python - 人口蒙特卡洛实现

Question

我正在尝试实现人口蒙特卡罗算法，如 this 中所述论文(参见第 78 页图 3)使用 Python 的一个参数的简单模型(参见函数 model())。不幸的是，该算法不起作用，我无法弄清楚出了什么问题。请参阅下面的实现。实际函数称为 abc()。所有其他功能都可以看作是辅助功能，并且似乎工作正常。

为了检查算法是否有效，我首先生成观测数据，并将模型的唯一参数设置为 param = 8。因此，ABC 算法产生的后验应该以 8 为中心。事实并非如此，我想知道为什么。

如有任何帮助或意见，我将不胜感激。

# imports

from math import exp
from math import log
from math import sqrt
import numpy as np
import random
from scipy.stats import norm


# globals
N = 300              # sample size
N_PARTICLE = 300      # number of particles
ITERS = 5            # number of decreasing thresholds
M = 10               # number of words to remember
MEAN = 7             # prior mean of parameter
SD = 2               # prior sd of parameter


def model(param):
  recall_prob_all = 1/(1 + np.exp(M - param))
  recall_prob_one_item = np.exp(np.log(recall_prob_all) / float(M))
  return sum([1 if random.random() < recall_prob_one_item else 0 for item in range(M)])

## example
print "Output of model function: \n" + str(model(10)) + "\n"

# generate data from model
def generate(param):
  out = np.empty(N)
  for i in range(N):
    out[i] = model(param)
  return out

## example

print "Output of generate function: \n" + str(generate(10)) + "\n"


# distance function (sum of squared error)
def distance(obsData,simData):
  out = 0.0
  for i in range(len(obsData)):
    out += (obsData[i] - simData[i]) * (obsData[i] - simData[i])
  return out

## example

print "Output of distance function: \n" + str(distance([1,2,3],[4,5,6])) + "\n"


# sample new particles based on weights
def sample(particles, weights):
  return np.random.choice(particles, 1, p=weights)

## example

print "Output of sample function: \n" + str(sample([1,2,3],[0.1,0.1,0.8])) + "\n"


# perturbance function
def perturb(variance):
  return np.random.normal(0,sqrt(variance),1)[0]

## example 

print "Output of perturb function: \n" + str(perturb(1)) + "\n"

# compute new weight
def computeWeight(prevWeights,prevParticles,prevVariance,currentParticle):
  denom = 0.0
  proposal = norm(currentParticle, sqrt(prevVariance))
  prior = norm(MEAN,SD)
  for i in range(len(prevParticles)):
    denom += prevWeights[i] * proposal.pdf(prevParticles[i])
  return prior.pdf(currentParticle)/denom


## example 

prevWeights = [0.2,0.3,0.5]
prevParticles = [1,2,3]
prevVariance = 1
currentParticle = 2.5
print "Output of computeWeight function: \n" + str(computeWeight(prevWeights,prevParticles,prevVariance,currentParticle)) + "\n"


# normalize weights
def normalize(weights):
  return weights/np.sum(weights)


## example 

print "Output of normalize function: \n" + str(normalize([3.,5.,9.])) + "\n"


# sampling from prior distribution
def rprior():
  return np.random.normal(MEAN,SD,1)[0]

## example 

print "Output of rprior function: \n" + str(rprior()) + "\n"


# ABC using Population Monte Carlo sampling
def abc(obsData,eps):
  draw = 0
  Distance = 1e9
  variance = np.empty(ITERS)
  simData = np.empty(N)
  particles = np.empty([ITERS,N_PARTICLE])
  weights = np.empty([ITERS,N_PARTICLE])

  for t in range(ITERS):
    if t == 0:
      for i in range(N_PARTICLE):
        while(Distance > eps[t]):
          draw = rprior()
          simData = generate(draw)
          Distance = distance(obsData,simData)

        Distance = 1e9
        particles[t][i] = draw
        weights[t][i] = 1./N_PARTICLE

      variance[t] = 2 * np.var(particles[t])
      continue


    for i in range(N_PARTICLE):
      while(Distance > eps[t]):
        draw = sample(particles[t-1],weights[t-1])
        draw += perturb(variance[t-1])
        simData = generate(draw)
        Distance = distance(obsData,simData)

      Distance = 1e9
      particles[t][i] = draw
      weights[t][i] = computeWeight(weights[t-1],particles[t-1],variance[t-1],particles[t][i])


    weights[t] = normalize(weights[t])  
    variance[t] = 2 * np.var(particles[t])

  return particles[ITERS-1]



true_param = 9
obsData = generate(true_param)
eps = [15000,10000,8000,6000,3000]
posterior = abc(obsData,eps)
#print posterior

Answer 1

我在寻找 PMC 算法的 pythonic 实现时偶然发现了这个问题，因为非常巧合的是，我目前正在将这篇确切论文中的技术应用到我自己的研究中。

你能发布你得到的结果吗？我的猜测是 1) 您使用的距离函数(和/或相似性阈值)选择不当，或者 2) 您没有使用足够的粒子。我在这里可能是错的(我不是很精通样本统计)，但你的距离函数暗示我随机抽取的顺序很重要。我必须更多地考虑这个以确定它是否真的对收敛特性有任何影响(可能不会)，但你为什么不简单地使用平均值或中位数作为你的样本统计量？

我使用 1000 个粒子和真实参数值为 8 运行您的代码，同时使用样本均值之间的绝对差作为我的距离函数，进行三次迭代，epsilons 为 [0.5, 0.3, 0.1]；我估计的后验分布的峰值似乎在每次迭代时都接近 8，同时总体方差也在减少。请注意，仍然存在明显的向右偏差，但这是因为模型的不对称性(8 或更小的参数值永远不会导致超过 8 个观察到的成功，而所有大于 8 的参数值都可以，导致向右分布中的偏态)。

这是我的结果图: