1. ## Discrete Probability Distribution

I have been unable to solve the following exercise from my Probability and Statistics book.

Under an insurance policy, a maximum of five claims may be filed per year by a policyholder. Let $\displaystyle p_n$ be the probability that a policyholder files n claims during a given year, where $\displaystyle n$ = 1,2,3,4,5. An actuary makes the following observations:

(i) $\displaystyle p_n \geq p_{n+1}$ for $\displaystyle n$ = 1,2,3,4.
(ii) The difference between $\displaystyle p_n$ and $\displaystyle p_{n+1}$ is the same for $\displaystyle n$ = 1,2,3,4.
(iii) Exactly 40% of the policyholders file fewer than three claims during a given year.

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

(A) 0.14 (B) 0.16 (C) 0.27 (D) 0.29 (E) 0.33
I succeeded in expressing observations (ii) and (iii) as mathematical statements:

(ii) $\displaystyle p_n - p_{n+1} = k$ for $\displaystyle n$ = 1,2,3,4.
(iii) $\displaystyle p_1 + p_2 = 0.4$

I also know the probabilities sum to 1, so

$\displaystyle p_1 + p_2 + p_3 + p_4 + p_5 = 1$

To solve the exercise, I need to find $\displaystyle p_4 + p_5$, but I have been unable to do so. I tried rewriting the sum of probabilities like so:

$\displaystyle p_1 + (p_1 - k) + (p_1 - 2k) + (p_1 - 3k) + p_5 = 1 \iff 4p_1 - 6k + p_5 = 1$

I have tried other approaches, but it seems I am aimlessly solving equations that are just restatements of each other. I believe if I could find $\displaystyle p_1$ and $\displaystyle k$, then I could find $\displaystyle p_4$ and $\displaystyle p_5$. If anyone could point me in the right direction, I would appreciate it.

By the way, the solutions manual says the answer is (C) 0.27.

2. ## Re: Discrete Probability Distribution

Originally Posted by NOX Andrew
I have been unable to solve the following exercise from my Probability and Statistics book.

I succeeded in expressing observations (ii) and (iii) as mathematical statements:

(ii) $\displaystyle p_n - p_{n+1} = k$ for $\displaystyle n$ = 1,2,3,4.
(iii) $\displaystyle p_1 + p_2 = 0.4$

I also know the probabilities sum to 1, so

$\displaystyle p_1 + p_2 + p_3 + p_4 + p_5 = 1$

To solve the exercise, I need to find $\displaystyle p_4 + p_5$, but I have been unable to do so. I tried rewriting the sum of probabilities like so:

$\displaystyle p_1 + (p_1 - k) + (p_1 - 2k) + (p_1 - 3k) + p_5 = 1 \iff 4p_1 - 6k + p_5 = 1$

I have tried other approaches, but it seems I am aimlessly solving equations that are just restatements of each other. I believe if I could find $\displaystyle p_1$ and $\displaystyle k$, then I could find $\displaystyle p_4$ and $\displaystyle p_5$. If anyone could point me in the right direction, I would appreciate it.

By the way, the solutions manual says the answer is (C) 0.27.

i dont understand the proposed solution.

$\displaystyle p_1 = p_2 = p_3 = p_4 = p_5= 0.2$ satisfies all the conditions and would give an answer of 0.4.

Moreover, the above solution is unique:

(ii)
$\displaystyle p_n - p_{n+1} = k$
=>$\displaystyle p_n =a + kn$

(iii)
$\displaystyle p_1 + p_2 = 0.4$
=>$\displaystyle (a+k) + (a + 2k) = 0.4$
=>$\displaystyle 2a+3k = 0.4~~~~~~~~~~:$ [1]

$\displaystyle p_1 + p_2 + p_3 + p_4 + p_5 = 1$
=>$\displaystyle 5a + 15k = 1 ~~~~~~~~~:$ [2]

The only solutions to [1] and [2] are a=0.2, k=0

and so $\displaystyle p_n = 0.2$

3. ## Re: Discrete Probability Distribution

Thank you for your reply. In [2], it appears you made the substitution $\displaystyle p_5 = a + 5k$, but $\displaystyle p_n - p_{n + 1} = k$ only applies for $\displaystyle n = 1,2,3,4$. Does this change anything?

Since it doesn't apply to $\displaystyle p_5$, a possible distribution satisfying the conditions is as follows $\displaystyle p_1 = 0.21$, $\displaystyle p_2 = 0.19$, $\displaystyle p_3 = 0.17$, $\displaystyle p_4 = 0.15$, $\displaystyle p_5 = 0.28$, in which case the answer is 0.43.

4. ## Re: Discrete Probability Distribution

the relationship $\displaystyle p_n = a + kn$ applies to n=1,2,3,4,5.

You already noted that $\displaystyle p_n - p_{n+1} = k$ (n=1,2,3,4)

use n=4, so:
$\displaystyle p_4 - p_5 = k$

and then use the same reasoning as before.

Remark
If you believe that the above isn't valid for $\displaystyle p_5$ then there are infinately many solutions to the problem(this isn't hard to show).

5. ## Re: Discrete Probability Distribution

Of course! It seems so obvious now. Thank you for understanding my question about applying it to p_5 and providing such a clear, concise explanation.

6. ## Re: Discrete Probability Distribution

FYI there is a mistake in my first post

when i said
$\displaystyle p_n - p_{n+1} = k$
$\displaystyle => p_n =a + kn$
I changed the sign of k. This doesn't affect the result