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# Monte Carlo sampling methods¶

This notebook is an element of the risk-engineering.org courseware. It can be distributed under the terms of the Creative Commons Attribution-ShareAlike licence.

Author: Eric Marsden eric.marsden@risk-engineering.org

This notebook contains an introduction to different sampling methods in Monte Carlo analysis (standard random sampling, latin hypercube sampling, and low discrepancy sequences such as that of Sobol’ and that of Halton). The notebook shows how to use Python, with the SciPy and SymPy libraries. See the associated course materials for more background information on stochastic simulation and to download this content as a Jupyter/Python notebook.

In [1]:
import math
import numpy
import scipy.stats
import matplotlib.pyplot as plt
plt.style.use("bmh")


Let’s start with a simple integration problem in 1D,

$\int_1^5 x^2$

This is easy to solve analytically, and we can use the SymPy library in case we’ve forgotten how to resolve simple integrals.

In [2]:
import sympy
# we’ll save results using different methods in this data structure, called a dictionary
result = {}
x = sympy.Symbol("x")
i = sympy.integrate(x**2)
result["analytical"] = float(i.subs(x, 5) - i.subs(x, 1))
print("Analytical result: {}".format(result["analytical"]))

Analytical result: 41.333333333333336


We can estimate this integral using a standard Monte Carlo method, where we use the fact that the expectation of a random variable is related to its integral

$\mathbb{E}(f(x)) = \int_I f(x) dx$

We will sample a large number $N$ of points in $I$ and calculate their average, and multiply by the range over which we are integrating.

In [3]:
N = 10_000
accum = 0
for i in range(N):
x = numpy.random.uniform(1, 5)
accum += x**2
volume = 5 - 1
result["MC"] = volume * accum / float(N)
print("Standard Monte Carlo result: {}".format(result["MC"]))

Standard Monte Carlo result: 41.50650693884985


If we increase $N$, the estimation will converge to the analytical value. (It will converge relatively slowly, following $1/\sqrt(N)$).

## Latin hypercube sampling¶

The LHS method consists of dividing the input space into a number of equiprobable regions, then taking random samples from each region. We can use it conveniently in Python thanks to the pyDOE library, which you will probably need to install on your computer, using a command such as

pip install pyDOE

or if you’re using a Google CoLaboratory notebook, execute a code cell containing

!pip install pyDOE

The lhs function in this library returns an “experimental design” consisting of points in the $[0, 1]^d$ hypercube, where $d$ is the dimension of your problem (it’s 2 in this simple example). You need to scale these points to your input domain.

In [4]:
# obtain the pyDOE library from https://pythonhosted.org/pyDOE/
from pyDOE import lhs

seq = lhs(2, N)
accum = 0
for i in range(N):
x = 1 + seq[i][0]*(5 - 1)
y = 1 + seq[i][1]*(5**2 - 1**2)
accum += x**2
volume = 5 - 1
result["LHS"] = volume * accum / float(N)
print("Latin hypercube result: {}".format(result["LHS"]))

Latin hypercube result: 41.333322067647465


Note that the error in this estimation is significantly lower than that obtained using standard Monte Carlo sampling (and if you repeat this experiment many times, you should find this is true in most cases).

## Halton’s low discrepency sequences¶

A low discrepancy (or quasi-random) sequence is a deterministic mathematical sequence of numbers that has the property of low discrepancy. This means that there are no clusters of points and that the sequence fills space roughly uniformly. The Halton sequence is a low discrepancy sequence that has useful properties for pseudo-stochastic sampling methods (also called “quasi-Monte Carlo” methods).

In [5]:
# adapted from https://mail.scipy.org/pipermail/scipy-user/2013-June/034744.html
def halton(dim: int, nbpts: int):
h = numpy.full(nbpts * dim, numpy.nan)
p = numpy.full(nbpts, numpy.nan)
P = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31]
lognbpts = math.log(nbpts + 1)
for i in range(dim):
b = P[i]
n = int(math.ceil(lognbpts / math.log(b)))
for t in range(n):
p[t] = pow(b, -(t + 1))

for j in range(nbpts):
d = j + 1
sum_ = math.fmod(d, b) * p[0]
for t in range(1, n):
d = math.floor(d / b)
sum_ += math.fmod(d, b) * p[t]

h[j*dim + i] = sum_
return h.reshape(nbpts, dim)

In [6]:
N = 1000
seq = halton(2, N)
plt.title("2D Halton sequence")
# Note: we use "alpha=0.5" in the scatterplot so that the plotted points are semi-transparent
# (alpha-transparency of 0.5 out of 1), so that we can see when any points are superimposed.
plt.scatter(seq[:,0], seq[:,1], marker=".", alpha=0.5);

In [7]:
N = 1000
plt.title("Pseudo-random sequence")
plt.scatter(numpy.random.uniform(size=N), numpy.random.uniform(size=N), marker=".", alpha=0.5);


Comparing the scatterplot of the 2D Halton sequence with that of a pseudo-random sequence (pseudo-random meaning “as close to randomness as we can get with a computer”), note that the Halton sequence looks less “random” and covers the space in a more regular manner. For this reason, a low discrepancy sequence gives, on average, better results for stochastic sampling problems than does a truly stochastic (really pseudo-random) sampling approach. Let’s test that on our integration problem:

In [8]:
seq = halton(2, N)
accum = 0
for i in range(N):
x = 1 + seq[i][0]*(5 - 1)
y = 1 + seq[i][1]*(5**2 - 1**2)
accum += x**2
volume = 5 - 1
result["QMCH"] = volume * accum / float(N)
print("Quasi-Monte Carlo result (Halton): {}".format(result["QMCH"]))

Quasi-Monte Carlo result (Halton): 41.21870562744141


Another quasi-random sequence commonly used for this purpose is the Sobol’ sequence.

