In [12]:
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns

Generating random numbers using the np.random module

We will be hammering the np.random module for the rest of this course and its sequel. Actually, you will probably call functions from this module more than any other while wearing your hacker statistician hat. Let's start by taking its simplest function, np.random.random() for a test spin. The function returns a random number between zero and one. Call np.random.random() a few times in the IPython shell. You should see numbers jumping around between zero and one.

In this exercise, we'll generate lots of random numbers between zero and one, and then plot a histogram of the results. If the numbers are truly random, all bars in the histogram should be of (close to) equal height.

You may have noticed that, in the video, Justin generated 4 random numbers by passing the keyword argument size=4 to np.random.random(). Such an approach is more efficient than a for loop: in this exercise, however, you will write a for loop to experience hacker statistics as the practice of repeating an experiment over and over again.

In [13]:
# Seed the random number generator

# Initialize random numbers: random_numbers
random_numbers = np.empty(100000)

# Generate random numbers by looping over range(100000)
for i in range(100000):
    random_numbers[i] = np.random.random()

# Plot a histogram
_ = plt.hist(random_numbers, ec='black')

# Show the plot

The np.random module and Bernoulli trials

You can think of a Bernoulli trial as a flip of a possibly biased coin. Specifically, each coin flip has a probability pp of landing heads (success) and probability 1−p, 1−p of landing tails (failure).

In this exercise, you will write a function to perform n Bernoulli trials, perform_bernoulli_trials(n, p), which returns the number of successes out of n Bernoulli trials, each of which has probability p of success. To perform each Bernoulli trial, use the np.random.random() function, which returns a random number between zero and one.

In [14]:
def perform_bernoulli_trials(n, p):
    """Perform n Bernoulli trials with success probability p
    and return number of successes."""
    # Initialize number of successes: n_success
    n_success = 0

    # Perform trials
    for i in range(n):
        # Choose random number between zero and one: random_number
        random_number = np.random.random()

        # If less than p, it's a success  so add one to n_success
        if random_number < p:
            n_success += 1

    return n_success

How many defaults might we expect?

Let's say a bank made 100 mortgage loans. It is possible that anywhere between 0 and 100 of the loans will be defaulted upon.

You would like to know the probability of getting a given number of defaults, given that the probability of a default is p = 0.05.

To investigate this, you will do a simulation. You will perform 100 Bernoulli trials using the perform_bernoulli_trials() function you wrote in the previous exercise and record how many defaults we get. Here, a success is a default.

(Remember that the word "success" just means that the Bernoulli trial evaluates to True, i.e., did the loan recipient default?) You will do this for another 100 Bernoulli trials. And again and again until we have tried it 1000 times. Then, you will plot a histogram describing the probability of the number of defaults.

In [45]:
# Seed random number generator

# Initialize the number of defaults: n_defaults
n_defaults = np.empty(1000)

# Compute the number of defaults
for i in range(1000):
    n_defaults[i] = perform_bernoulli_trials(100, 0.05)

# Plot the histogram with default number of bins; label your axes
_ = plt.hist(n_defaults, normed=True, ec='black')
_ = plt.xlabel('number of defaults out of 100 loans')
_ = plt.ylabel('probability')

# Show the plot
In [46]:
def ecdf(data):
    """Compute ECDF for a one-dimensional array of measurements."""

    # Number of data points: n
    n = len(data)

    # x-data for the ECDF: x
    x = np.sort(data)

    # y-data for the ECDF: y
    y = np.arange(1, n+1) / float(n)

    return x, y

Will the bank fail?

Plot the number of defaults you got from the previous exercise, in your namespace as n_defaults, as a CDF. The ecdf() function you wrote in the first chapter is available.

If interest rates are such that the bank will lose money if 10 or more of its loans are defaulted upon, what is the probability that the bank will lose money?

In [47]:
# Compute ECDF: x, y
x, y = ecdf(n_defaults)

# Plot the CDF with labeled axes
_ = plt.plot(x, y, marker='.', linestyle='none')
_ = plt.xlabel('number of defaults out of 100')
_ = plt.ylabel('CDF')

# Show the plot

# Compute the number of 100-loan simulations with 10 or more defaults: n_lose_money
n_lose_money = np.sum(n_defaults >= 10)

# Compute and print probability of losing money
print('Probability of losing money =', n_lose_money / float(len(n_defaults)))
('Probability of losing money =', 0.015)

calculation of P(m, n), C(m, n)

In [23]:
from scipy.special import comb, perm
In [38]:
comb(10,3), perm(10,3)
(120.0, 720.0)
In [40]:
comb(100, 10)*0.95**90*0.05**10
In [41]:
comb(100, 11)*0.95**89*0.05**11

Sampling out of the Binomial distribution

Compute the probability mass function for the number of defaults we would expect for 100 loans as in the last section, but instead of simulating all of the Bernoulli trials, perform the sampling using np.random.binomial(). This is identical to the calculation you did in the last set of exercises using your custom-written perform_bernoulli_trials() function, but far more computationally efficient. Given this extra efficiency, we will take 10,000 samples instead of 1000. After taking the samples, plot the CDF as last time. This CDF that you are plotting is that of the Binomial distribution.

