使用Simpson规则整合正态分布

Question

I'm trying to perform a simple integration using the scipy.integrate.simps function and I can't figure out the results it shows.

Here's a MWE:

import numpy as np
from scipy.integrate import simps

# Same normal function used by np.random.normal
def norm_func(x, mu, sigma):
    y = 1/(sigma*np.sqrt(2*np.pi))*np.exp(-(x-mu)**2/(2*sigma**2))
    return y

# Generate some random points from the normal distribution.
a = np.random.normal(1., 0.1, 1000)

# Integrate the evaluated values of these points.
print simps(norm_func(a, 1., 0.1), a)

I would expect that since I'm drawing random numbers from a normal distribution and then integrating their evaluation in an equivalent normal distribution, I should get the result of integrating said normal distribution which is 1 (or very close to it).

What I find instead is that the results appear to vary with the sample size of a . Even worst, if I set a value of 10000 in a = np.random.normal(1., 0.1, 10000) , the integration returns a negative value.

What am I doing wrong here?

Answer 1

Using your sample, just sort a first, since it should be an array of points to sample at, and it expects them to be in order to build the approximation. Simpson's rule uses

So it will be taking values for x from your array and evaluating the function. If they are in a random order you can see that the above formula makes little sense, as it will do integrals from one random point on the domain to another. It might be better to think of it as x , so I'll use that variable name:

x = np.random.normal(1., 0.1, 1000)
x.sort() # sorts in place
print simps(norm_func(x, 1., 0.1), x)
#0.999914876748

This also works for me:

s = np.sort(np.random.normal(1., 0.1, 10000))
print simps(norm_func(s, 1., 0.1), s)
#0.999943377731

Answer 2

Here is how you can integrate the standard normal pdf using simps :

In [37]: a = np.linspace(-20, 20)

In [38]: print simps(norm_func(a, 0., 1.), a)
1.0