Can Binomial Distribution Be Approximated to Poisson Distribution?

Click For Summary
SUMMARY

The discussion focuses on approximating the binomial distribution to the Poisson distribution, specifically through the expression N!/(N-n)!=N^n. The user attempts to simplify the factorial expression using various methods but finds it complicated. They mention the Stirling formula and its potential relevance to the approximation process. Ultimately, they seek clarification on the selective use of approximations and the application of the Stirling formula in this context.

PREREQUISITES
  • Understanding of binomial distribution and its formula: N!/n!(N-n)! p^n q^(N-n)
  • Familiarity with Poisson distribution and its relationship to binomial distribution
  • Knowledge of Stirling's approximation for factorials
  • Basic algebraic manipulation of factorial expressions
NEXT STEPS
  • Study the derivation and application of Stirling's approximation in probability theory
  • Explore the conditions under which the binomial distribution can be approximated by the Poisson distribution
  • Learn about the implications of using approximations in statistical calculations
  • Investigate examples of binomial to Poisson approximations in real-world scenarios
USEFUL FOR

Students and professionals in statistics, mathematicians, and anyone involved in probability theory who seeks to understand the relationship between binomial and Poisson distributions.

rsaad
Messages
76
Reaction score
0

Homework Statement



The question requires me to approximate binomial distribution to get poisson distribution.
Show that N!/(N-n)!=N^n.

Homework Equations



N!/n!(N-n)! p^n q^(N-n)=Binomial distribution



The Attempt at a Solution



I expanded N!/(N-n)! and got: (N-1)(N-2)(N-3)...(N-n+2)(N-n+1). This didn't help me in getting the required approximation. So, then I wrote it as follows:( N-(n-(n-1)) ) ( N-( n- (n-2) ) )...( N-(n-2) ) ( N-(n-1) ).
It seem to have further complicated the question.
A little help please.:redface:
Thank you.
 
Physics news on Phys.org
Look at how the Stirling formula is derived ... it should be quite helpful.
 
I looked at the stirling formula derivation but I don't know how it is helpful here.
So I have solved it the other way.
[N-(n-(n-0))] [N-(n-(n-1))] [N-(n-(n-2))] [N-(n-(n-3))]...[N-( n-(3) )][N-( n-(2) )][N-( n-(1) )]
For N>>n, using this approximation once, I get n terms:
[N-n] [N-n] [N-n] ... [N-n] [N-n] [N-n]
using the approximation again,
[N] [N] [N] ...[N] [N] [N] =N^n

My question is can I use this approximation selectively like I did in the above two steps.
Secondly, how was sterling formula derivation helpful? I used the x! formula and I get exponential(-n). Because n<<N, this term is big, making the entire answer zero.
 

Similar threads

Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
2
Views
2K
  • · Replies 16 ·
Replies
16
Views
4K
Replies
6
Views
3K
Replies
2
Views
2K
Replies
10
Views
2K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K