High Lambda Poisson approaches Normal Distribution

less than 1 minute read

Published: August 31, 2017

In my Generalized Linear Model class, we were tasked with proving that with high $\lambda$, the Poisson distribution approaches the standardized normal distribution. I went ahead and wrote up the proof for the extra credit and really enjoyed doing it. I’ll note that converting the exponential to a series was not my idea and had a little Stack Exchange inspiration. It reminded me how useful series expansions were, so that will hopefully prove useful in future research.

The Theorem

The limiting distribution of the Poisson$(\lambda)$ distribution as $\lambda \rightarrow \infty$ is normal.

The Proof

Let $X \sim Poisson(\lambda)$ which has the probability mass function

$f_{X}(x) = \frac{\lambda^xe^{-\lambda}}{x!} \quad \quad x = 0,1,2,...$

and moment generating function

$\text{M}_X(t) = e^{\lambda(e^t - 1)}$

We will specifically consider the standardized Poisson random variable $X$

$\frac{X-\lambda}{\sqrt{\lambda}}$

which has the Moment Generating Function

$\text{M}_{(X-\lambda)/\sqrt{\lambda}}(t) = \text{E}\left[\exp\left({t*\frac{X-\lambda}{\sqrt{\lambda}}}\right)\right]\\ =\exp(-t\sqrt{\lambda})*\text{E}\left[\exp\left(\frac{tX}{\sqrt{\lambda}}\right)\right]\\ =\exp\left(-t\sqrt{\lambda}+\lambda(e^{t/\sqrt{\lambda}} - 1)\right)$

Now, we take the limit as $\lambda$ approaches $\infty$ and utilize the taylor series expansion

$e^{t/\sqrt{\lambda}} = 1 + t\lambda^{-1/2} + \frac{t^2\lambda^{-1}}{2!} + \frac{t^3\lambda^{-3/2}}{3!} + ... + \frac{t^n\lambda^{-n/2}}{n!}$

Therefore, we have

$\exp\left(-t\sqrt{\lambda}+\lambda(e^{t/\sqrt{\lambda}} - 1)\right)\\ =\exp\left(-t\sqrt{\lambda}+\lambda \left[ 1 + t\lambda^{-1/2} + \frac{t^2\lambda^{-1}}{2!} + \frac{t^3\lambda^{-3/2}}{3!} + ... + \frac{t^n\lambda^{-n/2}}{n!} - 1\right] \right)\\ = \exp\left(-t\sqrt{\lambda} + t\sqrt{\lambda} + \frac{t^2}{2} + \frac{t^3\lambda^{-1/2}}{6} + ... + \frac{t^n\alpha^{-(n-1)/2}}{n!} \right)$

To find the limiting moment generating function, we take the limit of this moment generating function as $\lambda \rightarrow \infty$ which results in

$\exp\left(\frac{t^2}{2}\right)$

which is the moment generating function of N$(0,1)$.

Share on

Twitter Facebook Google+ LinkedIn

Ryan Honea

High Lambda Poisson approaches Normal Distribution

The Theorem

The Proof

Share on

You May Also Enjoy

Upcoming Individual Project

Simulations at a Bar

The Secretary Problem Part Two