# Normal Distributions and Claim Amounts

Suppose for company A, there is a $60 \%$ chance that no claim is made in the coming year. If one or more claims are made, then the total amount is normally distributed with mean $1000$ and standard deviation $2000$.

Suppose for company B, there is a $70 \%$ chance that no claim is made during the coming year. If one or more claims are made, then the total amount is normally distributed with mean $9000$ and standard deviation $2000$.

Assume the total claim amounts are independent. What is the probability that in the coming year, company B's total claim amount will exceed company A's total claim amount?

Let $X_A$ be company A's total claim amount and $X_B$ likewise. We want to find $P(X_B - X_A >0)$. Now $E[X_B-X_A] = 8000$ and $\text{Var}[X_B-X_A] = 2000^2+2000^2$. So what we want to find is:

$\displaystyle P \left[Z \geq \frac{-8000}{\sqrt{2000^2+2000^2}} \right]$?

I feel that we are given two conditional distributions: $P(X_A|I_A)$ and $P(X_B|I_B)$ where $I_A$ and $I_B$ indicate that more than one claim is made for company A and B respectively.

## 1 Answer

As the expected difference in claims is 2.8 standard deviations, there is little (how little?) chance that A's claims will exceed B's claims assuming they each get one. If we ignore it, and assume that the chance of a claim at A is independent of the chance of a claim at B (can we?) We have probability of no claim either place, so a tie, =60%*70%=42%. Probability of claim at A, none at B =40%*70%=28% chance of higher claim at A. Chance of higher claim at B=30%, as if there is a claim it will be higher than A's. These add to 100%, a good check. I'll let you figure the chance that they both get a claim and A's is higher than B's.

It is also a little hard to fathom a claim being made on A which is less than zero, but if you are only 1/2 standard deviation low, that will happen.

• I do state the question explicitly. – PrimeNumber Oct 26 '10 at 20:31
• You are right, sorry 'bout that. I have removed the remark. – Ross Millikan Oct 26 '10 at 20:46
• So what we want to find is the following: $P\{[I_{A}^{C} \cap I_B] \cup [(I_A \cap I_B) \cap (X_A < X_B)] \}$? – PrimeNumber Oct 26 '10 at 23:34
• Exactly. And the two sides of your union are nicely disjoint. – Ross Millikan Oct 27 '10 at 0:06