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INDEX
Bivariate Gaussian Random Variable
Uniform Random Variable
\(X\) 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 \(b > a\).
$$F_{X}(x) = \begin{cases} 0 & \quad x \leq a\\ (x - a)/(b - a) & \quad a < x \leq b\\ 1 & \quad x > b\\ \end{cases} $$
$$\text{Expected value} \quad E[X] = (b + a)/2.$$
$$\text{Variance} \quad Var[X] = (b - a)^2/12.$$
Exponential Random Variable
\(X\) 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 \(\lambda > 0\).
$$F_{X}(x) = \begin{cases} 1-e^{-\lambda x} & \quad x \geq 0\\ 0 & \quad \text{otherwise.}\\ \end{cases} $$
$$\text{Expected value} \quad E[X] = 1/\lambda.$$
$$\text{Variance} \quad Var[X] = 1/\lambda^2.$$
Erlang Random Variable
\(X\) 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.
$$\text{Expected value} \quad E[X] = \frac{n}{\lambda}.$$
$$\text{Variance} \quad Var[X] = \frac{n}{\lambda^2}.$$
Gaussian Random Variable
\(X\) 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 \(\sigma > 0\)
$$\text{Expected value} \quad E[X] = \mu.$$
$$\text{Variance} \quad Var[X] = \sigma^2.$$
Bivariate Gaussian Random Variable
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 \( -1 < \rho < 1\) .
Gaussian Random Vector
\(\mathbf{X}\) 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 \(det(\mathbf{C_X}) > 0\).
Beta Random Variable
\(X\) 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 \(\alpha, \beta > 0\).
\(\Gamma(z) = \int_0^{\infty}x^{z - 1}e^{-x}\mathrm{d}x \quad \quad (\Gamma(n) = (n - 1)! , \text{if n is positive integer}).\).
$$\text{Expected value} \quad E[X] = \frac{\alpha}{\alpha + \beta}.$$
$$\text{Variance} \quad Var[X] = \frac{\alpha \beta}{(\alpha + \beta )^2(\alpha + \beta + 1)}.$$
Proof - Gaussian
\(X\) 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
\( Var[X] = \int_{-\infty}^{\infty}(x - \mu)^2\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x - \mu)^2/2\sigma^2}\mathrm{d}x\\ \)
\( \text{Let}\quad \omega = \frac{x - \mu}{\sqrt{2\sigma^2}}, \mathrm{d}\omega = \frac{\mathrm{d}x}{\sqrt{2\sigma^2}}\\ \)
\( \begin{align} Var[X] = \int_{-\infty}^{\infty}2\sigma^2 \omega^2 \frac{1}{\sqrt{2\pi\sigma^2}}e^{-(x - \mu)^2/2\sigma^2}\mathrm{d}\omega\\ =\frac{2\sigma^2}{\sqrt{\pi}}\int_{-\infty}^{\infty}\omega^2 e^{-\omega^2}\mathrm{d}\omega\\ \end{align} \)
\( \text{Intergration by parts by}\quad u = \omega , \quad v = -\frac{1}{2}e^{-\omega^2}, u^{\prime} = 1, v^{\prime} = \omega e^{-\omega^2}\quad \quad \bigg( \int uv^{\prime} = uv - \int u^{\prime}v \bigg)\\ \)
\( \begin{align} Var[X] &= \frac{2\sigma^2}{\sqrt{\pi}}\bigg\{ \bigg[-\frac{1}{2}\omega e^{-\omega^2}\bigg]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} - \frac{1}{2} e^{- \omega^2} \mathrm{d} \omega \bigg\}\\ &= \frac{2\sigma^2}{\sqrt{\pi}}\bigg\{ \bigg[0 - 0\bigg] - \frac{1}{2}\sqrt{\pi} \bigg\}\\ &= \sigma^2 \end{align} \)
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