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Probability




1. Poisson distribution

If the probability of one success is p; the probabily to get exactly 
n success among N trials is: 

P(n/N) = C(n,N)pnqN-n = [N!/n!(N-n)!] pnqN-n	(1)

N!/(N-n)! = N x (N-1) x (N-2) x ... x (N- n+2) x (N - n+1)	(2)

We define the parent parameter λ as: λ = N x p Where N is the number of trials, and p is the parent probability to get success.
The expression (1) becomes: P(n/N) = [N x (N-1) x (N-2) x ... x (N-n+2) x (N-n+1)/n!] (λ/N)n(1 - λ/N)N-n = [N x (N-1) x (N-2) x ... x (N-n+2) x (N-n+1)/Nn] x [λn/n!] x (1 - λ/N)N x (1 - λ/N)-n (3) Now if N is large, the probability P(n/N) becomes: pλ(n) = lim p(n/N) when n → ∞ pλ(n) = 1 x [λn/n!] x (1 - λ/N)N x 1 (4) Using Taylor Series: ex = 1 + x + x2/2! + x3/3! +... (T1) (1 + x)n = 1 + nx + n(n - 1)/2! x2 + ... (T2) we have respectively: e- λ = 1 - λ + λ2/2! + ... (T11) (1 - λ/N)N = 1 - λ + N(N - 1)/2!N2 λ2 + ... (T22) When N → ∞ N(N - 1)/N2 → 1 ; the relationship (T22) becomes: (1 - λ/N)N = 1 - λ + λ2/2! + ... That is the relationship (T11) Thus, (1 - λ/N)N → e- λ Finally, the relationship (4) becomes: pλ(n) = [λn/n!] x e- λ
pλ(n) = λn e- λ/n!
More generally,
If we define the parent parameter λ as: λ = N x p; where N is a large number of trials, and p is the parent probability to get one success (one event E); the probability of observing X = x; where X is the random variable, a disctete value that describes the number of times a given outcome occurs is:
p(X = x) = λx e- λ/x!
The Poisson distribution is a discrete distribution that characterizes the occurence of rare events. It is associated with counting experiments; such as radioactive decay.
The parent probability p is associated with one (success) outcome E, we can then write p = E(X)/N, where E(X) is the expected value for the random variable X related to the event E. The relationship λ = N x p becomes: λ = E(X)
Using again the Taylor seriess: ex = 1 + x + x2/2! + x3/3! +... = ∑ xn/n! (T1) we sum the formula pλ(n) = λn e- λ/n! to obtain: ∑pλ(n) = ∑λn e- λ/n! = e- λ ∑λn/n! = e- λ eλ = 1

2. Poisson distribution Expectation value and variance

2.1. Expectation value


The expaectation value or mean for the Poisson distribution is:
E(x) = μ = ∑ x Po(x) [x: 0 → +∞]
= ∑ x μxe- μ/x! 	[x: 0 → +& infin;] 
= e- μ ∑ μx/(x - 1)! 	[x: 1 →+ ∞]
= e- μ ∑ μx+1/x! 	[x: 0 → + ∞]
= e- μ μ ∑ μx/x! 	[x: 0 → + ∞]

According to the expansion: 
eζ = ∑ ζx/x! 	[x: 0 → + ∞]
It follows:

E(x) = e- μ μ eμ = μ 

Poisson distribution: Po(x) = μxe- μ/x! Expectation = E(x) = μ

2.2. Variance

The variance of the Poisson distribution is:

σ2 = ∑ (x - μ)2 Po(x) [x: - ∞ → + ∞]
or 
σ2 = ∑ (x - μ)2 Po(x) [x: 0 → + n] 
for a large number n of measurements of x.
We have:
σ2 = ∑ x2 Po(x) - 2 μ ∑ x Po(x) +  ∑ x2 Po(x)
= ∑ x2 Po(x) - 2 μ2 + ∑2 
= ∑ x2 Po(x) -  μ2 
with x2 = x(x - 1) + x,
σ2  =  ∑ (x(x - 1) + x) Po(x) -  μ2 
=  ∑ (x(x - 1)) Po(x) + μ -  μ2 
=  ∑  μxe- μ/(x - 2)!) + μ -  μ2 
=  μ2 ∑  μx - 2e- μ/(x - 2)!) + μ -  μ2
=  μ2 ∑  μx - 2e- μ/(x - 2)!) + μ -  μ2 [x: 2 → + n] 

According to the expansion: 
eζ = ∑ ζx/x! 	[x: 0 → + ∞]
It follows:
∑  μx - 2/(x - 2)!) [x: 2 → + n]  
= ∑ μx /x! = 	eμ	[x: 0 → + ∞]

Then:
σ2 = μ2 e e + μ -  μ2 = μ

Poisson distribution: Po(x) = μxe- μ/x! Variance = σ2 = μ


  
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