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 $\displaystyle Y_{n} := {{ X_{n} - n } \over { \sqrt{n} }}$ are given $$ Y_{n} \overset{D}{\to} N(0,1) $$
- $N \left( \mu , \sigma^{2} \right)$ is a normal distribution with a mean of $\mu$ and a variance of $\sigma^{2}$.
- $\text{Poi} (\lambda)$ is a Poisson distribution with mean and variance of $\lambda$.
Explanation
Considering the approximation of the binomial distribution to the Poisson distribution, it is obvious that the standard normal distribution can also be derived from the Poisson distribution.
Derivation1
The moment generating function of $Y_{n}$, $M_{Y_{n}} (t)$, shows the convergence of the distribution.
Moment generating function of the Poisson distribution: $$ m(t) = \exp \left[ \lambda \left( e^{t} - 1 \right) \right] \qquad , t \in \mathbb{R} $$
Since $X_{n} \sim \text{Poi} (n)$, $$ \begin{align*} M_Y (t) =& E \left[ \text{exp} \left( Y_{n} t \right) \right] \\ =& E \left[ \text{exp} \left( {{ X_{n} - n } \over { \sqrt{n} }} t \right) \right] \\ =& E \left[ \text{exp} \left( {{ X_{n} } \over { \sqrt{n} }} t \right) \text{exp} ( -t \sqrt{n} ) \right] \\ =& \text{exp} ( -t \sqrt{n} ) E \left[ \text{exp} \left( X_{n} {{ t } \over { \sqrt{n} }} \right) \right] \\ =& \text{exp} ( -t \sqrt{n} ) \exp \left( n \left( e^{t/\sqrt{n}} - 1 \right) \right) \end{align*} $$ Through the second argument’s Taylor expansion, $$ \begin{align*} & \text{exp} \left( -t \sqrt{n} + n \left( 1 + {{t} \over {\sqrt{n}}} + {{1} \over {2!}} {{t^2} \over {n}} + {{1} \over {3!}} {{t^3} \over {n \sqrt{n} }} + \cdots - 1 \right) \right) \\ =& \text{exp} \left( -t \sqrt{n} + n \left( {{t} \over {\sqrt{n}}} + {{1} \over {2!}} {{t^2} \over {n}} + {{1} \over {3!}} {{t^3} \over {n \sqrt{n} }} + \cdots \right) \right) \\ =& \text{exp} \left( -t \sqrt{n} + t \sqrt{n} + {{t^2} \over {2!}} + {{1} \over {3!}} {{t^3} \over { \sqrt{n} }} + \cdots \right) \\ =& \text{exp} \left( {{t^2} \over {2!}} + {{1} \over {3!}} {{t^3} \over { \sqrt{n} }} + \cdots \right) \end{align*} $$ Therefore, $$ \lim_{n \to \infty} M_{Y_{n}} = e^{ t^2 \over 2 } $$
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