Countable Sample Space Workshop: Integer Outcomes

From Integer Weights to Event Probabilities

A countable outcome space can have infinitely many possible integer outcomes, so an event probability may require adding infinitely many point probabilities. The practical method is direct: identify the integers in the event, find the probability assigned to each, and evaluate the resulting series. This tutorial works through parity, threshold, and interval events in several integer-valued models.

The weights themselves are not automatically probabilities. They must be nonnegative and have a finite, positive total. Dividing each weight by that total gives point probabilities whose sum is one.

Definition: Let \(\Omega\) be a countable set of integers, and assign a weight \(w_k\geq 0\) to each \(k\in\Omega\). Suppose \(0<S<\infty\), where \(S=\sum_{k\in\Omega}w_k\). The normalized point probabilities are \(p_k=w_k/S\). For an event \(A\subseteq\Omega\), its probability is \(P(A)=\sum_{k\in A}p_k\).

The sum defining \(P(A)\) is taken over precisely the outcomes in \(A\). In particular, a threshold event such as \(A=\{k\in\Omega:k\geq m\}\) includes \(m\), while \(A=\{k\in\Omega:k>m\}\) does not. For an interval, the stated endpoints also determine which integer terms belong in the sum.

Worked Examples on Nonnegative Integers

Worked Example: Geometric Weights, Parity, and a Finite Interval

Let \(\Omega=\{0,1,2,\ldots\}\), with weight \(w_k=(1/2)^k\). Find the probabilities of an even outcome, an outcome at least 3, and an outcome from 2 through 5 inclusive.

First normalize. The geometric series gives

S=\sum_{k=0}^{\infty}(1/2)^k=\frac{1}{1-1/2}=2, \qquad p_k=\frac{(1/2)^k}{2}=2^{-(k+1)}.

For an even outcome, sum over \(k=0,2,4,\ldots\). The first point probability is \(p_0=1/2\), and each successive even-index term is one quarter of the preceding term:

P(\text{even})=\sum_{j=0}^{\infty}p_{2j} =\frac12\sum_{j=0}^{\infty}(1/4)^j =\frac12\cdot\frac{1}{1-1/4} =\frac23.

For the threshold \(k\geq 3\), the first included probability is \(p_3=1/16\), and subsequent terms have ratio \(1/2\):

P(k\geq 3)=\sum_{k=3}^{\infty}2^{-(k+1)} =\frac{1}{16}\cdot\frac{1}{1-1/2} =\frac18.

The inclusive interval from 2 through 5 contains exactly four outcomes, so this is a finite sum:

P(2\leq k\leq 5) =p_2+p_3+p_4+p_5 =\frac18+\frac{1}{16}+\frac{1}{32}+\frac{1}{64} =\frac{15}{64}.

As a check on the last result, the interval probability is less than one, and each of its four positive terms is included once.

Worked Example: Telescoping Weights and a Tail Probability

On \(\Omega=\{0,1,2,\ldots\}\), assign \(w_k=1/((k+1)(k+2))\). Find the normalizing total, \(P(k\geq 4)\), and \(P(2\leq k\leq 5)\).

Use the partial-fraction identity

\frac{1}{(k+1)(k+2)} =\frac{1}{k+1}-\frac{1}{k+2}.

Consequently, for an integer \(N\geq 0\),

\sum_{k=0}^{N}w_k =\sum_{k=0}^{N}\left(\frac{1}{k+1}-\frac{1}{k+2}\right) =1-\frac{1}{N+2}.

As \(N\) tends to infinity, the final term tends to zero, so the total weight is \(S=1\). The weights are already the point probabilities. For \(k\geq 4\), take the limit of the corresponding finite tail sum:

\sum_{k=4}^{N}\left(\frac{1}{k+1}-\frac{1}{k+2}\right) =\frac15-\frac{1}{N+2} \ \longrightarrow\ \frac15.

Thus \(P(k\geq 4)=1/5\). The inclusive interval from 2 through 5 telescopes as well:

P(2\leq k\leq 5) =\sum_{k=2}^{5}\left(\frac{1}{k+1}-\frac{1}{k+2}\right) =\frac13-\frac17 =\frac{4}{21}.

The tail begins with \(k=4\), whose weight is \(1/(5\cdot6)=1/30\); summing the entire tail gives \(1/5\), which is larger, as it should be.

Geometric Tails and Intervals

The same calculation as in the first example gives a useful general rule. Suppose the support is \(\{0,1,2,\ldots\}\) and the weights are \(w_k=r^k\), where \(0\leq r<1\). For \(r=0\), interpret the weights as \(w_0=1\) and \(w_k=0\) for \(k\geq1\). For \(0<r<1\), the geometric-series formula gives total weight \(1/(1-r)\), so \(p_k=(1-r)r^k\).

