From Outcomes to a Probability Law
On a countable sample space, a probability model can be specified by assigning a number to each individual outcome. The assignment must satisfy two conditions: no assigned number is negative, and the numbers over the entire sample space sum to one. Once those point probabilities are specified, an event’s probability is obtained by adding the probabilities of the outcomes it contains.
This representation is useful because many experiments have infinitely many possible outcomes: a count of attempts before success, an integer-valued measurement, or a waiting time recorded in discrete units. In such models, an event may contain infinitely many outcomes. The sum defining its probability must therefore be understood carefully, rather than treated as an ordinary finite addition.
For a finite set, these sums are ordinary finite sums. For a countably infinite set, the sum of nonnegative terms has a precise meaning. If \(C\) is a countable set of outcomes, define
In words, take the supremum of all finite partial sums. If the outcomes are listed as \(\omega_1,\omega_2,\ldots\), this is also the limit of the partial sums \(\sum_{i=1}^m p(\omega_i)\) as \(m\) increases. The terms are nonnegative, so those partial sums cannot decrease. The supremum definition also shows that the value does not depend on the order used to list the outcomes: every finite subset appears within some sufficiently long initial segment of any listing.
Basic Consequences of the Assignment
A valid assignment puts all the probability mass somewhere in \(\Omega\), and nowhere assigns a negative amount. These requirements immediately constrain individual and event probabilities. The following result records the finite-event properties we will use throughout the examples.
Proof. The empty sum is zero, so \(P(\varnothing)=0\); normalization gives \(P(\Omega)=1\). Since all terms are nonnegative, every finite subsum over \(A\) is at most the sum over \(\Omega\), which is one. Taking the supremum over those finite subsums gives \(0\leq P(A)\leq1\).
If \(A\cap B=\varnothing\), every finite subset of \(A\cup B\) splits into a finite subset of \(A\) and a finite subset of \(B\). Thus its sum is at most \(P(A)+P(B)\), giving \(P(A\cup B)\leq P(A)+P(B)\). Conversely, finite subsets of \(A\) and \(B\) can be combined into a finite subset of \(A\cup B\). Taking suprema over the two finite subsets gives \(P(A\cup B)\geq P(A)+P(B)\). Therefore equality holds. If \(A\subseteq B\), every finite subset of \(A\) is also a finite subset of \(B\), so \(P(A)\leq P(B)\). Finally, \(A\) and \(A^c\) are disjoint and have union \(\Omega\). Finite additivity and normalization yield \(P(A)+P(A^c)=1\), hence \(P(A^c)=1-P(A)\). \(\square\)
A point probability cannot exceed one: apply the event bound to the singleton \(\{\omega\}\). The assignment also determines every event probability uniquely, since the event is a set of outcomes and its probability is the sum of exactly their assigned masses. This is why checking the outcome-level assignment is central: an error in even one point probability can affect many event calculations.
Worked Examples
Worked Example: A Geometric Assignment on the Nonnegative Integers
Let \(\Omega=\{0,1,2,\ldots\}\), and assign \(p(k)=\frac14(\frac34)^k\) for \(k\geq0\). First check that this is a probability assignment, then find the probability that the outcome is at least two.
For \(0\leq r<1\), the finite geometric sum is \(\sum_{k=0}^m r^k=(1-r^{m+1})/(1-r)\). Since \(r^{m+1}\to0\), its infinite sum is \(1/(1-r)\). Therefore,
Every term is nonnegative, so the assignment is valid. For the event \(A=\{2,3,4,\ldots\}\), factor out the first term of the tail:
As a check, \(P(\{0,1\})=\frac14+\frac{3}{16}=\frac{7}{16}\), and the two events partition \(\Omega\). Their probabilities sum to \(\frac{7}{16}+\frac{9}{16}=1\).
Worked Example: A Probability Assignment on All Integers
Let \(\Omega=\mathbb Z\). Define \(p(0)=\frac12\), and for each integer \(j\geq1\), set \(p(j)=p(-j)=\frac18(\frac12)^{j-1}\). Verify the assignment and calculate \(P(X<0)\) and \(P(|X|\geq3)\), where \(X\) denotes the integer-valued outcome.
