## Monday, February 28, 2011

### Example 8.28: should we buy snowstorm insurance?

It's been a long winter so far in New England, with many a snow storm. In this entry, we consider a simulation to complement the analytic solution for a probability problem concerning snow.

Consider a company that buys a policy to insure its revenue in the event of major snowstorms that shut down business. The policy pays nothing for the first such snowstorm of the year and \$10,000 for each one thereafter, until the end of the year. The number of major snowstorms per year that shut down business is assumed to have a Poisson distribution with mean 1.5. What is the expected amount paid to the company under this policy during a one-year period?

Let SNOW be the number of snowstorms, and pay the amount paid out by the insurance. The following chart may be useful in discerning the patttern:
`SNOW    PAY     10000*(snow-1)0       0      -100001       0       02       10000   100003       20000   20000`

The analytic solution is straightforward, but involves a truncation of the first snowstorm. Since we can assume that the random variable SNOW ~ Poisson(1.5) we know that E[SNOW] = 1.5 and E[10000*(SNOW-1)] = 10000*E[snow] - 10000 = 15000 - 10000 = 5000.

E[PAY] is equal to E[10000*(SNOW-1]) + 10000*P(SNOW=0) so the exact answer is
`10000*P(snow=0) + 15000 - 10000 =10000*exp(-1.5) + 15000 - 10000 = \$7231`

Here the advantage of simulation is that it may provide a useful check on the results, as well as a ready measure of variability. In this situation, the code is quite simple, but the approach is powerful.

R
`numsim = 1000000snow = rpois(numsim, 1.5)pay = snow - 1      # subtract onepay[snow==0] = 0    # deal with the pesky P(snow=0)sim = mean(pay*10000)analytic = 10000*(dpois(0, 3/2) + 3/2 - 1)`

Yielding the following:
`> sim[1] 7249.55> analytic[1] 7231.302`

SAS
The simulation and analytic solutions are also straightforward in SAS. Here the analytic result is only calculated once

`data snow_insurance;do i = 1 to 1000000;  nsnow = ranpoi(0, 1.5);  payout = max(nsnow -1, 0) * 10000;  output;end;analytic = 10000 * (cdf("POISSON", 0, 1.5) + 1.5 -1);output;run;proc means data=snow_insurance mean;  var payout analytic;run;`

This results in the following output:
`Variable            Mean------------------------payout           7236.96analytic         7231.30------------------------`

Anonymous said...

Hi Nick

This is an interesting post and something that I'm interested in as an insurance actuary.

Some observations.

I don't think that you have enough data to deduce that the number of storms is Poi(1.5).

To determine this you would need to fit a distribution to a dataset which compares calendar years vs number of storms.

Putting this aside however, let's assume that Poi(1.5) is reasonable - certainly it sounds it based on recent weather in the northeast!

When I was studying, the situation you're describing was called a Compound Poisson process.

Putting aside this jargon, the key thing to remember with a Compound Poisson is that the average aggregate loss is given by:

ave freq * ave size of loss.

Again, I'm not sure that this will tie back to your calculation; because (as with yor frequency pick) I don't think you've got enough data to determine what distribution is suitable for describing the size of a snow loss (assuming there is a loss).

Hopefully I haven't gone too far off piste, as I'd be keen to understand your approach better.

Best regards,
dM/

Nick Horton said...

The rate parameter of 1.5 and the assumption of Poisson are highly suspect (since this question arises from one of the multiple choice questions on the published first actuarial exam). But I like it because it helps students to see the value of a simulation as a way of checking their analytical (closed form) solution. It's a simple problem, but more tricky than it first looks (at least to me!).