This is a question from SOA: 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.
It's old but none of the solutions to it make any sense to me. Here is all I can think of:
I know the following:
$X-1+Y-2<5$ so $X+Y<8$
It will not exceed 5 if $X<1$ and $Y<2$
I don't understand what else to do from here, the solutions already provided online make absolutely no sense to me as to how they get things like $X-1<5$ if $X>1$ and $Y\leq2$. Can anyone explain how this is the case? How can X be any number greater than 1 if Y is less than or equal to 2 and $X+Y$ be $<8$? This makes no sense.
Best Answer
Reality Check: Benefit paid by any particular insurer cannot be negative, irrespective of the deductable.
You seek the probability for: $$\{\max(X-1,0)+\max(Y-2,0)\leq 5\}$$
Partitioning on indication that $Y\leq 2$ or not:$$\{Y\leq 2\,,\max(X-1,0)\leq 5\}\cup\{Y>2\,, \max(X-1,0)+Y\leq 7\}$$
Partitioning again on indication that $X\leq 1$ or not:$${\{X\leq 1\,, Y\leq 2\}}\cup{\{1<X\leq 6\,, Y\leq 2\}}\\\cup{\{X\leq 1\,, 2<Y\leq 7\}}\cup{\{X>1\,, Y>2\,, X+Y\leq 8\}}$$
In the last part, we require $1<X$ and $2<Y$ and $X+Y\leq 8$, which is to say, the probability for $1<X\leq 6$ and $2<Y\leq 8-X$:
$$\mathsf P(1<X\leq 6\,, 2<Y\leq 8-X) = \int_1^6\int_2^{8-x}\tfrac 1{100}\mathop{\rm d}y\mathop{\rm d}x = \tfrac{1}{8}$$
Do similar for the other parts.