Theorem: Under these assumptions, for integers \(m\geq0\) and \(0\leq a\leq b\), the probabilities are \(P(k\geq m)=r^m\) and \(P(a\leq k\leq b)=r^a(1-r^{b-a+1})\) when \(0<r<1\). When \(r=0\), the tail probability is 1 if \(m=0\) and 0 if \(m\geq1\); the interval probability is 1 if it includes 0 and 0 otherwise.

Proof. For \(0<r<1\), normalization gives \(p_k=(1-r)r^k\). For a finite integer \(N\geq m\),

\sum_{k=m}^{N}(1-r)r^k =(1-r)r^m\frac{1-r^{N-m+1}}{1-r} =r^m(1-r^{N-m+1}).

Because \(0<r<1\), \(r^{N-m+1}\) tends to zero as \(N\) tends to infinity. The tail sum is therefore \(r^m\). For a finite interval, put \(j=k-a\) and sum the \(b-a+1\) terms:

\sum_{k=a}^{b}(1-r)r^k =(1-r)r^a\sum_{j=0}^{b-a}r^j =r^a(1-r^{b-a+1}).

When \(r=0\), all probability is at outcome 0, so the stated tail and interval cases follow by checking whether each event contains 0. \(\square\)

This result is useful because it separates the structure of the model from the particular numbers in an event. Once the geometric parameter is known, a threshold requires only its smallest included integer, and an inclusive finite interval requires its first and last included integers.

Worked Example: Symmetric Geometric Weights on All Integers

Worked Example: Parity and Events on the Integer Line

Let \(\Omega=\mathbb{Z}\), and assign \(w_k=(1/2)^{|k|}\). Find the probability of an even integer, the probability of \(k\geq 2\), and the probability of \(-1\leq k\leq 2\).

The total weight includes the outcome 0 once and each positive absolute value twice:

S=1+2\sum_{j=1}^{\infty}(1/2)^j =1+2\cdot 1 =3.

Thus \(p_k=(1/2)^{|k|}/3\). For even integers, include 0 and both signs of every positive even integer:

P(\text{even}) =\frac{1+2\sum_{j=1}^{\infty}(1/2)^{2j}}{3} =\frac{1+2\left(\frac{1/4}{1-1/4}\right)}{3} =\frac{1+2/3}{3} =\frac59.

For \(k\geq2\), only positive integers \(2,3,\ldots\) are included:

P(k\geq2) =\frac13\sum_{k=2}^{\infty}(1/2)^k =\frac13\cdot\frac{1/4}{1-1/2} =\frac16.

The interval \([-1,2]\) contains the four integers \(-1,0,1,2\), with unnormalized weights \(1/2,1,1/2,1/4\), respectively. Therefore,

P(-1\leq k\leq2) =\frac{1/2+1+1/2+1/4}{3} =\frac{9/4}{3} =\frac34.

This model is symmetric about zero: the probabilities of positive and negative integers with the same absolute value match. The interval calculation uses that symmetry in the weights, but it still lists exactly the outcomes in the requested interval.

Checks That Prevent Errors

An infinite support does not mean every event requires a difficult infinite sum. A geometric tail can be evaluated by the geometric-series formula, while weights written as differences may telescope. A finite interval, by contrast, is often safest to calculate by listing every included integer and its point probability.

Normalize before summing event weights. If the total weight is \(S\), divide the event's total weight by \(S\), unless \(S=1\) has been established.
Check the support. An event described by an interval contains only the integers in that interval that also belong to \(\Omega\).
Read endpoints literally. The event \(k\geq m\) includes \(m\); \(k>m\) starts at \(m+1\). Inclusive interval endpoints are both counted.
Do not confuse weights with probabilities. A weight such as \(r^k\) is not a point probability until it has been divided by the total weight.

A common mistake is to calculate a tail by adding the first few point probabilities and treating that partial sum as the whole tail. For geometric weights, the omitted terms are positive, so such a partial sum is too small. The correction is to use the infinite geometric-series formula, or to compute the finite partial sum and take its limit as shown in the proof. Another frequent error is to start a threshold at the wrong index: for \(k\geq m\), the first term is the probability at \(m\), not at \(m+1\).

Key takeaway: For integer outcomes, translate the event into its exact set of integers, use the normalized point probabilities, and evaluate the resulting sum with the appropriate series identity. A quick check of the support, normalization, and endpoints catches many errors.

Check Your Understanding

For each question, identify the outcomes included before calculating.

On \(\{0,1,2,\ldots\}\), weights are \(w_k=(1/3)^k\). What is the total weight, and what is the probability of an outcome at least 2?
Using the model in Question 1, calculate the probability of an even outcome.
Weights on \(\{0,1,2,\ldots\}\) are \(w_k=1/((k+1)(k+2))\). Find the probability of \(3\leq k\leq6\).
For weights \(w_k=(1/2)^{|k|}\) on \(\mathbb{Z}\), list the outcomes in \(-2\leq k<1\). Why is \(k=1\) excluded?
A student computes a threshold probability by adding only the first three terms of an infinite tail. What is missing from that calculation, and how can it be corrected for geometric weights?