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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
fX(x)={1/(b−a)a≤x≤b0otherwise
where the two parameters are b>a.
FX(x)={0x≤a(x−a)/(b−a)a<x≤b1x>b
Expected valueE[X]=(b+a)/2.
VarianceVar[X]=(b−a)2/12.
Exponential Random Variable
X is a exponential(λ) random variable if the PDF of X is
fX(x)={λe−λxx≥00otherwise.
where the parameter λ>0.
FX(x)={1−e−λxx≥00otherwise.
Expected valueE[X]=1/λ.
VarianceVar[X]=1/λ2.
Erlang Random Variable
X is a Erlang(n,λ) random variable if the PDF of X is
fX(x)={λnxn−1e−λx(n−1)!x≥00otherwise.
where the parameter λ>0and the parameter n≥1 is an integer.
Expected valueE[X]=nλ.
VarianceVar[X]=nλ2.
Gaussian Random Variable
X is a Gaussian(μ,σ) random variable if the PDF of X is
fX(x)=1√2πσ2e−(x−μ)2/2σ2
where the parameter μ can be any real number and the parameter σ>0
Expected valueE[X]=μ.
VarianceVar[X]=σ2.
Bivariate Gaussian Random Variable
Random variables X and Y have a bivariate Gaussian PDF with parameters μ1,σ1,μ2,σ2, and ρ if
fX,Y(x,y)=exp(−(x−μ1σ1)2−2ρ(x−μ1)(y−μ2)σ1σ2+(y−μ2σ2)22(1−ρ2))2πσ1σ2√1−ρ2
where μ1 and μ2 can be any real numbers, σ1>0,σ2>0, and −1<ρ<1 .
Gaussian Random Vector
X is the Gaussian(μX,CX) random vector with expected value μX and covariance CX if and only if
fX(x)=1(2π)n/2[det(CX)]1/2exp(−12(x−μX)TC−1X(x−μX))
where det(CX), the determinant of CX, satisfies det(CX)>0.
Beta Random Variable
X is a Beta(α,β) random variable if the PDF of X is
fX(x)=Γ(α+β)Γ(α)Γ(β)xα−1(1−x)β−1=1B(α,β)xα−1(1−x)β−1(∵
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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