The mass at zero is \(\frac12\). On the positive integers, the total mass is \(\frac18\sum_{j=1}^{\infty}(\frac12)^{j-1}=\frac18\cdot2=\frac14\). The negative integers receive the same total mass, also \(\frac14\). Hence the total is
All point probabilities are nonnegative, so the assignment is valid. The event \(X<0\) contains outcomes \(-1,-2,\ldots\), and its probability is the mass on the negative side:
For \(|X|\geq3\), both tails begin at absolute value three. On either side, the exponent \(j-1\) begins at two, so one tail has mass \[ \frac18\sum_{j=3}^{\infty}\left(\frac12\right)^{j-1} =\frac18\sum_{r=2}^{\infty}\left(\frac12\right)^r =\frac18\cdot\frac{\frac14}{1-\frac12} =\frac1{16}. \] The two tails are disjoint and have equal mass. Therefore \(P(|X|\geq3)=\frac1{16}+\frac1{16}=\frac18\).
Worked Example: A Telescoping Assignment
On \(\Omega=\{1,2,3,\ldots\}\), define \(p(n)=1/[n(n+1)]\). Verify that the point probabilities sum to one, then calculate \(P(X\geq4)\) and \(P(X\leq3)\).
The identity \(\frac{1}{n(n+1)}=\frac1n-\frac1{n+1}\) turns the finite partial sum into a telescoping sum:
As \(m\to\infty\), \(1/(m+1)\to0\), so \(\sum_{n=1}^{\infty}p(n)=1\). Each term is positive, which verifies the assignment.
For the tail beginning at four, the same cancellation gives
The first three point probabilities are \(p(1)=1/2\), \(p(2)=1/6\), and \(p(3)=1/12\). Thus \(P(X\leq3)=1/2+1/6+1/12=6/12+2/12+1/12=9/12=3/4\). The events \(X\leq3\) and \(X\geq4\) partition the sample space, and \(3/4+1/4=1\), as required.
Using the Assignment Carefully
The practical calculation has two parts. First identify exactly which outcomes belong to the event. Then add the point probabilities for those outcomes. When the event has infinitely many outcomes, organize the sum using a geometric series, a telescoping identity, or another appropriate series calculation. Writing down the event before manipulating the sum helps prevent missed terms and incorrect limits.
For instance, the event \(X\geq4\) in the telescoping example contains \(4,5,6,\ldots\), not just the outcomes observed in a finite initial segment. A partial sum such as \(\sum_{n=4}^{m}p(n)\) is not yet the probability of the whole event; it is the mass through \(m\). The event probability is the limit of those partial sums, with the limit justified by the explicit telescoping expression.
Another frequent error is to confuse a point probability with an event probability. The value \(p(3)\) is the probability of the singleton event \(\{3\}\), not automatically the probability of a broader event such as \(X\geq3\). That broader event includes every outcome \(3,4,5,\ldots\), so its probability requires the sum of all those point masses. Conversely, an event with just two outcomes, such as \(\{1,3\}\), has probability \(p(1)+p(3)\).
The countable sum definition is what makes an infinite event calculation well-defined. The corresponding series rule for disjoint events is a central consequence of countable additivity; it explains why the point-mass representation behaves consistently across countable partitions. Here the operational task is to specify the masses and calculate the event sum, as in the examples above.
Check Your Understanding
For each question, identify the outcomes included in the event before calculating its probability.
- On \(\{0,1,2,\ldots\}\), let \(p(k)=\frac13(\frac23)^k\). Verify normalization and find \(P(X\geq2)\).
- For the integer assignment with \(p(0)=1/2\) and \(p(\pm j)=\frac18(\frac12)^{j-1}\), what is \(P(X>0)\)? Show the series used.
- For \(p(n)=1/[n(n+1)]\) on the positive integers, calculate \(P(X\geq2)\) by a telescoping sum.
- Explain why assigning probability \(1/100\) to every integer outcome does not define a probability assignment on \(\mathbb Z\).
- If \(p(1)=0.2\), \(p(2)=0.3\), and \(p(3)=0.5\) on \(\{1,2,3\}\), find the probability of the event \(\{1,3\}\).