Families of Continuous Random Variables

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INDEX

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Uniform Random Variable

Exponential Random Variable

Erlang Random Variable

Gaussian Random Variable

Bivariate Gaussian Random Variable

Gaussian Random Vector

Beta Random Vector

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Uniform Random Variable

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$X$ is a $Uniform(a, b)$ random variable if the PDF of $X$ is

$$f_{X}(x) = \begin{cases} 1/(b-a) & \quad a\leq x \leq b\\ 0 & \quad \text{otherwise}\\ \end{cases}$$

where the two parameters are $b > a$.

$$F_{X}(x) = \begin{cases} 0 & \quad x \leq a\\ (x - a)/(b - a) & \quad a < x \leq b\\ 1 & \quad x > b\\ \end{cases}$$

$$\text{Expected value} \quad E[X] = (b + a)/2.$$

$$\text{Variance} \quad Var[X] = (b - a)^2/12.$$

PDF of Uniform Distribution (WIKIPEDIA)

CDF of Uniform Distribution (WIKIPEDIA)

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Exponential Random Variable

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$X$ is a $exponential(\lambda)$ random variable if the PDF of $X$ is

$$f_{X}(x) = \begin{cases} \lambda e^{-\lambda x} & \quad x \geq 0\\ 0 & \quad \text{otherwise.}\\ \end{cases}$$

where the parameter $\lambda > 0$.

$$F_{X}(x) = \begin{cases} 1-e^{-\lambda x} & \quad x \geq 0\\ 0 & \quad \text{otherwise.}\\ \end{cases}$$

$$\text{Expected value} \quad E[X] = 1/\lambda.$$

$$\text{Variance} \quad Var[X] = 1/\lambda^2.$$

PDF of Exponential Distribution (WIKIPEDIA)

CDF of Exponential Distribution (WIKIPEDIA)

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Erlang Random Variable

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$X$ is a $Erlang(n, \lambda)$ random variable if the PDF of $X$ is

$$f_{X}(x) = \begin{cases} \frac{\lambda^n x^{n-1} e^{-\lambda x}}{(n - 1)!} & \quad x \geq 0\\ 0 & \quad \text{otherwise.}\\ \end{cases}$$

where the parameter $\lambda > 0$and the parameter $n \geq 1$ is an integer.

$$\text{Expected value} \quad E[X] = \frac{n}{\lambda}.$$

$$\text{Variance} \quad Var[X] = \frac{n}{\lambda^2}.$$

PDF of Erlang Distribution (WIKIPEDIA)

CDF of Erlang Distribution (WIKIPEDIA)

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Gaussian Random Variable

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$X$ is a $Gaussian(\mu, \sigma)$ random variable if the PDF of $X$ is

$$f_X(x) = \frac{1}{\sqrt{2\pi \sigma^2}}e^{-(x - \mu)^2/2\sigma^2}$$

where the parameter $\mu$ can be any real number and the parameter $\sigma > 0$

$$\text{Expected value} \quad E[X] = \mu.$$

$$\text{Variance} \quad Var[X] = \sigma^2.$$

PDF of Normal Distribution (WIKIPEDIA)

CDF of Normal Distribution (WIKIPEDIA)

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Bivariate Gaussian Random Variable

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Random variables $X$ and $Y$ have a bivariate Gaussian PDF with parameters $\mu_1, \sigma_1, \mu_2, \sigma_2,$ and $\rho$ if

$$f_{X, Y}(x, y) = \frac{\exp(-\frac{\Big(\frac{x - \mu_1}{\sigma_1}\Big)^2 - \frac{2\rho (x - \mu_1)(y - \mu_2)}{\sigma_1 \sigma_2} + \Big(\frac{y - \mu_2}{\sigma_2}\Big)^2}{2(1 - \rho^2)})}{2\pi \sigma_1 \sigma_2 \sqrt{1 - \rho^2}}$$

where $\mu_1$ and $\mu_2$ can be any real numbers, $\sigma_1 >0, \sigma_2 >0$, and $-1 < \rho < 1$ .

