Poison distribution

1




Risk theory




Juma Clinton




SCHOOL OF MATHEMATICS AND ACTUARIAL SCIENCE.




JARAMOGI OGINGA ODINGA UNIVERSITY OF SCIENCE AND TECHNOLOGY


Course code. WAB 2208.




April ,24 th 2024.

, 2


A random variable has a poison distribution.
a) Derive the pdf of x.
Consider the occurrence or nonoccurrence of a change subinterval as a Bernoulli trial.
λ
So we have a sequence of n Bernoulli trials with probability p=n
λ
And q=(1 − n)

The binomial distribution tends towards the Poisson distribution as;

𝑛 → ∞ , 𝑃 → 0, 𝜆 = 𝑛𝑝 stays constant. This implies that;

𝑛 𝜆𝑥 ⅇ −𝜆
( ) 𝑃 𝑥 (1 − 𝑝)𝑛−𝑥
𝑥 𝑥!


𝑤𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 ;

𝑥 𝑛
𝑒 𝑥 = 𝑙𝑖𝑚 (1 + )
𝑛→𝑥 𝑛
𝑛 𝜆
( ) 𝑃 𝑥 (1 − 𝑝)𝑛−𝑥 where 𝜆 = 𝑛𝑝, 𝑃 = 𝑛
𝑥
𝑥
𝑛 𝜆 𝜆 𝑛−𝑥
( ) ( ) (1 − )
𝑥 𝑛 𝑛

𝑏𝑢𝑡:

𝑛 𝑛!
( ) = 𝑥(𝑛−𝑥)!
𝑥

𝑛! 𝜆 𝑥 𝜆 𝑛−𝑥
( ) (1 − 𝑛)
𝑥(𝑛−𝑥)! 𝑛


𝜆𝑥 𝑛! 1 𝜆 𝑛−𝑥
(1 − 𝑛)
𝑥(𝑛−𝑥)! 𝑛𝑥


𝜆𝑥 𝑛(𝑛−ⅈ)(𝑛−2)⋯(𝑛−𝑥+1)(𝑛−𝑥)! 𝜆 𝑛−𝑥
(𝑛−𝑥)!𝑛𝑥
(1 − 𝑛)
𝑥!


𝜆𝑥 1 2 𝑥+1 𝜆 𝑛 𝜆 −𝑥
(1 − 𝑛) (1 − 𝑛) … (1 − ) (1 − 𝑛) (1 − 𝑛)
𝑥! 𝑛

𝜆𝑥 1 2 𝑥+1 𝜆 𝑛 𝜆 −𝑥
𝑙𝑖𝑚 (1 − 𝑛) (1 − 𝑛) … (1 − ) X 𝑙𝑖𝑚 (1 − 𝑛) X 𝑙𝑖𝑚 (1 − 𝑛)
𝑥! 𝑛→∞ 𝑛 𝑛→∞ 𝑛→∞