%matplotlib notebook
import math
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

Monte Carlo integration examples

This notebook shows some examples of Monte Carlo integration which we'll look at in class.

If you are looking at the HTML export of this notebook, the source file is here: monte-carlo.ipynb and you can download it and run it in Jupyter Notebook if you have it. (If you'd like to get it, installing Anaconda is a quick way.)

Some 1D examples

Here is an integral we'd like to compute:

$$I = \int_0^1 1 - \sqrt{1 - x^4} dx$$

It doesn't have a closed form but the value (computed by Wolfram Alpha as an Elliptic Integral) is approximately 0.125981.

def f1(x):
    return 1 - np.sqrt(1 - x**4)

plt.figure()
xs = np.linspace(0, 1, 501)
plt.plot(xs, f1(xs));

Here is a strategy to estimate this integral using uniform sampling.

strategy1 = {
    'eval': lambda x: f1(x),
    'sample': lambda xi: xi,
    'pdf': lambda x: np.ones(x.shape)
}

def estimate(strategy, sampler, N):
    """Integrate a function using N independent samples drawn from a provided pdf."""
    xis = sampler(N)
    xs = strategy['sample'](xis)
    return np.mean(strategy['eval'](xs) / strategy['pdf'](xs))

def independent_sampler(N):
    return np.random.random_sample(N)

estimate(strategy1, independent_sampler, 10000) - 0.125981

0.0017636856234602705

Let's ses how the error depends on the number of samples...

def stdev_of(fn, M):
    """Compute the standard deviation of repeated calls to a function."""
    results = [fn() for i in range(M)]
    return np.std(results)

stdev_of(lambda: estimate(strategy1, independent_sampler, 1000000), 100)

0.00020658421995445867

Ns = np.int32(np.logspace(2, 5, 7))
stdevs = [stdev_of(lambda: estimate(strategy1, independent_sampler, N), 100) for N in Ns]
plt.figure()
plt.loglog(Ns, stdevs, '-')
plt.xlabel('# samples')
plt.ylabel('standard deviation');
plt.axis('equal');

This log-log plot comes out as a straight line, showing that the number of samples and the noise have a power-law relationship; and the slope of this line at $-1/2$ shows that the noise is inversely proportional to the square root of the number of samples, which is what we expected.

Importance sampling

Here is another sampling strategy that tries to mimic the integrand with something that is easy to integrate analytically.

strategy2 = {
    'eval': lambda x: f1(x),
    'sample': lambda xi: xi ** (1/5),
    'pdf': lambda x: 5 * x**4
}

A quick verification that this sampling strategy is internally consistent.

def histcheck(strategy, N):
    xis = np.random.random_sample(N)
    xs = strategy['sample'](xis)
    ps, edges, patches = plt.hist(xs, math.ceil(math.sqrt(N)/10), density=True)
    plt.plot(edges, strategy['pdf'](edges), 'k')

plt.figure()
histcheck(strategy2, 100000)

Ns = np.int32(np.logspace(2, 5, 7))
stdevs1 = [stdev_of(
               lambda: estimate(strategy1, independent_sampler, N), 100
           ) for N in Ns]
stdevs2 = [stdev_of(
               lambda: estimate(strategy2, independent_sampler, N), 100
           ) for N in Ns]
plt.figure()
plt.loglog(Ns, stdevs1, Ns, stdevs2)
plt.xlabel('# samples')
plt.ylabel('standard deviation');
plt.axis('equal');

From this plot we can see that importance sampling has significantly decreased the noise (by a factor of around 10 in this case) but since the slope is unchanged, the convergence rate is still the same ($1/\sqrt{N}$).

Stratified sampling

Another strategy for reducing variance is to change the set of random samples being fed into the process. We can do that in our example just by replacing the function that generates the samples with one that generates stratified samples.

def stratified_sampler(N):
    return (np.arange(N) + np.random.random_sample(N)) / N

Ns = np.int32(np.logspace(2, 5, 7))
stdevs1 = [stdev_of(
               lambda: estimate(strategy1, independent_sampler, N), 100
           ) for N in Ns]
stdevs2 = [stdev_of(
               lambda: estimate(strategy2, independent_sampler, N), 100
           ) for N in Ns]

stdevs1s = [stdev_of(
               lambda: estimate(strategy1, stratified_sampler, N), 100
           ) for N in Ns]
stdevs2s = [stdev_of(
               lambda: estimate(strategy2, stratified_sampler, N), 100
           ) for N in Ns]

plt.figure()
plt.loglog(Ns, stdevs1, color='tab:blue')
plt.loglog(Ns, stdevs2, color='tab:orange')
plt.loglog(Ns, stdevs1s, color='tab:blue', dashes=[4,2])
plt.loglog(Ns, stdevs2s, color='tab:orange', dashes=[4,2])
plt.loglog([1e1,1e6], [1e-2,1e-7], 'k:')
plt.xlabel('# samples')
plt.ylabel('standard deviation');
plt.axis('equal');

This plot shows that in this example stratification has a huge benefit. Comparing to the reference line at slope -1, you can see the slope of the dashed lines on that plot is about -1.5. Again this is as expected for a nice smooth 1D integrand.

Let's try a messier integrand to see if it takes more samples for stratification to start winning.

def f2(x):
    return np.cos(10*x**2) + np.cos(21*x-5) + np.cos(38.7*x+2) + np.cos(82.4*x-3) + np.cos(154.2*x)*2 + np.cos(301.1*x)*4

plt.figure()
xs = np.linspace(0,1,5001)
plt.plot(xs, f2(xs))