Examples of expectations of discrete random variables

Let’s revisit some of the distributions we encountered in Lecture 9 and calculate their expectations. We will do it both analytically, and using scipy.stats.

Example: Expectation of a Bernoulli random variable

Take a Bernoulli random variable:

\[ X \sim \text{Bernoulli}(\theta). \]

Then:

\[\begin{split} \begin{split} \mathbf{E}[X] &= \sum_x x p(x)\\ &= 0\cdot p(X=0) + 1\cdot p(X=1)\\ &= 0 \cdot (1-\theta) + 1\cdot \theta\\ &= \theta. \end{split} \end{split}\]

And here is how we can do it using scipy.stats:

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import matplotlib.pyplot as plt
%matplotlib inline
import seaborn as sns
sns.set(rc={"figure.dpi":100, 'savefig.dpi':300})
sns.set_context('notebook')
sns.set_style("ticks")
from IPython.display import set_matplotlib_formats
set_matplotlib_formats('retina', 'svg')
import numpy as np
import scipy.stats as st

theta = 0.7
X = st.bernoulli(theta)

Now that we have made the random variable we can get its expectation by X.expect():

print('E[X] = {0:1.2f}'.format(X.expect()))

E[X] = 0.70

Let’s visualize the PMF and the expectation on the same plot:

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fig, ax = plt.subplots()
xs = np.arange(2)
ax.vlines(xs, 0, X.pmf(xs), label='PMF of $X$')
ax.plot(X.expect(), 0, 'ro', label='$\mathbf{E}[X]$')
ax.set_xlabel('$x$')
ax.set_ylabel('$p(x)$')
ax.set_title(r'Bernoulli$(\theta={0:1.2f})$'.format(theta))
plt.legend(loc='upper left');

Example: Expectation of a Categorical random variable

Take a Categorical random variable:

\[ X \sim \text{Categorical}(0.1, 0.3, 0.4, 0.2). \]

The expectation is:

\[\begin{split} \begin{split} \mathbf{E}[X] &= \sum_x x p(x)\\ &= 0\cdot p(X=0) + 1 \cdot p(X=1) + 2\cdot p(X=2) + 3 \cdot p(X=3)\\ &= 0 \cdot 0.1 + 1 \cdot 0.3 + 2 \cdot 0.4 + 3 \cdot 0.2\\ &= 1.7. \end{split} \end{split}\]

Here is how we can find it with Python:

import numpy as np
# The values X can take
xs = np.arange(4)
print('X values: ', xs)
# The probability for each value
ps = np.array([0.1, 0.3, 0.4, 0.2])
print('X probabilities: ', ps)
# And the expectation in a single line
E_X = np.sum(xs * ps)
print('E[X] = {0:1.2f}'.format(E_X))

X values:  [0 1 2 3]
X probabilities:  [0.1 0.3 0.4 0.2]
E[X] = 1.70

Alternatively, we could use scipy.stats:

X = st.rv_discrete(name='X', values=(xs, ps))
print('E[X] = {0:1.2f}'.format(X.expect()))

E[X] = 1.70

And a visualization:

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fig, ax = plt.subplots()
ax.vlines(xs, 0, X.pmf(xs), label='PMF of $X$')
ax.plot(X.expect(), 0, 'ro', label='$\mathbf{E}[X]$')
ax.set_xlabel('$x$')
ax.set_ylabel('$p(x)$')
ax.set_title('Categorical$(0.1, 0.3, 0.4, 0.2)$'.format(theta))
plt.legend(loc='upper left');

Example: Expectation of a Binomial random variable

Take a Binomial random variable:

\[ X\sim \text{Binomial}(n, \theta). \]

The expectation is:

\[ \mathbf{E}[X] = n\theta. \]

This makes sense. Remember that \(X\) is the number of successes in a binary experiment that is repeated \(n\) times. Each binary experiment has probability of success equal to \(\theta\).

Here is how we can get it with scipy.stats:

n = 5       
theta = 0.6
X = st.binom(n, theta)
print('E[X] = {0:1.2f}'.format(X.expect()))
print('Compare to n * theta = {0:1.2f}'.format(n * theta))

E[X] = 3.00
Compare to n * theta = 3.00

Just like before, let’s visualize the PMF, the mean, and put two markers on two standard deviations below and above the mean.

fig, ax = plt.subplots()
xs = np.arange(n+1)
ax.vlines(xs, 0, X.pmf(xs), label='PMF of $X$')
ax.plot(X.expect(), 0, 'ro', label='$\mathbf{E}[X]$')
ax.set_xlabel('$x$')
ax.set_ylabel('$p(x)$')
ax.set_title(r'Binomial$(n={0:d}, \theta={1:1.2f})$'.format(n, theta))
plt.legend(loc='upper left');

Questions

• Rerun the case of the Binomial with \(n=50\). Does the shape of the PMF you get look familiar?

Example: Expectation of a Poisson random variable

Take Poisson random variable:

\[ X\sim \operatorname{Poisson}(\lambda). \]

The expectation is:

\[\begin{split} \begin{split} \mathbf{E}[X] &= \sum_{k=0}^\infty k p(X=k)\\ &= \sum_{k=0}^\infty k \frac{\lambda^ke^{-\lambda}}{k!}. \end{split} \end{split}\]

Again, with quite a bit of algebra you can show that:

\[ \mathbf{E}[X] = \lambda. \]

Let’s also do it in scipy.stats:

lam = 2.0
X = st.poisson(lam)
print('E[X] = {0:1.2f}'.format(X.expect()))

E[X] = 2.00

And let’s visualize the PMF and the expectation together:

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fig, ax = plt.subplots()
xs = np.arange(X.ppf(0.9999)) # I will explain this later
ax.vlines(xs, 0, X.pmf(xs), label='PMF of $X$')
ax.plot(X.expect(), 0, 'ro', label='$\mathbf{E}[X]$')
ax.set_xlabel('$x$')
ax.set_ylabel('$p(x)$')
ax.set_title(r'Poisson$(\lambda={0:1.2f})$'.format(lam))
plt.legend(loc='upper right');

Question

• Rerun the case for the Poisson with a rate parameter \(\lambda = 50\). Does the shape look familiar?