8.2.1. Constant#

This little example is worth writing out because it gets used all the time. Suppose a random variable $X$ is actually a constant $c$, that is, suppose $P(X=c)=1$. Then the distribution of $X$ puts all its mass on the single value $c$, and $E(X)=c\cdot 1=c$. We just write $E(c)=c$.

8.2.2. Bernoulli and Indicators#

If $X$ has the Bernoulli $(p)$ distribution, then $P(X=1)=p$ and $P(X=0)=1-p$. So

As you saw earlier, zero/one valued random variables are building blocks for other variables and are called indicators.

Let $A$ be any event. Then the indicator of $A$ is the random variable $I_A$ that is 1 if $A$ occurs and 0 if $A$ doesn’t occur. Thus $I_A$ is has the Bernoulli $(P(A))$ distribution and

by our calculation above. Thus every probability is an expectation. We will use this heavily in later sections.

x = [0, 1]
qp = [0.75, 0.25]
bern_1_3 = Table().values(x).probabilities(qp)
Plot(bern_1_3, show_ev=True)
plt.title('Bernoulli (0.25)');

8.2.3. Uniform on an Interval of Integers#

Let $a$ and $b$ be two integers such that $a<b$. If $X$ has the uniform distribution on the integers $a,a+1,a+2,\dots ,b$, then by symmetry, $E(X)$ is half-way between $a$ and $b$. That’s where the probability histogram will balance. So

For example, if $X$ has the uniform distribution on $1,2,\dots ,n$, then

An instance of this is if $X$ is the number of spots on one roll of a die. Then $E(X)=3.5$.

If instead $X$ is uniform on $0,1,2,\dots ,n$, then

x = np.arange(10)
probs = 0.1*np.ones(10)
unif_10 = Table().values(x).probabilities(probs)
Plot(unif_10, show_ev=True)
plt.title('Uniform on Integers 0, 1, 2, ..., 9');

8.2.4. Poisson#

Let $X$ have the Poisson $(\mu )$ distribution. Then

We now have an important new interpretation of the parameter of the Poisson distribution. We saw earlier it was close to the mode; now we know that it is also the balance point or expectation of the distribution. The notation $\mu$ was chosen to stand for “mean”.

k = np.arange(15)
poi_2_probs = stats.poisson.pmf(k, 2)
dist_poi_2 = Table().values(k).probabilities(poi_2_probs)
Plot(dist_poi_2, show_ev=True)
plt.title('Poisson (2)');

8.2.5. Tail Sum Formula#

To find the expectation of a non-negative integer valued random variable it is sometimes quicker to use a formula that uses only the right hand tail probabilities $P(X>k)=1-F(k)$ where $F$ is the cdf of $X$ and $k\ge 0$.

For non-negative integer valued $X$,

Rewrite this as

Add the terms along each column on the right hand side to get the tail sum formula for the expectation of a non-negative integer valued random variable.

This formula comes in handy if a random variable has tail probabilities that are easy to find and also easy to sum.

8.2.6. Geometric#

In a sequence of i.i.d. Bernoulli $(p)$ trials, let $X$ be the number of trials till the first success. We will use the word “till” to mean “up to and including”.

Let $q=1-p$. The distribution of $X$ is given by

This is called the geometric $(p)$ distribution on $\{1,2,3,\dots \}$ because the probabilities are terms in a geometric series.

The right tails of $X$ are simple because for each $k\ge 1$,

The formula is also true for $k=0$ because $P(X>0)=1=q^0$.

By the tail sum formula,