In time and space suppose that we are dealing with

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Unformatted text preview: npredictable pp 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 cavalry-unit-years Number of deaths per unit per year Number of unit-years 0 1 2 3 4 >4 109 65 22 3 1 0 Bortkiewicz showed that the number of deaths per cavalry-unityear 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 unit-years Observed unit-years 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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