蒙特卡洛法—非均匀随机数的产生

流过昼夜 提交于 2020-02-06 13:57:08

1.反变换法

设需产生分布函数为F(x)的连续随机数X。若已有[0,1]区间均匀分布随机数R,则产生X的反变换公式为:

F(x)=r, 即x=F-1(r)

反函数存在条件:如果函数y=f(x)是定义域D上的单调函数,那么f(x)一定有反函数存在,且反函数一定是单调的。分布函数F(x)为是一个单调递增函数,所以其反函数存在。从直观意义上理解,因为r一一对应着x,而在[0,1]均匀分布随机数R≤r的概率P(R≤r)=r 因此,连续随机数X≤x的概率P(X≤x)=P(R≤r)=r=F(x)

即X的分布函数为F(x)。

例子:下面的代码使用反变换法在区间[0, 6]上生成随机数,其概率密度近似为P(x)=e-x  

 1 import numpy as np
 2 import matplotlib.pyplot as plt
 3 
 4 # probability distribution we're trying to calculate
 5 p = lambda x: np.exp(-x)
 6 
 7 # CDF of p
 8 CDF = lambda x: 1-np.exp(-x)
 9 
10 # invert the CDF
11 invCDF = lambda x: -np.log(1-x)
12 
13 # domain limits
14 xmin = 0 # the lower limit of our domain
15 xmax = 6 # the upper limit of our domain
16 
17 # range limits
18 rmin = CDF(xmin)
19 rmax = CDF(xmax)
20 
21 N = 10000 # the total of samples we wish to generate
22 
23 # generate uniform samples in our range then invert the CDF
24 # to get samples of our target distribution
25 R = np.random.uniform(rmin, rmax, N)
26 X = invCDF(R)
27 
28 # get the histogram info
29 hinfo = np.histogram(X,100)
30 
31 # plot the histogram
32 plt.hist(X,bins=100, label=u'Samples');
33 
34 # plot our (normalized) function
35 xvals=np.linspace(xmin, xmax, 1000)
36 plt.plot(xvals, hinfo[0][0]*p(xvals), 'r', label=u'p(x)')
37 
38 # turn on the legend
39 plt.legend()
40 plt.show()

 一般来说,直方图的外廓曲线接近于总体X的概率密度曲线。

 2.舍选抽样法(Rejection Methold)

用反变换法生成随机数时,如果求不出F-1(x)的解析形式或者F(x)就没有解析形式,则可以用F-1(x)的近似公式代替。但是由于反函数计算量较大,有时也是很不适宜的。另一种方法是由Von Neumann提出的舍选抽样法。下图中曲线w(x)为概率密度函数,按该密度函数产生随机数的方法如下:

基本的rejection methold步骤如下:

1. Draw x uniformly from [xmin  xmax]

2. Draw x uniformly from [0, ymax]

3. if y < w(x),accept the sample, otherwise reject it

4. repeat

即落在曲线w(x)和X轴所围成区域内的点接受,落在该区域外的点舍弃。

例子:下面的代码使用basic rejection sampling methold在区间[0, 10]上生成随机数,其概率密度近似为P(x)=e-x  

 1 # -*- coding: utf-8 -*-
 2 '''
 3 The following code produces samples that follow the distribution P(x)=e^−x  
 4 for x=[0, 10] and generates a histogram of the sampled distribution.
 5 '''
 6 import numpy as np
 7 import matplotlib.pyplot as plt
 8 
 9 
10 P = lambda x: np.exp(-x)
11 
12 # domain limits
13 xmin = 0  # the lower limit of our domain
14 xmax = 10 # the upper limit of our domain
15 
16 # range limit (supremum) for y
17 ymax = 1
18 
19 N = 10000    # the total of samples we wish to generate
20 accepted = 0 # the number of accepted samples
21 samples = np.zeros(N)
22 count = 0    # the total count of proposals
23 
24 # generation loop
25 while (accepted < N):
26     
27     # pick a uniform number on [xmin, xmax) (e.g. 0...10)
28     x = np.random.uniform(xmin, xmax)
29     
30     # pick a uniform number on [0, ymax)
31     y = np.random.uniform(0,ymax)
32     
33     # Do the accept/reject comparison
34     if y < P(x):
35         samples[accepted] = x
36         accepted += 1
37     
38     count +=1
39     
40 print count, accepted
41 
42 # get the histogram info
43 # If bins is an int, it defines the number of equal-width bins in the given range 
44 (n, bins)= np.histogram(samples, bins=30) # Returns: n-The values of the histogram,n是直方图中柱子的高度
45 
46 # plot the histogram
47 plt.hist(samples,bins=30,label=u'Samples')   # bins=30即直方图中有30根柱子
48 
49 # plot our (normalized) function
50 xvals=np.linspace(xmin, xmax, 1000)
51 plt.plot(xvals, n[0]*P(xvals), 'r', label=u'P(x)')
52 
53 # turn on the legend
54 plt.legend()
55 plt.show()

