Proof of Normal approximation to Poisson.

Click For Summary
SUMMARY

The discussion focuses on the proof that the Poisson distribution approximates a Normal distribution for large parameter lambda. The proof utilizes Stirling's approximation, specifically n! ~ sqrt(2*pi*n) * (n/e)^n, and involves taking the natural logarithm of the Poisson distribution. By defining a new variable y = x - μ and applying the Maclaurin series to ln(1 + y/μ), the proof simplifies to demonstrate that the resulting distribution is Normal with mean and variance μ.

PREREQUISITES
  • Understanding of Poisson distribution and its properties
  • Familiarity with Normal distribution characteristics
  • Knowledge of Stirling's approximation
  • Basic calculus, specifically Taylor and Maclaurin series
NEXT STEPS
  • Study the application of Stirling's approximation in probability theory
  • Learn about the Central Limit Theorem and its implications for distributions
  • Explore the derivation of the Normal distribution from the Poisson distribution
  • Investigate advanced topics in asymptotic analysis in statistics
USEFUL FOR

Statisticians, mathematicians, and students studying probability theory who seek to understand the relationship between Poisson and Normal distributions.

Helper
Messages
1
Reaction score
0
I have been looking for a proof of the fact that for a large parameter lambda, the Poisson distribution tends to a Normal distribution. I know the classic proof using the Central Limit Theorem, but I need a simpler one using just limits and the corresponding probability density functions. I was told this was really easy using Stirling's approximation:

n! ~ sqrt(2*pi*n) * (n/e)^n

but I just don't see it. Anyone knows this proof?
 
Physics news on Phys.org
First you take the natural logarithm to the Poisson distribution and then apply Stirlings approximation. Then define a new variable

y=x-\mu

and assume that y is much smaller than \mu
By doing this you will end up with a term

\ln\left(1+\frac{y}{\mu}\right)

which can be approximated by looking at the Maclaurin series

\ln\left(1+\frac{y}{\mu}\right) \approx \frac{y}{\mu} - \frac{y^{2}}{2\mu^{2}}.

Now any term with a power of \mu greater than 2 in the denominator may be approximated as zero due to the assumption that y is much smaller than \mu. When the algebra is done you just takt the exponential function on both sides and you end up with a normal distribution with mean and variance \mu.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
5K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K