For Company A there is 60% chance that no claim is made during the coming year. If one or more claim are made, the goal claim amount is normally distributed with mean 10,000 and standard deviation 2,000. For Company B there is a 70% chance that no claim is made during the coming year. If one or more claims are made, the total claim amount is normally distributed with mean 9,000 and standard deviation 2,000. Assume that the total claim amounts of the two companies are in dependent. What is the prob. that, in the coming year, Company B's total claim amount will exceed Company A's total claim amount?

Thank you in advance.


First note that according to the model used, negative claims are possible, although very unlikely. We could take account of such negative claims, or else we can tacitly assume that they will really not happen. We take the second approach.

There are two ways that B's total claim amount can exceed A's:

(i) B makes a claim and A doesn't. This has probability $(0.3)(0.6)$.

(ii) They both make claims, but B's is bigger. The probability they both make claims is $(0.4)(0.3)$. We must multiply this by the probability B's claim is bigger than A's, given they both made a claim. We now go after this probability.

Let $X$ be A's claim, given it made a claim, and let $Y$ be the corresponding claim fo $B$. We want the probability that $Y-X\gt 0$.

Let $W=Y-X$, Then $W$ is normal, mean $9000-10000$, and variance $(2000)^2+(2000)^2$, so standard deviation $\sqrt{2}(2000)$.

So we want the probability that a normal of mean $-1000$ and standard deviation $\sqrt{2}(2000)$ is greater than $0$. This is a standard calculation using standard normal tables or software.

Now one needs to put the pieces together.


Here I will let $C_{i}=1$ be event that company i has a claim maid in coming year , $C_{i}=0$ be event that company i has no claim maid in coming year, $A$, $B$ be random variables of total claim amount of company $A$ and $B$ respectively.

Now the we want to know $P(B>A)=P(B-A>0)$. Well a route of doing this is by using law of total probability where we are pretty much going to take a waited probability of all combinations of event of whether A or B make a claim or not. To ensure this is valid notice $$1) P\left(\bigcup_{i,j\in\{1,0\}}\left(C_{A}=i\bigcap C_{B}=j\right)\right)=1$$ $$2)\left(C_{A}=i_{1} \bigcap C_{B}=j_{1}\right)\bigcap \left(C_{A}=i_{2} \bigcap C_{B}=j_{2}\right)=\emptyset,\textrm{ for }i_{1},j_{1},i_{2},j_{2}\in\{0,1\}$$

Thus the events of (1) are a valid partition. Thus we have $$P(B-A>0)=\sum_{i,j\in\{1,0\}}P\left(B-A>0\bracevert C_{A}=i \bigcap C_{B}=j\right)P\left(C_{A}=i \bigcap C_{B}=j\right)=\sum_{i,j\in\{1,0\}}P\left(B-A>0\bracevert C_{A}=i \bigcap C_{B}=j\right)P\left(C_{A}=i\right)\left(C_{B}=j\right)$$ where this above in addition to properties of linear combination of normal distributions should be more obtainable for information you are given.

  • $\begingroup$ Also good luck on Exam P, I will be taking it this month too $\endgroup$ – Kamster Jul 13 '14 at 2:56
  • $\begingroup$ Note as Andre Nicolas pointed out, events with $C_{A}=0$ or $C_{B}=0$ have about $P(B-A>0)$ practically equal to 0, so for ease during test you could just ignore those $\endgroup$ – Kamster Jul 13 '14 at 3:18

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