# Proof of Normal approximation to Poisson.

1. Dec 5, 2009

### Helper

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?

2. May 4, 2011

### DanskFysiker

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 $$\mu$$.