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Maximum Likelihood Estimates (MLEs)

By Delaney Granizo-Mackenzie and Andrei Kirilenko developed as part of the Masters of Finance curriculum at MIT Sloan.

In this tutorial notebook, we'll do the following things:

  1. Compute the MLE for a normal distribution.
  2. Compute the MLE for an exponential distribution.
  3. Fit a normal distribution to asset returns using MLE.

First we need to import some libraries

In [1]:
import math
import matplotlib.pyplot as plt
import numpy as np
import scipy
import scipy.stats

Normal Distribution

We'll start by sampling some data from a normal distribution.

In [2]:
TRUE_MEAN = 40
TRUE_STD = 10
X = np.random.normal(TRUE_MEAN, TRUE_STD, 1000)

Now we'll define functions that, given our data, will compute the MLE for the $\mu$ and $\sigma$ parameters of the normal distribution.

Recall that

$$\hat\mu = \frac{1}{T}\sum_{t=1}^{T} x_t$$$$\hat\sigma = \sqrt{\frac{1}{T}\sum_{t=1}^{T}{(x_t - \hat\mu)^2}}$$
In [3]:
def normal_mu_MLE(X):
    # Get the number of observations
    T = len(X)
    # Sum the observations
    s = sum(X)
    return 1.0/T * s

def normal_sigma_MLE(X):
    T = len(X)
    # Get the mu MLE
    mu = normal_mu_MLE(X)
    # Sum the square of the differences
    s = sum( np.power((X - mu), 2) )
    # Compute sigma^2
    sigma_squared = 1.0/T * s
    return math.sqrt(sigma_squared)

Now let's try our functions out on our sample data and see how they compare to the built-in np.mean and np.std

In [4]:
print("Mean Estimation")
print(normal_mu_MLE(X))
print(np.mean(X))
print("Standard Deviation Estimation")
print(normal_sigma_MLE(X))
print(np.std(X))
Mean Estimation
40.0277237219
40.0277237219
Standard Deviation Estimation
9.863010482653552
9.86301048265

Now let's estimate both parameters at once with scipy's built in fit() function.

In [5]:
mu, std = scipy.stats.norm.fit(X)
print("mu estimate:",  str(mu))
print("std estimate:", str(std))
mu estimate: 40.0277237219
std estimate: 9.86301048265

Now let's plot the distribution PDF along with the data to see how well it fits. We can do that by accessing the pdf provided in scipy.stats.norm.pdf.

In [6]:
pdf = scipy.stats.norm.pdf
# We would like to plot our data along an x-axis ranging from 0-80 with 80 intervals
# (increments of 1)
x = np.linspace(0, 80, 80)
plt.hist(X, bins=x, density='true')
plt.plot(pdf(x, loc=mu, scale=std))
plt.xlabel('Value')
plt.ylabel('Observed Frequency')
plt.legend(['Fitted Distribution PDF', 'Observed Data', ]);

Exponential Distribution

Let's do the same thing, but for the exponential distribution. We'll start by sampling some data.

In [7]:
TRUE_LAMBDA = 5
X = np.random.exponential(TRUE_LAMBDA, 1000)

numpy defines the exponential distribution as $$\frac{1}{\lambda}e^{-\frac{x}{\lambda}}$$

So we need to invert the MLE from the lecture notes. There it is

$$\hat\lambda = \frac{T}{\sum_{t=1}^{T} x_t}$$

Here it's just the reciprocal, so

$$\hat\lambda = \frac{\sum_{t=1}^{T} x_t}{T}$$
In [8]:
def exp_lamda_MLE(X):
    T = len(X)
    s = sum(X)
    return s/T
In [9]:
print("lambda estimate:", str(exp_lamda_MLE(X)))
lambda estimate: 4.85890096817
In [10]:
# The scipy version of the exponential distribution has a location parameter
# that can skew the distribution. We ignore this by fixing the location
# parameter to 0 with floc=0
_, l = scipy.stats.expon.fit(X, floc=0)
In [11]:
pdf = scipy.stats.expon.pdf
x = range(0, 80)
plt.hist(X, bins=x, density='true')
plt.plot(pdf(x, scale=l))
plt.xlabel('Value')
plt.ylabel('Observed Frequency')
plt.legend(['Fitted Distribution PDF', 'Observed Data', ]);

MLE for Asset Returns

Now we'll fetch some real returns and try to fit a normal distribution to them using MLE.

In [12]:
from quantrocket.master import get_securities
from quantrocket import get_prices

aapl_sid = get_securities(symbols="AAPL", vendors='usstock').index[0]

prices = get_prices('usstock-free-1min', data_frequency='daily', sids=aapl_sid, fields='Close', start_date='2014-01-01', end_date='2015-01-01')
prices = prices.loc['Close'][aapl_sid]

# This will give us the number of dollars returned each day
absolute_returns = np.diff(prices)
# This will give us the percentage return over the last day's value
# the [:-1] notation gives us all but the last item in the array
# We do this because there are no returns on the final price in the array.
returns = absolute_returns/prices[:-1]

Let's use scipy's fit function to get the $\mu$ and $\sigma$ MLEs.

In [13]:
mu, std = scipy.stats.norm.fit(returns)
pdf = scipy.stats.norm.pdf
x = np.linspace(-1,1, num=100)
h = plt.hist(returns, bins=x, density='true')
l = plt.plot(x, pdf(x, loc=mu, scale=std))

Of course, this fit is meaningless unless we've tested that they obey a normal distribution first. We can test this using the Jarque-Bera normality test. The Jarque-Bera test will reject the hypothesis of a normal distribution if the p-value is under a c.

In [14]:
from statsmodels.stats.stattools import jarque_bera
jarque_bera(returns)
Out[14]:
(866.6489261082892,
 6.450254076453388e-189,
 -0.14337787515083958,
 12.098604328561194)
In [15]:
jarque_bera(np.random.normal(0, 1, 100))
Out[15]:
(1.9029474229453722,
 0.38617149853205984,
 -0.10078189081497935,
 2.3549578126650013)

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