}[\text{G}]+\text{Pr}[\text{N}=0|\text{B}]\cdot\text{Pr}[\text{B}=1|\text{B}]\cdot\text{Pr}[\text{N}=2|\text{B}]\cdot\text{Pr}[\text{B}]\\ &=0.7\cdot0.2\cdot0.1\cdot0.75+0.5\cdot0.3\cdot0.2\cdot0.25\\ &=0.01800.\\ \end{aligned}$ $$ Note that each number of claim is i.i.d. , then the probability that the third observation $\text{N}_3\ge1$ given that $\text{N}_1=0$ and $\text{N}_2=1$ is $$ $\begin{aligned} \text{Pr}[\text{N}_3\ge1|\text{N}_1=0\cap\text{N}_2=1]&=\text{Pr}[\text{N}_3=1|\text{N}_1=0\cap\text{N}_2=1]+\text{Pr}[\text{N}_3=2|\text{N}_1=0\cap\text{N}_2=1]\\ &=\frac{\text{Pr}[\text{N}_1=0\cap\text{N}_2=1\cap\text{N}_3=1]}{\text{Pr}[\text{N}_1=0\cap\text{N}_2=1]}+\frac{\text{Pr}[\text{N}_1=0\cap\text{N}_2=1\cap\text{N}_3=2]}{\text{Pr}[\text{N}_1=0\cap\text{N}_2=1]}\\ &=\frac{0.03225}{0.1425}+\frac{0.01800}{0.1425}\\ &=\frac{0.05025}{0.1425}\\ &=\frac{67}{190}. \end{aligned}$ $$ Another way to solve this $$ $\begin{aligned} \text{Pr}[\text{N}_3\ge1|\text{N}_1=0\cap\text{N}_2=1]&=1-\text{Pr}[\text{N}_3<1|\text{N}_1=0\cap\text{N}_2=1]\\ &=1-\text{Pr}[\text{N}_3=0|\text{N}_1=0\cap\text{N}_2=0], \end{aligned}$ $$ which yields the same result with the previous approach. Thus, $p+q=67+190=\boxed{\color{#3D99F6}{257}}$ . $$

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$\text{\# }\mathbb{Q.E.D.}\text{ \#}$

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Solve the problem intuitively by considering a pool of 4000* statistically idealized drivers.

3/4 of them, or 3000, are good drivers. Multiply by 0.7 to find that 2100 of them file zero claims in the first year. Multiply that by 0.2 to see that 420 of that group file one claim the next year, and finally by 0.3 to get 126 filing one or more claims in the third year.

Likewise, 1000 are bad drivers, 500 of which have zero claims in year 1, 150 of those having one claim in year 2, and finally 75 of those that have one or more in the last year.

The policyholder in question is one of the 420 + 150 = 570 who have zero then one claims, and we're asked to find whether she is also one of the 126 + 75 = 201 who file one or more in the third year as well.

201/570 = 67/190, so the answer is 257. No big scary formulas.

(* we get this number by thinking ahead and multiplying all the denominators of the probabilities we're going to be dealing with)