(This article was first published on

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. **SAS and R**, and kindly contributed to R-bloggers)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 -10000

1 0 0

2 10000 10000

3 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 = 1000000

snow = rpois(numsim, 1.5)

pay = snow - 1 # subtract one

pay[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 7236.96

analytic 7231.30

------------------------

To

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