In [9]:
# you will need to install this library with something like
#
#    pip install sobol_seq
import sobol_seq

seq = sobol_seq.i4_sobol_generate(2, N)
plt.title("2D Sobol sequence")
plt.scatter(seq[:,0], seq[:,1], marker=".", alpha=0.5);

In [10]:
seq = sobol_seq.i4_sobol_generate(2, N)

accum = 0
for i in range(N):
x = 1 + seq[i][0]*(5 - 1)
y = 1 + seq[i][1]*(5**2 - 1**2)
accum += x**2
volume = 5 - 1
result["QMCS"] = volume * accum / float(N)
print("Quasi-Monte Carlo result (Sobol): {}".format(result["QMCS"]))

Quasi-Monte Carlo result (Sobol): 41.31674468994141


## Comparison (1 dimensional)¶

We can compare the error of our different estimates (each used the same number of runs):

In [11]:
for m in ["MC", "LHS", "QMCH", "QMCS"]:
print("{:4} result: {:.8}  Error: {:E}".format(m, result[m], abs(result[m]-result["analytical"])))

MC   result: 41.506507  Error: 1.731736E-01
LHS  result: 41.333322  Error: 1.126569E-05
QMCH result: 41.218706  Error: 1.146277E-01
QMCS result: 41.316745  Error: 1.658864E-02


Note that in practice, it’s possible to combine Latin Hypercube sampling with low discrepancy sequences.

## A higher dimensional integral¶

Let us now analyze an integration problem in dimension 4, the Ishigami function. This is a well-known function in numerical optimization and stochastic analysis, because it is very highly non-linear.

In [12]:
def ishigami(x1, x2, x3) -> float:
return numpy.sin(x1) + 7*numpy.sin(x2)**2 + 0.1 * x3**4 * numpy.sin(x1)


We want to resolve the integral over $[-\pi, \pi]^3$. We start by resolving the problem analytically, using SymPy.

In [13]:
x1 = sympy.Symbol("x1")
x2 = sympy.Symbol("x2")
x3 = sympy.Symbol("x3")
expr = sympy.sin(x1) + 7*sympy.sin(x2)**2 + 0.1 * x3**4 * sympy.sin(x1)
res = sympy.integrate(expr,
(x1, -sympy.pi, sympy.pi),
(x2, -sympy.pi, sympy.pi),
(x3, -sympy.pi, sympy.pi))
# Note: we use float(res) to convert res from symbolic form to floating point form
result["analytical"] = float(res)


Result from a standard Monte Carlo sampling method:

In [14]:
N = 10_000
accum = 0
for i in range(N):
xx1 = numpy.random.uniform(-numpy.pi, numpy.pi)
xx2 = numpy.random.uniform(-numpy.pi, numpy.pi)
xx3 = numpy.random.uniform(-numpy.pi, numpy.pi)
accum += numpy.sin(xx1) + 7*numpy.sin(xx2)**2 + 0.1 * xx3**4 * numpy.sin(xx1)
volume = (2 * numpy.pi)**3
result["MC"] = volume * accum / float(N)


Using latin hypercube sampling:

In [15]:
seq = lhs(3, N)
accum = 0
for i in range(N):
xx1 = -numpy.pi + seq[i][0] * numpy.pi * 2
xx2 = -numpy.pi + seq[i][1] * numpy.pi * 2
xx3 = -numpy.pi + seq[i][2] * numpy.pi * 2
accum += numpy.sin(xx1) + 7*numpy.sin(xx2)**2 + 0.1 * xx3**4 * numpy.sin(xx1)
volume = (2 * numpy.pi)**3
result["LHS"] = volume * accum / float(N)


A low-discrepancy Halton sequence, for a quasi-Monte Carlo approach:

In [16]:
seq = halton(3, N)
accum = 0
for i in range(N):
xx1 = -numpy.pi + seq[i][0] * numpy.pi * 2
xx2 = -numpy.pi + seq[i][1] * numpy.pi * 2
xx3 = -numpy.pi + seq[i][2] * numpy.pi * 2
accum += numpy.sin(xx1) + 7*numpy.sin(xx2)**2 + 0.1 * xx3**4 * numpy.sin(xx1)
volume = (2 * numpy.pi)**3
result["QMC"] = volume * accum / float(N)


Comparing the results of the three estimation methods:

In [17]:
for m in ["MC", "LHS", "QMC"]:
print("{:3} result: {:.8} Error: {:E}".format(m, result[m], abs(result[m]-result["analytical"])))

MC  result: 877.05452 Error: 8.878774E+00
LHS result: 873.06303 Error: 4.887285E+00
QMC result: 868.23893 Error: 6.318098E-02

In [ ]: