An insurance policy is written to cover a loss $X$ where $X$ has density function:
$f(x) = \frac38\cdot x^2$, in the interval $0\leq x\leq2$, and $f(x) = 0$, otherwise.
The time (in hours) to process a claim of size $x$, where $0\leq x\leq2$, is uniformly distributed on the interval $[x,2x]$.
Calculate the probability that a randomly chosen claim on this policy is processed in three hours or more.
In my perspective, it is necessary to use the Law of Total Probability.
I am not clear about how to find what is in the formula that is below as P(A / X=x). I also found the formula of conditional density function, but I do not know how to apply it to this problem. My guess is that as density is not equal to probability, I need to integrate that conditional density function to get the probability, but I am not sure. I would appreciate explanations about how to find that conditional density probability, the associate conditional probability, how to apply the Law of Total Probability to this problem, and in general, any way to solve this problem. My goal is to have clear those concepts.
Best Answer
Given an arbitrary $x$, the probability density function for claim processing time is $$f(z)= \begin{cases} 0 & z< x \\ \frac1x & x\leq z\leq 2x \\ 0 & 2x< z \end{cases} $$
Note that when $x=1.5$, the area under $f(z)$ where $z\geq3$ is $0$, while for $x=2$, the area under $f(z)$ is $1$. Since the function is linear, we can define the area under the curve as $$\textrm{Area under the curve}= \begin{cases} 0 & z< 1.5 \\ 2z-3 & 1.5\leq z\leq 2 \\ 0 & 2< z \end{cases} $$
So, the probability of an arbitrary $z$ having claim time over 3 hours is $$\frac{f(z)}{\textrm{length of }[z,2z]}=\frac{2z-3}z=2-\frac3z$$Obviously only when $1.5\leq z\leq2$, as otherwise, $f(z)=0$.
So, we integrate this function times the main probability distribution. This gives us $$\int_{-\infty}^\infty \frac38x^2\cdot (2-\frac3x)\textrm{dx}=\int_{1.5}^2 \frac38x^2\cdot (2-\frac3x)\textrm{dx}=\frac38\cdot\bigg(\frac{2x^3}3-\frac{3x^2}2\bigg)\bigg\vert_{1.5}^2=\color{red}{\frac{11}{64}}$$
Edit
In some sense, this problem involves two random variables, and an aggregation of their results. However, I think the following approach, while being less general than others, may serve as a good approach for many problems of this kind.
Suppose we have to find $P(B)$. We note that $$P(B)=\int_{-\infty}^\infty P(A)P(B|A) dA$$In intuitive terms, the meaning of this integral, is that we find the probability of each event in $A$, multiply it by the probability of $B$ given the event in $A$ happened, and then sum over all such events.
The tricky part of the computation is finding a general expression for $P(A)$ and $P(B|A)$. In this problem, $$P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{2x-3}x$$ and $P(A)=\frac38x^2$.