Note: For this exercise and all going forward, the random number generator is pre-seeded for you (with np.random.seed(7)) to save you typing that each time.

In [53]:
# Take 10,000 samples out of the binomial distribution: n_defaults
n_defaults = np.random.binomial(n=100, p=0.05, size=10000)

print n_defaults.shape
# Compute CDF: x, y
x, y = ecdf(n_defaults)

# Plot the CDF with axis labels
_ = plt.plot(x, y, marker='.', linestyle='none')
_ = plt.xlabel('number of defaults out of 100 loans')
_ = plt.ylabel('CDF')

# Show the plot

Plotting the Binomial PMF

Probability Mass Function

As mentioned in the video, plotting a nice looking PMF requires a bit of matplotlib trickery that we will not go into here. Instead, we will plot the PMF of the Binomial distribution as a histogram with skills you have already learned. The trick is setting up the edges of the bins to pass to plt.hist() via the bins keyword argument. We want the bins centered on the integers. So, the edges of the bins should be -0.5, 0.5, 1.5, 2.5, ... up to max(n_defaults) + 1.5. You can generate an array like this using np.arange() and then subtracting 0.5 from the array.

You have already sampled out of the Binomial distribution during your exercises on loan defaults, and the resulting samples are in the NumPy array n_defaults.

In [58]:
# Compute bin edges: bins
bins = np.arange(0, max(n_defaults) + 1.5) - 0.5

# Generate histogram
_ = plt.hist(n_defaults, normed=True, bins=bins, ec='black')

# Set margins

# Label axes
_ = plt.xlabel('number of defaults out of 100 loans')
_ = plt.ylabel('PMF')

# Show the plot
In [59]:
from scipy import stats

Relationship between Binomial and Poisson distributions

You just heard that the Poisson distribution is a limit of the Binomial distribution for rare events. This makes sense if you think about the stories. Say we do a Bernoulli trial every minute for an hour, each with a success probability of 0.1. We would do 60 trials, and the number of successes is Binomially distributed, and we would expect to get about 6 successes. This is just like the Poisson story we discussed in the video, where we get on average 6 hits on a website per hour. So, the Poisson distribution with arrival rate equal to npnp approximates a Binomial distribution for nn Bernoulli trials with probability pp of success (with nn large and pp small). Importantly, the Poisson distribution is often simpler to work with because it has only one parameter instead of two for the Binomial distribution.

Let's explore these two distributions computationally. You will compute the mean and standard deviation of samples from a Poisson distribution with an arrival rate of 10. Then, you will compute the mean and standard deviation of samples from a Binomial distribution with parameters nn and pp such that np=10np=10.

In [83]:
np.random.poisson(20, size=32)
array([14, 19, 18, 20, 12, 19, 21, 22, 15, 25, 14, 15, 16, 25, 27, 25, 21,
       19, 16, 11, 17, 30, 27, 27, 24, 19, 22, 21, 15, 21, 23, 11])
In [84]:
# Draw 10,000 samples out of Poisson distribution: samples_poisson
samples_poisson = np.random.poisson(10, size=10000)

# Print the mean and standard deviation
print('Poisson:     ', np.mean(samples_poisson),

# Specify values of n and p to consider for Binomial: n, p
n = [20, 100, 1000]
p = [0.5, 0.1, 0.01]

# Draw 10,000 samples for each n,p pair: samples_binomial
for i in range(3):
    samples_binomial = np.random.binomial(n[i], p[i], size=10000)

    # Print results
    print('n =', n[i], 'Binom:', np.mean(samples_binomial),
('Poisson:     ', 10.0007, 3.157229720815386)
('n =', 20, 'Binom:', 10.0306, 2.23782565004515)
('n =', 100, 'Binom:', 9.9555, 2.9712825092878665)
('n =', 1000, 'Binom:', 10.0556, 3.1745721979504578)

How many no-hitters in a season?

In baseball, a no-hitter is a game in which a pitcher does not allow the other team to get a hit. This is a rare event, and since the beginning of the so-called modern era of baseball (starting in 1901), there have only been 251 of them through the 2015 season in over 200,000 games. The ECDF of the number of no-hitters in a season is shown to the right. Which probability distribution would be appropriate to describe the number of no-hitters we would expect in a given season?

Note: The no-hitter data set was scraped and calculated from the data sets available at (license).

Was 2015 anomalous?

1990 and 2015 featured the most no-hitters of any season of baseball (there were seven). Given that there are on average 251/115 no-hitters per season, what is the probability of having seven or more in a season?

In [86]:
# Draw 10,000 samples out of Poisson distribution: n_nohitters
n_nohitters = np.random.poisson(251/115, size=10000)

# Compute number of samples that are seven or greater: n_large
n_large = np.sum(n_nohitters >= 7)

# Compute probability of getting seven or more: p_large
p_large = n_large / 10000.0

# Print the result
print('Probability of seven or more no-hitters:', p_large)
('Probability of seven or more no-hitters:', 0.0041)