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events that occurs, on average, λ times.
The number of occurrences follows a Poisson distribution.
The probability of having k occurrences is given by e − λ λk
P ( k occurrences ) =
k!
Where k = 0, 1, 2, … NUS/FOS/DSAP for λ > 0 6 GEM2900 Semester 1, 2009/1010 Rare Events & Poisson Distribution
A random variable, X, has a Poisson distribution, i.e.
X ~ Poisson(λ). The p.m.f is
Th e − λ λk
P( X = k ) =
k!
Where k = 0, 1, 2, … for λ > 0 The expected value and the variance for a Poisson
random variable: E[ X ] = Var[ X ] = λ NUS/FOS/DSAP 7 GEM2900 Semester 1, 2009/1010 Rare Events & Poisson Distribution
Example: Death by Horse Kick in Prussian Army...
Once in a while, someone in cavalry units would be killed by a
horse kick. A statistician, Bortkiewicz took the record of such death
from 10 cavalry units in Prussian army over a period of 20 years.
That is, he has information on 200 cavalryunityears
Number of deaths
per unit per year
Number of unityears 0 1 2 3 4 >4 109 65 22 3 1 0 Bortkiewicz showed that the number of deaths per cavalryunityear followed a Poisson distribution. NUS/FOS/DSAP 8 GEM2900 Semester 1, 2009/1010 Rare Events & Poisson Distribution
Example: Death by Horse Kick in Prussian Army...
Therefore, the parameter of the distribution would be expected
number of deaths per unit per year if the death number follows a
Poisson distribution
di
Total number of death… 0 × 109 + 1 × 65 + 2 × 22 + 3 × 3 + 4 × 1 = 122
Therefore, the parameter, λ, can be estimated from data: ˆ
λ= 122
= 0.61
200 We expect, on average, 0.61 deaths per unit per year NUS/FOS/DSAP 9 GEM2900 Semester 1, 2009/1010 Rare Events & Poisson Distribution
Example: Death by Horse Kick in Prussian Army...
Let X be the number of deaths per unit per year
If X ~ Poisson (0.61)
Poisson
x P(X = x) Expected
unityears Observed
unityears O–E 0 0.543 108.67 109 0.23 1 0.331 66.29 65 –1.29 2 0.101 20.22 22 1.78 3 0.0206 4.11 3 – 1.11 0.00356 0.71 1 0.29 ≥4 NUS/FOS/DSAP 10 GEM2900 Semester 1, 2009/1010 Rare Events & Poisson Distribution
Example: Death by Horse Kick in Prussian Army...
Comparing ‘the expected number of deaths under the
assumption of X~Poisson(0.61)’ with observed number of deaths
It seems the expected number of deaths compares favorably
with the observed number of deaths
How to do the comparisons statistically?
… Just look at the squared deviation
look at
square ∑ (Oi − E i )2
Ei … whether this value is small or suspiciously large NUS/FOS/DSAP 11 GEM2900 Semester 1, 2009/1010 Rare Events & Poisson Distribution
Example: Death by Horse Kick in Prussian Army... χ2 = ∑ (Oi − E i )2
Ei 0.23 2 ( −1.29) 2 1.78 2 ( −1.11) 2 (0.29) 2
=
+
+
+
+
≈ 0.5999
108.67
66.29
20.22
4.11
0.71
… If this value is considered small, then the assumption of X~ Poisson(0.61) is supported by data … If this value is considered suspiciously large, then the assumption of X~ Poisson(0.61) is NOT...
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 Fall '10
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