Derivation of the Standard Normal Distribution as the Limiting Distribution of the Poisson Distribution
Theorem
If $X_{n} \sim \text{Poi} \left( n \right)$ and $\displaystyle Y_{n} := {{ X_{n} - n } \over { \sqrt{n} }}$, then $$ Y_{n} \overset{D}{\to} N(0,1) $$
- $N \left( \mu , \sigma^{2} \right)$ is the normal distribution with mean $\mu$ and variance $\sigma^{2}$.
- $\text{Poi} (\lambda)$ is the Poisson distribution with mean and variance $\lambda$.
Explanation
Considering the Poisson approximation of the binomial distribution, it is only natural that the standard normal distribution can also be derived from the Poisson distribution.
Derivation1
We show convergence in distribution via the moment generating function $M_{Y_{n}} (t)$ of $Y_{n}$.
Moment generating function of the Poisson distribution: $$ m(t) = \exp \left[ \lambda \left( e^{t} - 1 \right) \right] \qquad , t \in \mathbb{R} $$
Since $X_{n} \sim \text{Poi} (n)$, $$ \begin{align*} M_Y (t) =& E \left[ \text{exp} \left( Y_{n} t \right) \right] \\ =& E \left[ \text{exp} \left( {{ X_{n} - n } \over { \sqrt{n} }} t \right) \right] \\ =& E \left[ \text{exp} \left( {{ X_{n} } \over { \sqrt{n} }} t \right) \text{exp} ( -t \sqrt{n} ) \right] \\ =& \text{exp} ( -t \sqrt{n} ) E \left[ \text{exp} \left( X_{n} {{ t } \over { \sqrt{n} }} \right) \right] \\ =& \text{exp} ( -t \sqrt{n} ) \exp \left( n \left( e^{t/\sqrt{n}} - 1 \right) \right) \end{align*} $$ By the Taylor expansion of the second factor, $$ \begin{align*} & \text{exp} \left( -t \sqrt{n} + n \left( 1 + {{t} \over {\sqrt{n}}} + {{1} \over {2!}} {{t^2} \over {n}} + {{1} \over {3!}} {{t^3} \over {n \sqrt{n} }} + \cdots - 1 \right) \right) \\ =& \text{exp} \left( -t \sqrt{n} + n \left( {{t} \over {\sqrt{n}}} + {{1} \over {2!}} {{t^2} \over {n}} + {{1} \over {3!}} {{t^3} \over {n \sqrt{n} }} + \cdots \right) \right) \\ =& \text{exp} \left( -t \sqrt{n} + t \sqrt{n} + {{t^2} \over {2!}} + {{1} \over {3!}} {{t^3} \over { \sqrt{n} }} + \cdots \right) \\ =& \text{exp} \left( {{t^2} \over {2!}} + {{1} \over {3!}} {{t^3} \over { \sqrt{n} }} + \cdots \right) \end{align*} $$ Therefore $$ \lim_{n \to \infty} M_{Y_{n}} = e^{ t^2 \over 2 } $$
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