基本命令、蒲丰投针与随机数生成.ppt

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基本命令、蒲丰投针与随机数生成

Proof: Exponential Distribution Density function Distribution Function Inverse of the distribution function Generate exponential distribution If u is uniformly distributed on (0,1) Then Is exponentially distributed with parameter lambda Easy way to go When u is uniformly distributed 1-u is also another uniformly distributed random variable on interval (0,1) So Is also the exponential distribution we want Program to generate exponential distribution clear set obs 10000 gen exp = -log(uniform()) histogram x, bin(50) norm Generate normal distribution from exponential Suppose (x,y) are i.i.d. normal Polar coordinate transformation Jacobian of the polar coordinate transformation Joint Density function The inverse of the Jacobian Uniform (0, 2Pi) Exponential distribution Box-Muller transformation Program to generate normal distribution clear set obs 10000 gen u1 = uniform() gen u2 = uniform() gen x= sqrt(-2*log(u1))*cos(2*_pi*u2) gen y= sqrt(-2*log(u1))*sin(2*_pi*u2) histogram x, bin(50) norm histogram y, bin(50) norm Box-Muller is inefficient Use the Box-Muller transformation to generate normal distribution is computationally inefficient because of the sine and cosine trigonometric functions Avoid trigonometric v1 v2 R 1 -1 1 -1 Inefficient? Let v1 , v2 i.i.d.~ u(-1, 1) and v1^2+v2^2=1 Let It is easy to proof that Avoid trigonometric Program clear set obs 10000 gen u = uniform() gen v1 = uniform()*2-1 gen v2 = uniform()*2-1 drop if v1^2+v2^2=1 gen x= sqrt(-2*log(u))*v1/sqrt(v1^2+v2^2) gen y= sqrt(-2*log(u))*v2/sqrt(v1^2+v2^2) Normal distribution As computers are getting faster and faster, we do not use the above polar method to generate normal distributions in practice We use the simple inverse function table saved in computer gen x=invnorm(uniform()) Chi-2 distribution Sum of n i. i.d. standard normal distribution clear set obs 10000 forval i = 1(1) 10 { gen x`i=invnorm(uniform()) gen x`i_sq=x`i^2 } keep *_sq egen chi_sq = rowto

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