>>> 

99552 10000

3.推广的舍取抽样法

从上图中可以看出,基本的rejection methold法抽样效率很低,因为随机数x和y是在区间[xmin  xmax]和区间[0  ymax]上均匀分布的,产生的大部分点不会落在w(x)曲线之下(曲线e-x的形状一边高一边低,其曲线下的面积占矩形面积的比例很小,则舍选抽样效率很低)。为了改进简单舍选抽样法的效率,可以构造一个新的密度函数q(x)(called a proposal distribution from which we can readily draw samples),使它的形状接近p(x),并选择一个常数k使得kq(x)≥w(x)对于x定义域内的值都成立。对应下图,首先从分布q(z)中生成随机数z0,然后按均匀分布从区间[0   kq(z0)]生成一个随机数u0。 if u> p(z0) then the sample is rejected,otherwise uis retained.  即下图中灰色区域内的点都要舍弃。可见,由于随机点u0只出现在曲线kq(x)之下,且在q(x)较大处出现次数较多,从而大大提高了采样效率。显然q(x)形状越接近p(x),则采样效率越高。 

 

根据上述思想,也可以表达采样规则如下:

1. Draw x from your proposal distribution q(x)

2. Draw y uniformly from [0  1]

3. if y < p(x)/kq(x) , accept the sample, otherwise reject it

4. repeat

下面例子中选择函数p(x)=1/(x+1)作为proposal distribution,k=1。曲线1/(x+1)的形状与e-x相近。

 1 import numpy as np
 2 import matplotlib.pyplot as plt
 3 
 4 p = lambda x: np.exp(-x)         # our distribution
 5 g = lambda x: 1/(x+1)            # our proposal pdf (we're choosing k to be 1)
 6 CDFg = lambda x: np.log(x +1)    # generates our proposal using inverse sampling
 7 
 8 # domain limits
 9 xmin = 0  # the lower limit of our domain
10 xmax = 10 # the upper limit of our domain
11 
12 # range limits for inverse sampling
13 umin = CDFg(xmin)
14 umax = CDFg(xmax)
15 
16 N = 10000 # the total of samples we wish to generate
17 accepted = 0 # the number of accepted samples
18 samples = np.zeros(N)
19 count = 0 # the total count of proposals
20 
21 # generation loop
22 while (accepted < N):
23     
24     # Sample from g using inverse sampling
25     u = np.random.uniform(umin, umax)
26     xproposal = np.exp(u) - 1
27 
28     # pick a uniform number on [0, 1)
29     y = np.random.uniform(0, 1)
30     
31     # Do the accept/reject comparison
32     if y < p(xproposal)/g(xproposal):
33         samples[accepted] = xproposal
34         accepted += 1
35     
36     count +=1
37     
38 print count, accepted
39 
40 # get the histogram info
41 hinfo = np.histogram(samples,50)
42 
43 # plot the histogram
44 plt.hist(samples,bins=50, label=u'Samples');
45 
46 # plot our (normalized) function
47 xvals=np.linspace(xmin, xmax, 1000)
48 plt.plot(xvals, hinfo[0][0]*p(xvals), 'r', label=u'p(x)')
49 
50 # turn on the legend
51 plt.legend()
52 plt.show()

>>> 

24051 10000

可以对比基本的舍取法和改进的舍取法的结果,前者产生符合要求分布的10000个随机数运算了99552步,后者运算了24051步,可以看到效率明显提高。

 

参考:

http://iacs-courses.seas.harvard.edu/courses/am207/blog/lecture-3.html

http://blog.csdn.net/xianlingmao/article/details/7768833

http://blog.sina.com.cn/s/blog_60b44d6a0101l45z.html

http://www.ruanyifeng.com/blog/2015/07/monte-carlo-method.html

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!