- Random number generators and hacker statistics
- Probability distributions and stories: The Binomial distribution
- Poisson processes and the Poisson distribution
import pandas as pd import numpy as np import matplotlib.pyplot as plt import seaborn as sns %matplotlib inline sns.set() df = pd.read_csv('./dataset/iris.csv') renamed_columns = ['sepal length (cm)', 'sepal width (cm)', 'petal length (cm)', 'petal width (cm)', 'species'] df.columns = renamed_columns versicolor_petal_length = df[df['species'] == 'Versicolor']['petal length (cm)'] setosa_petal_length = df[df['species'] == 'Setosa']['petal length (cm)'] virginica_petal_length = df[df['species'] == 'Virginica']['petal length (cm)'] versicolor_petal_width = df[df['species'] == 'Versicolor']['petal width (cm)']
- Uses simulated repeated measurements to compute probabilities
- E.g. Coin Flips
- Suite of functions based on random number generation
np.random.random(): draw a number between 0 and 1
- An experiment that has two options, "success" (True) and "failure" (False).
Random number seed
- Integer fed into random number generating algorithm
- Manually seed random number generator if you need reproducibility
- Specified using
Hacker stats probabilities
- Determine how to simulate data
- Simulate many many times
- Probability is approximately fraction of trials with the outcome of interest
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
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.
np.random.seed(42) # 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)
You can think of a Bernoulli trial as a flip of a possibly biased coin. Specifically, each coin flip has a probability p of landing heads (success) and probability 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.
def perform_bernoulli_trials(n, p): """Perform n Bernoulli trials with success probability p and return number of successes.""" # Initialize number of successes: n_successes 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
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.
np.random.seed(42) # 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, density=True) _ = plt.xlabel('number of defaults out of 100 loans') _ = plt.ylabel('probability')
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) / n return x, y
x, y = ecdf(n_defaults) # Plot the ECDF with labeled axes _ = plt.plot(x, y, marker='.', linestyle='none') _ = plt.xlabel('x') _ = plt.ylabel('y') # 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 / len(n_defaults))
Probability of losing money = 0.022
- Probability mass function (PMF)
- The set of probabilities of discrete outcomes
- Probability distribution
- A mathmatical description of outcomes
- Binomial distribution
- The number r of successes in n Bernoulli trials with probability p of success, is Binomially distributed
- The number r of heads in 4 coin flips with probability 0.5 of heads, is Binomially distributed
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(42)) to save you typing that each time.
n_defaults = np.random.binomial(n=100, p=0.05, size=10000) # Compute CDF: x, y x, y = ecdf(n_defaults) # Plot the CDF with axis labels _ = plt.plot(x, y, marker='.', linestyle='none') _ = plt.xlabel('x') _ = plt.ylabel('y')
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
bins = np.arange(0, max(n_defaults) + 2) - 0.5 # Generate histogram _ = plt.hist(n_defaults, bins=bins, density=True) # Label axes _ = plt.xlabel('x') _ = plt.ylabel('y') plt.savefig('../images/bin-dist.png')
- Poisson process
- The timing of the next event is completely independent of when the previous event happened
- Natural births in a given hospital
- Hit on a website during a given hour
- Meteor strikes
- Molecular collisions in a gas
- Aviation incidents
- Buses in Poissonville
- Poisson Distribution
- The number r of arrivals of a Poisson process in a given time interval with average rate of? arrivals per interval is Poisson distributed.
- The number r of hits on a website in one hour with an average hit rate of 6 hits per hour is Poisson distributed.
- Limit of the Binomial distribution for low probability of success and large number of trials
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 np approximates a Binomial distribution for n Bernoulli trials with probability p of success (with n large and p 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 n and p such that np=10.
samples_poisson = np.random.poisson(10, 10000) # Print the mean and standard deviation print('Poisson: ', np.mean(samples_poisson), np.std(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), np.std(samples_binomial))
Poisson: 9.9895 3.176159591393354 n = 20 Binom: 10.0621 2.240277569856021 n = 100 Binom: 10.0432 2.988567175085747 n = 1000 Binom: 10.0106 3.1406508306400442
n_nohitters = np.random.poisson(251/115, 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 # Print the result print('Probability of seven or more no-hitters:', p_large)
Probability of seven or more no-hitters: 0.0071