Proof of Normal approximation to Poisson.

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 9K views
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

[tex]y=x-\mu[/tex]

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

[tex]\ln\left(1+\frac{y}{\mu}\right)[/tex]

which can be approximated by looking at the Maclaurin series

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

Now any term with a power of [tex]\mu[/tex] greater than 2 in the denominator may be approximated as zero due to the assumption that y is much smaller than [tex]\mu[/tex]. 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 [tex]\mu[/tex].