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Gaussian Random Vector

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$\mathbf{X}$ is the $Gaussian(\mathbf{\mu_X, C_X})$ random vector with expected value $\mathbf{\mu_X}$ and covariance $\mathbf{C_X}$ if and only if

$$f_{\mathbf{X}}(\mathbf{x}) = \frac{1}{(2\pi)^{n/2}[det(\mathbf{C_X})]^{1/2}}\exp\bigg(-\frac{1}{2}(\mathbf{x}-\mathbf{\mu_X})^T \mathbf{C_X^{-1}} (\mathbf{x} - \mathbf{\mu_X}) \bigg)$$

where $det(\mathbf{C_X})$, the determinant of $\mathbf{C_X}$, satisfies $det(\mathbf{C_X}) > 0$.

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Beta Random Variable

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$X$ is a $Beta(\alpha, \beta)$ random variable if the PDF of $X$ is

$$f_{X}(x) = \frac{\Gamma (\alpha + \beta)}{\Gamma (\alpha) \Gamma (\beta)}x^{\alpha - 1}(1 - x)^{\beta - 1} = \frac{1}{B(\alpha, \beta)}x^{\alpha - 1}(1 - x)^{\beta - 1} \quad \quad (\because B(\alpha, \beta) = \frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma (\alpha + \beta)})$$

where the parameter $0 \leq x \leq 1$, and shape parameters $\alpha, \beta > 0$.

$\Gamma(z) = \int_0^{\infty}x^{z - 1}e^{-x}\mathrm{d}x \quad \quad (\Gamma(n) = (n - 1)! , \text{if n is positive integer}).$.

$$\text{Expected value} \quad E[X] = \frac{\alpha}{\alpha + \beta}.$$

$$\text{Variance} \quad Var[X] = \frac{\alpha \beta}{(\alpha + \beta )^2(\alpha + \beta + 1)}.$$

PDF of Beta Distribution (WIKIPEDIA)

CDF of Beta Distribution (WIKIPEDIA)

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Proof - Gaussian

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$X$ is a $Gaussian(\mu, \sigma)$ random variable if the PDF of $X$ is

$$f_X(x) = \frac{1}{\sqrt{2\pi \sigma^2}}e^{-(x - \mu)^2/2\sigma^2}$$

Expected Value

$\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x - \mu)^2/2\sigma^2}\mathrm{d}x\\ $

$\text{Let}\quad \omega = \frac{x - \mu}{\sqrt{2\sigma^2}}, \mathrm{d}\omega = \frac{\mathrm{d}x}{\sqrt{2\sigma^2}}\\ $

$\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\omega^2}\sqrt{2\sigma^2}\mathrm{d}\omega\\ =\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-\omega^2}\mathrm{d}\omega\\ $

$\begin{align} \bigg\{\int_{-\infty}^{\infty}e^{-\omega^2}\mathrm{d}\omega\bigg\}^2 &= \int_{-\infty}^{\infty}e^{-x^2}\mathrm{d}x \cdot \int_{-\infty}^{\infty}e^{-y^2}\mathrm{d}y\\ &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2 + y^2)}\mathrm{d}x\mathrm{d}y\\ \end{align}$

$\text{Let}\quad x = r\cos{\theta}, \quad y = r\sin{\theta}\\ $

$\begin{align} \int_{0}^{2\pi}\int_{0}^{\infty}e^{-r^2}det(J(r,\theta))\mathrm{d}r\mathrm{d}\theta &= \int_{0}^{2\pi}\int_{0}^{\infty}e^{-r^2}r\mathrm{d}r\mathrm{d}\theta\\ &= 2\pi\int_{0}^{\infty}re^{-r^2}r\mathrm{d}r\\ \end{align}$

