# Is there a problem with this old Exam P actuarial probability question?

This is a problem from an old Society of Actuaries Exam P (probability) exam:

A family buys two policies from the same insurance company. Losses under the two policies are independent and have continuous uniform distributions on the interval from 0 to 10. One policy has a deductible of 1 and the other has a deductible of 2. The family experiences exactly one loss under each policy.

Calculate the probability that the total benefit paid to the family does not exceed 5.

I don't understand this language: the second sentence says

$$X:=\text{Loss}_{\text{Policy 1}}\sim\text{Unif}[0,10]$$ $$Y:=\text{Loss}_{\text{Policy 2}}\sim\text{Unif}[0,10]$$

where $X$ and $Y$ are independent. So the losses are uniformly distributed. But nothing is said explicitly about the benefit each policy will pay.

Am I supposed to assume that each policy will pay the entire loss after the deductible? Otherwise there simply is not enough information to answer the question.

But it seems absurd to me for the Society of Actuaries to require candidates to make an assumption that is implausible in practice: insurance policies often don't pay the entire loss after a deductible. There is usually some upper bound you know in advance.

Call the random losses under each policy $L_1$, $L_2$, respectively. We are told that $$L_1 \sim \operatorname{Uniform}(0,10), \quad L_2 \sim \operatorname{Uniform}(0,10),$$ with $L_1$, $L_2$ independent. Call the benefit random variables under each policy $X_1$, $X_2$; these represent the money that is paid out when $L_1$ and $L_2$ are observed. Since we are also told that the first policy has an ordinary deductible of $1$ and the second has an ordinary deductible of $2$, it follows that $$X_1 = \max(0, L_1 - 1), \quad X_2 = \max(0, L_2 - 2).$$ That is to say, the payout is $0$ if the loss does not exceed the deductible; otherwise, the payout is the loss minus the deductible. We are then asked to determine $$\Pr[X_1 + X_2 \le 5].$$ To this end, it seems natural to want to compute the complementary probability: $$\Pr[X_1 + X_2 > 5].$$ We have three cases; either

1. $X_1 > 5$ and $X_2 = 0$
2. $X_1 = 0$ and $X_2 > 5$
3. $X_1, X_2 > 0$ and $X_1 + X_2 > 5$.

In the first case, we have from independence $\Pr[X_1 > 5] = \Pr[L_1 > 6] = 0.4$, and $\Pr[X_2 = 0] = \Pr[L_2 \le 2] = 0.2$, thus $$\Pr[(X_1 > 5) \cap (X_2 = 0)] = (0.4)(0.2) = 0.08.$$

The second case is handled similarly; we get $$\Pr[(X_1 = 0) \cap (X_2 > 5)] = (0.1)(0.3) = 0.03.$$

The third case is the trickiest. We can see that it comprises the outcomes for which $$(1 < L_1 \le 10) \cap (2 < L_2 \le 10) \cap (L_1 + L_2 > 8).$$ If we think of this as inequalities plotted in a coordinate plane, this set of inequalities corresponds to a pentagonal region with vertices $$(L_1, L_2) \in \{(6,2), (10, 2), (10, 10), (1, 10), (1,7) \}.$$ The area of such a region is $(8)(9) - (5)(5)/2 = 119/2$, thus the desired probability of the third case is $$\Pr[(0 < X_1) \cap (0 < X_2) \cap (X_1 + X_2 > 5)] = \frac{119}{2(10^2)} = 0.595.$$ It follows that the desired total probability is $$1 - (0.08 + 0.03 + 0.595) = 0.295.$$

Regarding the interpretation of the terminology used in the question: it is standard terminology that when the word "deductible" is used without any qualifier, that what this means is an ordinary deductible: losses not exceeding the deductible do not generate a payment on the claim, and losses above the deductible generate a payment of the loss minus the deductible. This is in contrast to, for example, a franchise deductible, in which losses below the deductible still generate no payment, but losses at or above the deductible are paid in full.

• @BruceET I don't believe my calculation was for a discrete uniform distribution. It works for the continuous uniform. – heropup Mar 10 '18 at 8:59
• Got it.... Sorry. – BruceET Mar 10 '18 at 9:01

Comment and Diagram: I'm not familiar with this actuarial terminology. Informed by @heropup's excellent Answer, I have revised my graphical presentation of the problem. See heropup's answer for details of the computation. (Perhaps the major assumption is that the company will pay anything at all.)

As an answer to the exam question, I believe that you want the probability (area/100) of the red region in the diagram below. Points are possible total losses (covered by the two policies except for deductibles). The joint probability density of the total losses takes the value $0.01$ above the square with vertices at $(0,0)$ and $(10,10).$

Policy 1 will pay nothing to the left of the vertical blue line; but Policy 2 may pay in part of this region. Similarly, for the region below the horizontal blue line. In the triangular region above and to the right of $(1,2)$ the diagonal line bounds the total loss to 8 (so that the total payout will not exceed 5).

Considering the red region altogether, the probability is as in @heroup's answer. (The points plotted to make the figure amount to a simulation with 100,000 iterations; the proportion of red points is about $0.297 \pm 0.003.$ More points would have given a more accurate approximation, but a less attractive figure.)