Poisson Distribution (Continued)

As stated in the previous post,

Let

$\lambda = n \cdot p$ $Binomial(n, p) \approx Poisson(n p) = Poisson(\lambda)$

For Binomial Distribution,

$P(x) = C_n^x p^x q^{n-x} = C_n^x \cdot (\frac{\lambda}{n})^x \cdot (1 - \frac{\lambda}{n})^{n -x}$

Thus,

$\lim_{n \rightarrow \infty} P(x) = \lim_{n \rightarrow \infty} \frac{n \cdot (n - 1) ... \cdot (n - x + 1)}{x!} \cdot \frac{\lambda^x}{n^x} \cdot (1 - \frac{\lambda}{n})^n \cdot (1 - \frac{\lambda}{n})^{-x}$ $= \lim_{n \rightarrow \infty} \frac{n \cdot (n - 1) ... \cdot (n - x+ 1)}{n^x} \cdot \frac{\lambda^x}{x !} \cdot \lim_{n \rightarrow \infty} (1 - \frac{\lambda}{n})^n (1 - \frac{\lambda}{n})^{-x}$ $= 1 \cdot \frac{\lambda^x}{x!} \cdot e^{-\lambda} \cdot 1 = e^{-\lambda} \cdot \frac{\lambda^k}{k!}$

The positive real number $\lambda$ is equal to the expected value and also its variance.

$\lambda = \mu = \sigma^2$

Gaussian Distribution (Normal Distribution)

Probability Density Function

Given $\mu = E(X)$ and $\sigma^2 = Var(X)$ ,

$f(x \mid \mu, \sigma^2) = \frac{1}{\sqrt{2\pi\sigma^2}} \; e^{-\frac{(x-\mu)^2}{2\sigma^2}}$

A standard normal distribution follows $\mu = 1$ and $\sigma^2 = 1$

Cumulative Distribution Function

Cumulative Distribution Function, or $F(x)$ is the probability that $X \leq x$ .

$F(x \mid \mu, \sigma^2) = \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{x} e^{-\frac{(t-\mu)^2}{2\sigma^2}}dt$