Latest revision as of 11:27, 30 November 2010

1.6 Continuous Random Variables

From the ECE600 Pre-requisites notes of Sangchun Han, ECE PhD student.

1.6.1 Gaussian distribution (normal distribution) $ \mathcal{N}\left(\mu,\sigma^{2}\right) $

$ f_{\mathbf{X}}(x)=\frac{1}{\sqrt{2\pi}\sigma}\cdot e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}} $

$ \Phi_{\mathbf{X}}\left(\omega\right)=e^{i\mu\omega}e^{-\frac{1}{2}\sigma^{2}\omega^{2}} $

$ \phi_{\mathbf{X}}\left(s\right)=e^{\mu s}e^{\frac{1}{2}\sigma^{2}s^{2}} $

$ E\left[\mathbf{X}\right]=\mu $

$ Var\left[\mathbf{X}\right]=\sigma^{2} $

1.6.2 Log-normal distribution $ \ln\mathcal{N}\left(\mu,\sigma^{2}\right) $

For log-normal distribution, its logaritm is normally distributed. If \mathbf{X} is a random variable with log-normal distribution, then $ \mathbf{Y}=\ln\mathbf{X} $ is a random variable with Gaussian distribution. This distribution is characterized with two parameters; $ \mu $ and $ \sigma $ (that are mean and standard deviation of $ \mathbf{Y} $ rather than $ \mathbf{X} $ ).

$ f_{\mathbf{X}}\left(x\right)=\frac{1}{x\sqrt{2\pi}\sigma}\cdot e^{-\frac{(\ln x-\mu)^{2}}{2\sigma^{2}}} $.

$ E\left[\mathbf{X}\right]=e^{\mu+\sigma^{2}/2} $.

$ Var\left[\mathbf{X}\right]=\left(e^{\sigma^{2}}-1\right)e^{2\mu+\sigma^{2}} $.

MLE of log-normal distribution

$ \hat{\mu}\frac{\sum_{k}\ln x_{k}}{n} $.

$ \hat{\sigma}^{2}=\frac{\sum_{k}\left(\ln x_{k}-\hat{\mu}\right)^{2}}{n} $.

1.6.3 Exponential distribution with mean $ \mu $

It is about the time until success in Poisson process. It has the characteristic of memoryless.

$ F_{\mathbf{X}}\left(x\right)=\left(1-e^{-\frac{x}{\mu}}\right)\cdot\mathbf{1}_{\left[0,\infty\right)}\left(x\right) $.

$ f_{\mathbf{X}}\left(x\right)=\frac{1}{\mu}e^{-\frac{x}{\mu}}\cdot\mathbf{1}_{\left[0,\infty\right)}\left(x\right) $.

$ \Phi_{\mathbf{X}}\left(\omega\right)=\int_{0}^{\infty}\frac{1}{\mu}e^{-\frac{x}{\mu}}\cdot e^{i\omega x}dx=\frac{1}{\mu}\int_{0}^{\infty}e^{x\left(i\omega-\frac{1}{\mu}\right)}dx=\frac{1}{\mu}\cdot\frac{e^{x\left(i\omega-\frac{1}{\mu}\right)}}{i\omega-\frac{1}{\mu}}\Biggl|_{0}^{\infty}=\frac{\frac{1}{\mu}}{\frac{1}{\mu}-i\omega}=\frac{1}{1-i\mu\omega} $.

$ E\left[\mathbf{X}\right]=\mu $.

$ Var\left[\mathbf{X}\right]=\mu^{2} $.

Consider Poisson process

$ f_{\mathbf{X}}\left(t\right)dt $ In fact, $ \lambda=1/\mu $ .

Moment generating function

$ \phi_{\mathbf{X}}\left(s\right)=E\left[e^{s\mathbf{X}}\right]=\int_{0}^{\infty}e^{sx}\cdot\lambda e^{-\lambda x}dx=\lambda\int_{0}^{\infty}e^{x\left(s-\lambda\right)}dx=\frac{\lambda}{s-\lambda}e^{x\left(s-\lambda\right)}\biggl|_{0}^{\infty}=\frac{\lambda}{\lambda-s}=\frac{1}{1-s\cdot\mu} $.

1.6.4 Erlang distribution

If $ \mathbf{X}_{1},\mathbf{X}_{2},\cdots $ are i.i.d. exponential random variables, then Erlang random variable is defined as $ \mathbf{Z}_{m}=\mathbf{X}_{1}+\mathbf{X}_{2}+\cdots+\mathbf{X}_{m} $. This is about the the time until m -th success. $ f_{\mathbf{Z}_{m}}\left(t\right)=\frac{\left(\lambda t\right)^{m-1}e^{-\lambda t}}{\left(m-1\right)!}\cdot\lambda,\quad t\geq0 $.

$ E\left[\mathbf{Z}_{m}\right]=E\left[\mathbf{X}_{1}+\mathbf{X}_{2}+\cdots+\mathbf{X}_{m}\right]=\sum_{k=1}^{m}E\left[\mathbf{X}_{k}\right]=m\cdot\frac{1}{\lambda}=\frac{m}{\lambda} $.

$ Var\left[\mathbf{Z}_{m}\right]=Var\left[\mathbf{X}_{1}+\mathbf{X}_{2}+\cdots+\mathbf{X}_{m}\right]=\sum_{k=1}^{m}Var\left[\mathbf{X}_{k}\right]=m\cdot\frac{1}{\lambda^{2}}=\frac{m}{\lambda^{2}} $. There are one important relationship that is useful to solve several problems.$ P\left(\mathbf{Z}_{m}\leq t\right)=P\left(\mathbf{N}\left(t\right)\geq m\right) $.

Moment generating function

$ \phi_{\mathbf{Z}_{m}}\left(s\right)=E\left[e^{s\mathbf{Z}_{m}}\right]=E\left[e^{s\left(\mathbf{X}_{1}+\mathbf{X}_{2}+\cdots+\mathbf{X}_{m}\right)}\right]=\left(\phi_{\mathbf{X}}\left(s\right)\right)^{m}=\left(\frac{\lambda}{\lambda-s}\right)^{m} $.

Gamma distribution

The $ m $ is the natural number in Erlang distribution. If $ m $ is the positive real number, then it becomes Gamma distribution. But we also change $ \left(m-1\right) $! into Gamma function $ \Gamma\left(m\right)=\int_{0}^{\infty}e^{x}x^{m-1}dx $ .

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