I am developing a simulation of the integration of montecarlo (dx) and I find myself wondering which is the best way to determine the exact number of samples (N) in the Monte Carlo method to approximate the solution to a definite integral.
This is a simple code of implementation:
import math
import random
class Montecarlo:
def __init__ (self):
print ("Inicializa..")
def fy(self, ri, a, b):
res = math.pow((b-a)*ri+a, 2.0)+math.sqrt((b-a)*ri+a)
return res
def integral (self, a, b, N):
suma = 0.0
ri = 0.0
for i in range (N):
ri = random.random()
suma+=self.fy(ri,a,b)
res=((b-a)/N)*suma
return res
if __name__ == "__main__":
monte = Montecarlo()
res = monte.integral(10.0,27.0,N)
print("Res: ", res)
Win Monte Carlo, you could compute statistical error (stddev) of the simulation. It goes down as 1/sqrt(N). You could set your goal - say, make error below 2% - and easily compute now many samples (N) you need.
I modified your code and added calculation of second momentum, sigma and simulation error
import math
import random
class Montecarlo:
def __init__(self):
print ("Inicializa..")
def fy(self, ri, a, b):
res = math.pow((b-a)*ri+a, 2.0) + math.sqrt((b-a)*ri+a)
return res
def integral (self, a, b, N):
sum = 0.0
var = 0.0
for i in range(N):
ri = random.random()
v = self.fy(ri, a, b)
sum += v
var += v*v
sum /= float(N)
var /= float(N)
sigma = ( var - sum*sum ) * float(N)/float(N-1)
error = sigma / math.sqrt(N)
return ((b-a) * sum, (b-a)*error)
if __name__ == "__main__":
N = 100000
monte = Montecarlo()
res, err = monte.integral(10.0, 27.0, N)
print("Res: {0}, Err: {1}".format(res, err))