Calculate the probability for the total insurance benefit -- solution verification

Question

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 $5$.

This question has been discussed here and here. I've read them both but couldn't figure out why my solution is incorrect based on those discussions.

My attempt: Let $P$ denote the loss amount under the first policy and let $Q$ denote that under the second policy. Let $X$ denote the benefits received under the first policy and let $Y$ denote that under the second policy. Then $f(p) = \frac{1}{10}$ when $0\le p\le 10$ and $f(q) = \frac{1}{10}$ when $0\le q\le 10$ and it follows that $f(p, q) = \dfrac{1}{100}$ when $0\le p, q\le 10$. Also,

$$X = \begin{cases} 0 & 0\le P<1\\ P-1 & 1\le P\le 10 \end{cases}$$

$$Y = \begin{cases} 0 & 0\le Q<2\\ Q-2 & 2\le Q\le 10 \end{cases}$$

We need to find $\Pr(X+Y\le 5 | X, Y>0)$. I am not sure what distributions $X$ and $Y$ follow (do they necessarily follow the uniform distribution?), so I tried to express the desired probability in terms of the variables whose joint distribution I derived. $$\Pr(X+Y\le 5| X, Y>0)= \Pr(P+1+Q+2\le 5| P>1, Q>2) =\Pr(P+Q<8 | P>1, Q>2) $$ $$= \int_1^8 \int_2^{8-p} \frac{1}{100} \; \text{d}Q \, \text{d}P$$

This, however, did not lead to the correct response. Where have I gone wrong? Did I compute the conditional probability correctly? Should I divide $\int_1^8 \int_2^{8-p} \frac{1}{100} \; \text{d}Q \, \text{d}P$ by the product of the marginal distributions of $P$ and $Q$?

Answer 1

You've mixed up your random variables. $P$ and $Q$ are the ground-up loss variables, and $X$ and $Y$ are the claim payment variables. $P, Q$ are uniform, but $X, Y$ are not. You defined $X = P - 1$ if $P > 1$, yet when you wrote $$\Pr[X + Y \le 5 \mid X, Y > 0] = \Pr[P + 1 + Q + 2 \le 5 \mid P > 1, Q > 2] + \cdots,$$ you are now saying $X = P+1$ and $Y = Q + 2$. This is not correct. Your probability statement should instead read: $$\begin{align} \Pr[X + Y \le 5] &= \Pr[P-1 + Q-2 \le 5 \mid P > 1, Q > 2]\Pr[P > 1, Q > 2] \\ &\quad + \Pr[P-1 \le 5 \mid P > 1, Q \le 2]\Pr[P > 1, Q \le 2] \\ &\quad + \Pr[Q-2 \le 5 \mid P \le 1, Q > 2]\Pr[P \le 1, Q > 2] \\ &\quad + \Pr[0 \le 5 \mid P \le 1, Q \le 2]\Pr[P \le 1, Q \le 2]. \end{align}$$

This further simplifies to

$$\begin{align} \Pr[X + Y \le 5] &= \Pr[P + Q \le 8 \mid P > 1, Q > 2]\frac{72}{100} + \Pr[P \le 6 \mid P > 1]\frac{18}{100} \\ &\quad + \Pr[Q \le 7 \mid Q > 2]\frac{8}{100} + \frac{2}{100}. \end{align}$$ I leave it as an exercise to conclude that $$\Pr[P + Q \le 8 \mid P > 1, Q > 2] = \frac{25}{2(8)(9)} = \frac{1}{8}, \\ \Pr[P \le 6 \mid P > 1] = \frac{5}{9}, \\ \Pr[Q \le 7 \mid Q > 2] = \frac{5}{8},$$ hence $$\Pr[X + Y \le 5] = \frac{59}{200}.$$

Answer 2

The deductibles mean that payout distributions are mixtures of a discrete value at $0$ and a continuous value up to $9$ or $8.$ So you have to be careful finding the distribution of the sum. The following simulation in R is not a solution, but it may help you to find the correct solution (or to see mistakes in an incorrect one). With a million iterations of such families, you can expect two (maybe three) place accuracies for probabilities approximated by simulation.

set.seed(107)
L1 = runif(10^6, -1, 9);  L2 = runif(10^6, -2, 8)
x = L1;  x[L1 < 0]=0
summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  0.000   1.500   4.004   4.052   6.506   9.000 
mean(x==0)
[1] 0.099936      # aprx P(X = 0) = 0.1

y = L2;  y[L2 < 0]=0  # 0 payout if below deductible 
summary(y)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 0.0000  0.4955  3.0001  3.2007  5.5062  8.0000 
mean(y==0)
[1] 0.200588      # aprx P(Y = 0) = 0.2

t = x + y
summary(t)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  0.000   4.417   7.250   7.253   9.932  16.987 
mean(t <= 5)
[1] 0.294548      # aprx P(T <= 5)
2*sd(t <=5)/1000
[1] 0.0009116791  # 95% margin of sim error in above

Because of the possibility of $0$ payouts for one or both policies, histograms (below) have to be interpreted with care.

par(mfrow=c(1,3))
 hist(x, prob=T, br=50, ylim=c(0,1.2), col="wheat", main="First Policy")
 hist(y, prob=T, br=50, ylim=c(0,1.2), col="wheat", main="Second Policy")
 hist(t, prob=T, br=50, col="skyblue2", main="Sum of Two")
  abline(v=5, col="red", lwd=2)
par(mfrow=c(1,1))