# -*- coding: utf-8 -*-
import numpy as np
import matplotlib.pyplot as plt

def f(x):
  return x**2 + 4*x*np.sin(x) 
def intf(x): 
  return x**3/3.0+4.0*np.sin(x) - 4.0*x*np.cos(x)
a = 2;  
b = 3; 
# use N draws 
N= 10000
X = np.random.uniform(low=a, high=b, size=N) # N values uniformly drawn from a to b 
Y =f(X)  # CALCULATE THE f(x) 
# 蒙特卡洛法計(jì)算定積分：面積=寬度*平均高度
Imc= (b-a) * np.sum(Y)/ N;
exactval=intf(b)-intf(a)
print "Monte Carlo estimation=",Imc, "Exact number=", intf(b)-intf(a)
# --How does the accuracy depends on the number of points(samples)? Lets try the same 1-D integral 
# The Monte Carlo methods yield approximate answers whose accuracy depends on the number of draws.
Imc=np.zeros(1000)
Na = np.linspace(0,1000,1000)
exactval= intf(b)-intf(a)
for N in np.arange(0,1000):
  X = np.random.uniform(low=a, high=b, size=N) # N values uniformly drawn from a to b 
  Y =f(X)  # CALCULATE THE f(x) 
  Imc[N]= (b-a) * np.sum(Y)/ N;   
plt.plot(Na[10:],np.sqrt((Imc[10:]-exactval)**2), alpha=0.7)
plt.plot(Na[10:], 1/np.sqrt(Na[10:]), 'r')
plt.xlabel("N")
plt.ylabel("sqrt((Imc-ExactValue)$^2$)")
plt.show()

# -*- coding: utf-8 -*-
# Example: Calculate ∫sin(x)xdx

# The function has a shape that is similar to Gaussian and therefore
# we choose here a Gaussian as importance sampling distribution.
from scipy import stats
from scipy.stats import norm
import numpy as np
import matplotlib.pyplot as plt
mu = 2;
sig =.7;
f = lambda x: np.sin(x)*x
infun = lambda x: np.sin(x)-x*np.cos(x)
p = lambda x: (1/np.sqrt(2*np.pi*sig**2))*np.exp(-(x-mu)**2/(2.0*sig**2))
normfun = lambda x: norm.cdf(x-mu, scale=sig)

plt.figure(figsize=(18,8)) # set the figure size
# range of integration
xmax =np.pi 
xmin =0
# Number of draws 
N =1000
# Just want to plot the function
x=np.linspace(xmin, xmax, 1000)
plt.subplot(1,2,1)
plt.plot(x, f(x), 'b', label=u'Original $x\sin(x)$')
plt.plot(x, p(x), 'r', label=u'Importance Sampling Function: Normal')
plt.xlabel('x')
plt.legend()
# =============================================
# EXACT SOLUTION 
# =============================================
Iexact = infun(xmax)-infun(xmin)
print Iexact
# ============================================
# VANILLA MONTE CARLO 
# ============================================
Ivmc = np.zeros(1000)
for k in np.arange(0,1000):
  x = np.random.uniform(low=xmin, high=xmax, size=N)
  Ivmc[k] = (xmax-xmin)*np.mean(f(x))
# ============================================
# IMPORTANCE SAMPLING 
# ============================================
# CHOOSE Gaussian so it similar to the original functions

# Importance sampling: choose the random points so that
# more points are chosen around the peak, less where the integrand is small.
Iis = np.zeros(1000)
for k in np.arange(0,1000):
  # DRAW FROM THE GAUSSIAN: xis~N(mu,sig^2)
  xis = mu + sig*np.random.randn(N,1);
  xis = xis[ (xis<xmax) & (xis>xmin)] ;
  # normalization for gaussian from 0..pi
  normal = normfun(np.pi)-normfun(0)   # 注意:概率密度函數(shù)在采樣區(qū)間[0 pi]上的積分需要等于1
  Iis[k] =np.mean(f(xis)/p(xis))*normal  # 因此,此處需要乘一個(gè)系數(shù)即p(x)在[0 pi]上的積分
plt.subplot(1,2,2)
plt.hist(Iis,30, histtype='step', label=u'Importance Sampling');
plt.hist(Ivmc, 30, color='r',histtype='step', label=u'Vanilla MC');
plt.vlines(np.pi, 0, 100, color='g', linestyle='dashed')
plt.legend()
plt.show()