从混合正态分布生成随机变量

机器算法验证 r 随机生成 混合分布
2022-02-05 06:23:18

如何从混合分布中采样,尤其是正态分布的混合R例如,如果我想从中采样:

0.3×N(0,1)+0.5×N(10,1)+0.2×N(3,.1)

我怎么能那样做?

4个回答

出于性能原因,避免for循环是一种很好的做法。R利用这一事实的替代解决方案rnorm是矢量化的:

N <- 100000

components <- sample(1:3,prob=c(0.3,0.5,0.2),size=N,replace=TRUE)
mus <- c(0,10,3)
sds <- sqrt(c(1,1,0.1))

samples <- rnorm(n=N,mean=mus[components],sd=sds[components])

通常,从混合分布中采样的最简单方法之一如下:

算法步骤

1) 生成随机变量UUniform(0,1)

2) 若区间,其中对应于混合模型的分量的概率,然后由第分量的分布生成U[i=1kpk,i=1k+1pk+1)pkkthkth

3) 重复步骤 1) 和 2),直到从混合物分布中获得所需数量的样本

R现在使用上面给出的通用算法,您可以使用以下代码从示例法线混合中采样:

#The number of samples from the mixture distribution
N = 100000                 

#Sample N random uniforms U
U =runif(N)

#Variable to store the samples from the mixture distribution                                             
rand.samples = rep(NA,N)

#Sampling from the mixture
for(i in 1:N){
    if(U[i]<.3){
        rand.samples[i] = rnorm(1,0,1)
    }else if(U[i]<.8){
        rand.samples[i] = rnorm(1,10,1)
    }else{
        rand.samples[i] = rnorm(1,3,.1)
    }
}

#Density plot of the random samples
plot(density(rand.samples),main="Density Estimate of the Mixture Model")

#Plotting the true density as a sanity check
x = seq(-20,20,.1)
truth = .3*dnorm(x,0,1) + .5*dnorm(x,10,1) + .2*dnorm(x,3,.1)
plot(density(rand.samples),main="Density Estimate of the Mixture Model",ylim=c(0,.2),lwd=2)
lines(x,truth,col="red",lwd=2)

legend("topleft",c("True Density","Estimated Density"),col=c("red","black"),lwd=2)

生成:

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作为健全性检查:

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从概念上讲,您只是选择一种分布(来自k可能性)以一定的概率,然后从该分布生成伪随机变量。R中,这将是(例如):

set.seed(8)               # this makes the example reproducible
N     = 1000              # this is how many data you want
probs = c(.3,.8)          # these are *cumulative* probabilities; since they 
                          #   necessarily sum to 1, the last would be redundant
dists = runif(N)          # here I'm generating random variates from a uniform
                          #   to select the relevant distribution

# this is where the actual data are generated, it's just some if->then
#   statements, followed by the normal distributions you were interested in
data = vector(length=N)
for(i in 1:N){
  if(dists[i]<probs[1]){
    data[i] = rnorm(1, mean=0, sd=1)
  } else if(dists[i]<probs[2]){
    data[i] = rnorm(1, mean=10, sd=1)
  } else {
    data[i] = rnorm(1, mean=3, sd=.1)
  }
}

# here are a couple of ways of looking at the results
summary(data)
#    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
# -3.2820  0.8443  3.1910  5.5350 10.0700 13.1600 

plot(density(data))

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已经给出了完美的答案,所以对于那些想要在 Python 中实现这一点的人来说,这是我的解决方案:

import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline

mu = [0, 10, 3]
sigma = [1, 1, 1]
p_i = [0.3, 0.5, 0.2]
n = 10000

x = []
for i in range(n):
    z_i = np.argmax(np.random.multinomial(1, p_i))
    x_i = np.random.normal(mu[z_i], sigma[z_i])
    x.append(x_i)

def univariate_normal(x, mean, variance):
    """pdf of the univariate normal distribution."""
    return ((1. / np.sqrt(2 * np.pi * variance)) * 
            np.exp(-(x - mean)**2 / (2 * variance)))

a = np.arange(-7, 18, 0.01)
y = p_i[0] * univariate_normal(a, mean=mu[0], variance=sigma[0]**2) + p_i[1] * univariate_normal(a, mean=mu[1], variance=sigma[0]**2)+ p_i[2] * univariate_normal(a, mean=mu[2], variance=sigma[0]**2)

fig, ax = plt.subplots(figsize=(8, 4))

ax.hist(x, bins=100, density=True)
ax.plot(a, y)

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