P(X=k)=e-0.45(0.45)k/k!In this formula, e is the base for natural logarithms (e is about 2.71) and k! is the product k*(k-1)*(k-2)*....*2*1. 0! and 1! are special cases: both are defined to be 1.
The parameter 0.45 is of course critical---it defines the probabilities as above. It also has a natural physical meaning; it equals the average number of hurricanes per year (the expectation of the random variable X). It is also the variance of X.
To answer the original question (one or more hurricanes this year), compute the probability of the complementary event (no hurricane) and subtract from 1:
P(at least one hurricane)=1-P(no hurricane)=1-e-0.45=.36
Mr Jianxin Wang estimated the probability of tropical cyclones coming within 250 nautical
miles of Honolulu, in a May 1997 M. S. thesis in meteorology, regardless of whether the cyclone reached hurricane status.
[He also estimated probabilities for various strengths of these storms.] Mr. Wang's estimate is based upon 47 years
of data, 1949-1955, during which time 26 tropical cyclones came within 250 nautical miles of Honolulu.
Below is a summary table of the data:
The Poisson estimates are quite good, especially given the major uncertainties in the data. For example, in 1969 satellite observations of the North Pacific greatly increased the comprehensiveness of the survey (and introduced its own technical difficulties, of identifying storms and measuring their strengths from optical data of cloud patterns). Where does the parameter .5532 come from?
Matching means. The parameter of the Poisson distribution just happens to also be the mean of the Poisson distribution and .5532=26/47 is the average number of cyclones per year from 1949 to 1995.Alternatively, one can make a maximum likelihood estimate . Let L be an arbitrary parameter for the Poisson distribution and for each value of L compute W(L)=P(29,12,4,2 | L) [the probability of 29 years of no cyclones, 12 years with 1, 4 years with 2 and 2 years with three]. Find a value of L such that W(L) is a maximum. In this case, viewing the sequence of years as independent rolls of an infinitely sided die (with 0 cylones, 1 cyclone, etc. on the faces of the die), W(L) is a multinomial probability:
W(L)=(47-choose-29,12,4,2) [e-L]29 * [e-L*L]12 * [e-L *L2/2] 4 *[e-L*L3/6]2This simplifies to
W(L)=(47-choose-29,12,4,2)*e-47LL26/576To find a maximum value for W, set dW/dL = 0 and solve for L. One obtains L=0 and L=26/47; the first derivative test will show that L=26/47 provides a maximum.
I revisited Mr. Wang's data for the years from 1969 to 1995, when satellite technology played a role in
identifying cyclones. Here is the data, with a mean of .7037=19/27.
Given that the parameter is "really" .5523, how likely is it that we have the observed pattern of cyclones of the 1969--1995 period? One way to estimate this is to create a new random variable, X, for the number of hurricanes in the period from 1969 to 1995. X is again Poisson, with parameter 27*.5532=14.94. What is the probability that X is at least 19?
P(X > 18)=1-P(X< =18) = 1-[Sum of e-L Lk/k!, 0 < = k < = 18].Thus, P(X > 18)=1-.82395=0.17605. So, the pattern of storms in the 1969--1995 is slightly unusual but not extremely so and certainly not strong enough to reject 0.5523 as the "real" parameter in the Poisson distribution.
On the other hand, suppose the parameter is "really" .7037. How likely is it to have only 7 cyclones during the 1949 to 1968 period? To analyze this, let Y be the number of cyclones in a 20 year period. Y is again a Poisson random variable, but with parameter M equal to 20*.7037=14.62. What is the probability that Y is at most 7?
P(Y< = 7)=Sum of e-M Mk/k!, 0 < = k < = 7.Thus, P(Y < =7)=.0224257. This is a low probability, and serves as evidence against 0.7037 as the "real" parameter [at a 95% confidence level for a two sided exact confidence interval analysis].
This raises the interesting question of what would be a 90% confidence interval for this parameter? Let us use all 47 years of data and let Z be the number of cyclones in a 47-year period. For what intervals of L, where L is the parameter for the occurrence of cyclones in a single year, will P(Z < 27) > = .05?
P(Z < 27)=e-47*L* [Sum of (47*L)^k/k!, 0 < = k < = 26].For L > .7675, this probability is less than .05. Next, consider the intervals of L for which P(Z > 25) > = .05?
P(Z > 25)=1-e-47*L* [Sum of (47*L)^k/k!, 0 < = k < = 25].For L < .3877, this probability is less than .05. So a 90% confidence interval could be [.3877,.7675], meaning that for parameters outside this interval the observed number of storms (26) has a low probability. It is truly unfortunate that this interval is so wide [it must make insurance companies very nervous about Hawaii].