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An auto insurance company is implementing a new bonus system. In each month the policyholder does not have an accident, he ore she will receive 5.00 cash back bonus from the insurer. Among the 1000 policyholders of the auto insurance company, 400 are low risk drivers and 600 are high risk. The probability of a high risk driver having a accident in a given month is 0.2 and for low risk is 0.1. What is the expeccted bonus payment from the insurer to the 1000 policyholders?

I saw the solution and it gave no explanation as to why they used a poisson distribution. I cannot see why they did that a little insight would help. When should I consider using each type of distribution?

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  • $\begingroup$ Poisson often fits well the number of "accidents" in time $t$. But it may very well be, depending on wording, that we can do the expectation calculation knowing very little about the distribution. What does $0.2$ represent? Probability of at least one accident in a year? In a month? $\endgroup$ – André Nicolas Dec 20 '13 at 20:03
  • $\begingroup$ in a month sorry I didnt catch that $\endgroup$ – adam Dec 20 '13 at 20:24
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The distributions of numbers of drivers having one or more accidents in each month are Binomial in the case of both the low risk group of drivers and the high risk group of drivers. When N is large and p is small the Binomial can be approximated by a Poisson distribution with same expected value (Np), so total number of drivers having accidents in a month can be approximated as the sum of two variables each drawn from a Poisson distribution and which can be assumed to be independent. The sum of two independent Poisson variables is also Poisson distributed (with expected value of the sum equal to the sum of the expected values).

The Poisson approximation to the Binomial and the distribution of the sum of independent Poisson variables are both standard results which you should be able to find in any decent introductory text on statistical theory (or via Google).

The probability value of 0.2 for the high-risk group is a bit on the high side for using the Poisson approximation and for this problem an alternative would be to use the Normal approximation to the Binomial instead. In the second step the distribution of the total numbers of drivers having accidents would be the sum of two independent Normal variables, so also Normally distributed. Again, the Normal approximation and the ditribution of the sum are both standard results.

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  • $\begingroup$ so since $n=600 $and$ n=400 $we use a poisson distribution. We multiply by 12 to fit the interval of it being in one year. Ok I understand, so I just need to find my $n$ and determine if it is large enough to be approximated with the poisson distribution. $\endgroup$ – adam Dec 20 '13 at 20:18
  • $\begingroup$ The answer was based on the monthly distribution of drivers with accidents (Poisson with mean equal to 400*0.1 + 600*0.2). However, as we can assume that the distributions in successive months are independent then distribution across the year of driver-months where driver had an accident would also be Poisson with mean of 12 times the monthly mean. $\endgroup$ – DMM Dec 20 '13 at 20:35
  • $\begingroup$ @AndreNicolas - agree with your comments about independence and not needing to know the distribution if only interested in the annual expected values. However, the original question did ask about why a Poisson distribution might have been used - hence my answer. $\endgroup$ – DMM Dec 20 '13 at 20:47
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If the probability of at least one accident in a calendar month is $0.2$ for a high-risk driver, then the probability of no accident is $0.8$. Thus the expected number of accident-free months is $(0.8)(12)$. Add up over all high risk drivers. We get that the expected total bonuses to high risk drivers is $(5)(600)(0.8)(12)$.

There is a similar expression for the expected total bonuses to low risk drivers. Add.

Please note that we did not need information about the distribution. In particular, We do not need to assume independence, by the linearity of expectation.

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