$\text{Let}\quad s = -r^2, \quad \mathrm{d}s = -2r\mathrm{d}r\\ $

$\begin{align} 2\pi\int_{0}^{-\infty}-\frac{1}{2}e^{s}\mathrm{d}s &= 2\pi\int_{-\infty}^{0}\frac{1}{2}e^{s}\mathrm{d}s\\ &= \pi\int_{-\infty}^{0}e^{s}\mathrm{d}s\\ &=\pi e^s \Big|_{-\infty}^{0}\\&=\pi \end{align}$

$\begin{align} \bigg\{\int_{-\infty}^{\infty}e^{-\omega^2}\mathrm{d}\omega\bigg\}^2 = \pi, \quad \int_{-\infty}^{\infty}e^{-\omega^2}\mathrm{d}\omega = \sqrt{\pi} \end{align}$

$\begin{align} \frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-\omega^2}\mathrm{d}\omega = 1 \end{align}$

$\begin{align} \therefore \int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x - \mu)^2/2\sigma^2}\mathrm{d}x = 1 \end{align}$

$\begin{align} E[X] &= \int_{-\infty}^{\infty}x\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x - \mu)^2/2\sigma^2}\mathrm{d}x \\ &= \int_{-\infty}^{\infty}[x - \mu + \mu]\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x - \mu)^2/2\sigma^2}\mathrm{d}x\\ &= \int_{-\infty}^{\infty}(x - \mu)\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x - \mu)^2/2\sigma^2}\mathrm{d}x + \mu\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x - \mu)^2/2\sigma^2}\mathrm{d}x\\ &= \int_{-\infty}^{\infty}(x - \mu)\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x - \mu)^2/2\sigma^2}\mathrm{d}x + \mu \quad \quad \bigg( \because \int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x - \mu)^2/2\sigma^2}\mathrm{d}x = 1 \bigg) \\ &= \frac{-\sigma^2}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty}-\frac{(x - \mu)}{\sigma^2}e^{-(x - \mu)^2/2\sigma^2}\mathrm{d}x + \mu\\ &= \frac{-\sigma^2}{\sqrt{2\pi\sigma^2}}\bigg[ e^{-(x - \mu)^2/2\sigma^2}\bigg]_{-\infty}^{\infty} + \mu\\ &= \frac{-\sigma^2}{\sqrt{2\pi\sigma^2}}\bigg[ 0 - 0\bigg] + \mu\\ &= \mu \end{align}$

Variance

$Var[X] = \int_{-\infty}^{\infty}(x - \mu)^2\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x - \mu)^2/2\sigma^2}\mathrm{d}x\\ $

$\text{Let}\quad \omega = \frac{x - \mu}{\sqrt{2\sigma^2}}, \mathrm{d}\omega = \frac{\mathrm{d}x}{\sqrt{2\sigma^2}}\\ $

$\begin{align} Var[X] = \int_{-\infty}^{\infty}2\sigma^2 \omega^2 \frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x - \mu)^2/2\sigma^2}\mathrm{d}\omega\\ =\frac{2\sigma^2}{\sqrt{\pi}}\int_{-\infty}^{\infty}\omega^2 e^{-\omega^2}\mathrm{d}\omega\\ \end{align}$

$\text{Intergration by parts by}\quad u = \omega , \quad v = -\frac{1}{2}e^{-\omega^2}, u^{\prime} = 1, v^{\prime} = \omega e^{-\omega^2}\quad \quad \bigg( \int uv^{\prime} = uv - \int u^{\prime}v \bigg)\\ $

$\begin{align} Var[X] &= \frac{2\sigma^2}{\sqrt{\pi}}\bigg\{ \bigg[-\frac{1}{2}\omega e^{-\omega^2}\bigg]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} - \frac{1}{2} e^{- \omega^2} \mathrm{d} \omega \bigg\}\\ &= \frac{2\sigma^2}{\sqrt{\pi}}\bigg\{ \bigg[0 - 0\bigg] - \frac{1}{2}\sqrt{\pi} \bigg\}\\ &= \sigma^2 \end{align}$