# Estimating the exponent of discrete power law data

Suppose you have data from a discrete power law with exponent α. That is, the probability of an outcome n is proportional to n. How can you recover α?

A naive approach would be to gloss over the fact that you have discrete data and use the MLE (maximum likelihood estimator) for continuous data. That does a very poor job . The discrete case needs its own estimator.

To illustrate this, we start by generating 5,000 samples from a discrete power law with exponent 3 in the following Python code.

   import numpy.random

alpha = 3
n = 5000
x = numpy.random.zipf(alpha, n)


The continuous MLE is very simple to implement:

    alpha_hat = 1 + n / sum(log(x))


Unfortunately, it gives an estimate of 6.87 for alpha, though we know it should be around 3.

The MLE for the discrete power law distribution satisfies Here ζ is the Riemann zeta function, and xi are the samples. Note that the left side of the equation is the derivative of log ζ, or what is sometimes called the logarithmic derivative.

There are three minor obstacles to finding the estimator using Python. First, SciPy doesn’t implement the Riemann zeta function ζ(x) per se. It implements a generalization, the Hurwitz zeta function, ζ(x, q). Here we just need to set q to 1 to get the Riemann zeta function.

Second, SciPy doesn’t implement the derivative of zeta. We don’t need much accuracy, so it’s easy enough to implement our own. See an earlier post for an explanation of the implementation below.

Finally, we don’t have an explicit equation for our estimator. But we can easily solve for it using the bisection algorithm. (Bisect is slow but reliable. We’re not in a hurry, so we might as use something reliable.)

    from scipy import log
from scipy.special import zeta
from scipy.optimize import bisect

xmin = 1

def log_zeta(x):
return log(zeta(x, 1))

def log_deriv_zeta(x):
h = 1e-5
return (log_zeta(x+h) - log_zeta(x-h))/(2*h)

t = -sum( log(x/xmin) )/n
def objective(x):
return log_deriv_zeta(x) - t

a, b = 1.01, 10
alpha_hat = bisect(objective, a, b, xtol=1e-6)
print(alpha_hat)


We have assumed that our data follow a power law immediately from n = 1. In practice, power laws generally fit better after the first few elements. The code above works for the more general case if you set xmin to be the point at which power law behavior kicks in.

The bisection method above searches for a value of the power law exponent between 1.01 and 10, which is somewhat arbitrary. However, power law exponents are very often between 2 and 3 and seldom too far outside that range.

The code gives an estimate of α equal to 2.969, very near the true value of 3, and much better than the naive estimate of 6.87.

Of course in real applications you don’t know the correct result before you begin, so you use something like a confidence interval to give you an idea how much uncertainty remains in your estimate.

The following equation  gives a value of σ from a normal approximation to the distribution of our estimator. So an approximate 95% confidence interval would be the point estimate +/- 2σ.

    from scipy.special import zeta
from scipy import sqrt

def zeta_prime(x, xmin=1):
h = 1e-5
return (zeta(x+h, xmin) - zeta(x-h, xmin))/(2*h)

def zeta_double_prime(x, xmin=1):
h = 1e-5
return (zeta(x+h, xmin) -2*zeta(x,xmin) + zeta(x-h, xmin))/h**2

def sigma(n, alpha_hat, xmin=1):
z = zeta(alpha_hat, xmin)
temp = zeta_double_prime(alpha_hat, xmin)/z
temp -= (zeta_prime(alpha_hat, xmin)/z)**2
return 1/sqrt(n*temp)

print( sigma(n, alpha_hat) )


Here we use a finite difference approximation for the second derivative of zeta, an extension of the idea used above for the first derivative. We don’t need high accuracy approximations of the derivatives since statistical error will be larger than the approximation error.

In the example above, we have α = 2.969 and σ = 0.0334, so a 95% confidence interval would be [2.902, 3.036]. Using the continuous MLE with discrete data is not so bad when the minimum output xmin is moderately large. But here, where xmin = 1 it’s terrible.

 Equation 3.6 from Power-law distributions in empirical data by Aaron Clauset, Cosma Rohilla Shalizi, and M. E. J. Newman.

## 9 thoughts on “Estimating the exponent of discrete power law data”

1. def log_zeta(x):
return log(zeta(x, xmin))

def log_deriv_zeta(x):
h = 1e-5
return (log_zeta(x+h, xmin) – log_zeta(x-h, xmin))/(2*h)

log_deriv_zeta(x) calls log_zeta(x+h, xmin) with two arguments but log_zeta(x) only takes one argument. When I try to run I get “TypeError: log_zeta() takes exactly 1 argument (2 given)”

2. I initially had xmin as an argument, then decided to simplify the call by pulling it out and I missed a couple edits.

3. Sjoerd Visscher

3.306 should be 3.036, I guess?

4. David Tate

Is this an opportunity to use the subtractionless numerical derivative based on F(x+ih)? How does python handle complex arguments?

5. Yes it is! I mentioned that in the earlier post. SciPy does support complex arguments for some (most?) special functions, but unfortunately not zeta. Maybe Sage has a zeta with complex arguments.

6. George Paci

Interesting. Also, when I looked for an image of the zeta function, one of the top hits was a 3D visualization…. in Minecraft.

Minor typos: “left size” should be “left side” (all those zetas primed this mistake); “Shalzi” should be “Shalizi” (maybe reading science fiction?).

7. Andrew Follmann

Very informative post, thanks for it! Some of the more concise and explicit writing I have found on fitting data to a zeta distribution.

One question though in the third code block (on estimating alpha) – how does setting xmin reflect masking the top values of the data? When running this code with most other values where xmin > 1, the sign of the t parameter flips, and so makes the bisect optimization fail.
Would sorting the data, and then taking only the top xmin values work in this case?
“
x = numpy.random.zipf(alpha, n)
x.sort()
x = x[::-1]
t = -sum( log(x[xmin:]) )/n
“

8. John Mauer

Any ideas on the fitting of zipf to integer data where alpha is less than two, so that no moments exist?

9. John,

The point estimate doesn’t depend on the existence of moments. The confidence interval uses a normal approximation, which would be dubious if you don’t have a second moment.

You might do some sort of bootstrap where you compute a bunch of point estimates based on random samples of the data, and base a confidence interval on these values.