📂Probability Distribution

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 $$Y_{n} \overset{D}{\to} N(0,1)$$

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• $N \left( \mu , \sigma^{2} \right)$ 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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