Introduction

Aside from the aforementioned Uniform Distribution, there are many types of distributions for continuous random variables.

The Exponential Distribution

Parameter $ \lambda \quad (\lambda > 0) $

$X \sim Exp(\lambda)$ $c.d.f \quad F(x) = 1 - e^{-\lambda x} \quad (x > 0)$ $p.d.f \quad f(x) = \frac{d F(x)}{dx} = -e^{-\lambda x} \cdot -\lambda = \lambda e^{-\lambda x} \quad (x > 0)$ $\mu = \int_{0}^{\infty} xf(x)dx = \lambda \int_{0}^{\infty} xe^{-\lambda x}dx$ $u = x, \quad v = \frac{e^{-\lambda x}}{-\lambda}$ $\int xe^{-\lambda x}dx = uv - \int u'v = \frac{xe^{-\lambda x}}{-\lambda} - \frac{-1}{\lambda} \int e^{-\lambda x} dx$ $w = -\lambda x, \quad \frac{dw}{dx} = -\lambda, \quad dx = \frac{dw}{-\lambda}$ $\int e^{-\lambda x}dx = \int e^{w} \cdot \frac{dw}{-\lambda} = \frac{1}{-\lambda} e^{w} = \frac{1}{-\lambda} e^{-\lambda x}$ $\int xe^{-\lambda x}dx = \frac{xe^{-\lambda x}}{-\lambda} - \frac{-1}{\lambda} \frac{1}{-\lambda} e^{-\lambda x} = -\frac{xe^{-\lambda x}}{\lambda} - \frac{e^{-\lambda x}}{\lambda^2}$ $\mu = \lambda [-\frac{xe^{-\lambda x}}{\lambda} - \frac{e^{-\lambda x}}{\lambda^2}]_0^\infty = \lambda (0 + \frac{1}{\lambda^2}) = \frac{1}{\lambda}$ $\sigma^2 = \int_{0}^{\infty} x^2f(x) dx - \mu^2 = \lambda \int_{0}^{\infty} x^2 e^{-\lambda x} dx - \frac{1}{\lambda^2}$ $u = x^2, \quad v = \frac{e^{-\lambda x}}{-\lambda}$ $\int x^2 e^{-\lambda x} dx = uv - \int u'v = \frac{x^2 e^{-\lambda x}}{-\lambda} - 2\int x \cdot \frac{e^{-\lambda x}}{-\lambda}dx = \frac{-x^2 e^{-\lambda x}}{\lambda} + \frac{2}{\lambda} \int x \cdot e^{-\lambda x}dx$ $\int x^2 e^{-\lambda x} dx = \frac{-x^2 e^{-\lambda x}}{\lambda} + \frac{2}{\lambda} ( -\frac{xe^{-\lambda x}}{\lambda} - \frac{e^{-\lambda x}}{\lambda^2}) = \frac{-x^2 \cdot e^{-\lambda x}}{\lambda} - \frac{2xe^{-\lambda x}}{\lambda^2} -\frac{2 e^{-\lambda x}}{\lambda^3}$ $\sigma^2 = \lambda [\frac{-x^2 \cdot e^{-\lambda x}}{\lambda} - \frac{2xe^{-\lambda x}}{\lambda^2} -\frac{2 e^{-\lambda x}}{\lambda^3}]^{\infty}_{0} - \frac{1}{\lambda^2} = \lambda (0 + \frac{2}{\lambda^3}) - \frac{1}{\lambda^2} = \frac{1}{\lambda^2}$

The Normal Distribution

Parameter $ \mu, \sigma^2 $

$X \sim N(\mu, \sigma^2)$ $p.d.f \quad f(x) = \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{(x - \mu)^2}{2 \sigma^2}}, \quad (-\infty < x < \infty)$ $c.d.f \quad F(x) = \int_{-\infty}^{x} f(x)dx = \int_{-\infty}^{x} \frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{(x - \mu)^2}{2 \sigma^2}} dx = \frac{1}{\sigma \sqrt{2 \pi}} \int_{-\infty}^{x} e^{\frac{- x^2 + 2\mu x - \mu^2}{2 \sigma^2}}$ $F(x) = \frac{1}{\sigma \sqrt{2 \pi}} \cdot \int_{-\infty}^{x} e^{-\frac{1}{2\sigma^2} x^2} \cdot \int_{-\infty}^{x} e^{\frac{\mu}{\sigma^2} x} \cdot e^{\frac{-\mu^2}{2\sigma^2}} = \frac{1}{\sigma \sqrt{2 \pi}} \cdot \frac{\sqrt{\pi}\sigma \cdot erf(\frac{x}{\sqrt{2}\sigma})}{\sqrt{2}} \cdot \frac{\sigma^2 e^{\frac{\mu x}{\sigma^2}}}{\mu} \cdot e^{\frac{-\mu^2}{2 \sigma^2}}$ $F(x) = \frac{\sigma^2 \cdot erf(\frac{x}{\sqrt{2}\sigma}) \cdot e^{\frac{2\mu x - \mu^2}{2\sigma^2}}}{2\mu}$ $E(X) = \mu$ $Var(X) = \sigma^2$

Central Limit Theorem

The sample mean of a large independent random sample (from any distribution) is approximately normally distributed.

$\bar{X} = \frac{1}{n} \Sigma_{i = 1}^{n} X_i$

$ \bar{X} $ is for large $n$, normal distributed with mean $\mu$ and variance $\frac{\sigma^2}{n}$.

$\Sigma_{i = 1}^{n} X_i \sim N(n\mu, n\sigma^2)$

Standard Normal Distribution

Let $ X \sim N(\mu, \sigma^2) $ and let $ Z = \frac{X - \mu}{\sigma} $

$E(Z) = \frac{1}{\sigma}E(X) - \frac{\mu}{\sigma} = 0$ $Var(Z) = \frac{1}{\sigma^2}Var(X) + 0 = 1$

$ Z \sim N(0, 1) $ is standard normal distribution.