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Under an insurance policy, a maximum of five claims may be filed per year by a policy holder. Let $p_n$ be the probability that a policy holder files $n$ claims during a given year, where $n = 0, 1, 2, 3, 4, 5.$

An actuary makes the following observations:

(i) $p_n\geq p_{n+1}$ for $0\leq n \leq 4$

(ii) The difference between $p_n$ and $p_{n+1}$ is the same for $0 \leq n \leq 4$

(iii) Exactly $40\%$ of policyholders file fewer than two claims during a given year.

Calculate the probability that a random policyholder will file more than three claims during a given year.

Source: Marcel B. Finan's A Probability Course for the Actuaries

My thoughts: The goal is to find $P(n > 3)$. We are given that $P(n<2)=.4$, which means that $P(n\geq2)=.6$. $P(n>3) = p_4 + p_5$. From this, we know that $p_0 + p_1 = .4$, and $p_2 + p_3 + p_4 + p_5 = .6$; so, $p_2 + p_3 + P(n > 3) = .6$. But I don't know how to solve for $p_2$ or $p_3$ to find $P(n > 3)$. Any help would be greatly appreciated.

Community
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Andrew Elkouri
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2 Answers2

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Let the constant difference $p_n-p_{n+1}$ be $d$.

We have $p_1=p_0-d$, $p_2=p_1-d=p_0-2d$, and similarly $p_3=p_0-3d$, $p_4=p_0-4d$, $p_5=p_0-5d$.

We are told that $p_0+p_1=0.4$. Thus $$2p_0-d=0.4.$$

Also, the sum of all the probabilities is $1$. Thus $$6p_0-15d=1.$$

Solve these two linear equations for $p_0$ and $d$. Now you can compute anything.

Remark: It would fit in better with standard notation for arithmetic progressions if we called $p_0$ by the name $a$, and let $d=p_{n+1}-p_n$.

André Nicolas
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Hint: Look at condition ii), the difference between any $p_n$ and $p_{n+1}$ is the same. Thus, $p_0-p_1=p_1-p_2\Rightarrow{p_0}-2p_1=p_2....$

Eleven-Eleven
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