Probability & statistics

Gaussian distribution or normal distribution can be 
derived from Poisson distribution with large mean value μ

 We have in this case: σ2 = μ

pμ(x) = μx e- μ/x!

Let's take the ln:
ln pμ(x) = x ln μ - μ - ln x!
Using Stirling's formula: ln x! ≈ (x + 1/2) ln x  - x + (1/2) ln (2π), 
we get:
ln pμ(x) = x ln μ - μ - (x + 1/2) ln x  + x - (1/2) ln (2π)

μ is large, we write x = μ + δ with δ << &mu:. 
We have: x/μ = 1 + δ/μ
And, by using the expansion:
ln (1 + ξ) = ξ - ξ2/2 + ξ3/3 - ... + , we get:

ln x  = ln (μ + δ) = ln μ (1 + δ/μ) 
≈ ln μ + δ/μ - (1/2) (δ/μ)2

Hence:
ln pμ(δ) = (μ + δ) ln μ - μ - (μ + δ + 1/2) [ln μ + δ/μ - (1/2) (&delta/μ)2] 
+ (μ + δ) - (1/2) ln (2π)

We neglect the terms:  - δ/2μ,  (δ)2/4μ2, and 
(δ)3/2μ2; the dominant term is:
- (δ)2/2μ

Then:
ln pμ(δ) = - (δ)2/2μ  - (1/2) ln(2πμ)

Hence:

pμ(δ) = exp {- (δ)2/2μ - (1/2) ln(2πμ)} = 
exp {- (δ)2/2μ } x (2πμ)- 1/2
= (1/2πμ)1/2 exp {- (δ)2/2μ } 

pμ(x) = (1/2πμ)1/2 exp {- (x - μ)2/2μ }