Building a CDF on a Finite Outcome Space

From Possible Values to a CDF

When a random variable can take only finitely many values, its cumulative distribution function can be built directly from a short list. The main work is to identify the distinct numerical values the variable can take, determine the probability of each value, put those values in increasing order, and accumulate the probabilities.

The order and the word “distinct” matter. Several outcomes in the underlying experiment may produce the same value of \(X\). Their probabilities must first be combined. Then the CDF at a threshold \(x\) includes the probability of every value of \(X\) that is less than or equal to \(x\).

Here \(q_j\) is the probability of a value of \(X\), not necessarily the probability of a single elementary outcome. To find \(q_j\), collect every outcome \(\omega\) for which \(X(\omega)=x_j\) and add the probabilities of those outcomes. In a finite probability model, this is a finite addition.

A Reliable Construction

Use the following procedure whenever the underlying outcome space is finite. If the experiment’s outcomes have probabilities \(P(\{\omega\})\), group them according to the value \(X(\omega)\). Sort the resulting distinct values as \(x_1<\cdots<x_k\), and set \(q_j=P(X=x_j)\). The accumulated totals are \(Q_j=q_1+\cdots+q_j\).

For an ordered list \(x_1<\cdots<x_k\), the piecewise form is

This expression is simply a compact way to record which listed values are at or below \(x\). Between two consecutive possible values, changing the threshold does not add any new probability. At a listed value, its probability is included.

Proof. The event \(\{X=x_j\}\) consists precisely of the outcomes \(\omega_i\) whose assigned value is \(x_j\). These single-outcome events are disjoint, so finite additivity gives \(P(X=x_j)=\sum_{\{i:X(\omega_i)=x_j\}}p_i\). Likewise, \(\{X\leq x\}\) consists precisely of the outcomes whose assigned values are at most \(x\). The single-outcome events in this set are disjoint, so its probability is the sum of their probabilities: \(F(x)=P(X\leq x)=\sum_{\{i:X(\omega_i)\leq x\}}p_i\). Grouping the terms in this sum by their common value \(x_j\) gives the accumulated-probability formula. \(\square\)

Why the Cumulative Total Is the Right One

The event defining the CDF at \(x\) is \(\{X\leq x\}\), not just \(\{X=x\}\). When the possible values are ordered, this event is the union of all value-events at or below \(x\). Those events are disjoint: \(X\) cannot equal two different values at once. Consequently, their probabilities add. This explains both why the accumulation works and why the order of the values determines the piecewise intervals.

For \(x<x_1\), no possible value is at or below \(x\), so the sum is empty and \(F(x)=0\). For \(x\geq x_k\), every possible value is at or below \(x\), so the sum is \(q_1+\cdots+q_k=1\). These cases complete the finite piecewise description.

Common Mistake: Forgetting to Group or Include Equality

A frequent error is to put one row in the CDF table for every elementary outcome, even when several outcomes give the same value of \(X\). For example, in the two-toss example, counting the outcome \(HT\) but not \(TH\) as the value 1 would incorrectly assign probability \(1/4\), rather than \(1/2\), to \(X=1\). The correction is to collect all outcomes with that value and add their probabilities before forming cumulative totals.

Another error is to evaluate \(F(1)\) as \(P(X<1)\). That is the wrong event: by definition, \(F(1)=P(X\leq 1)\). In the two-toss example, \(P(X<1)=1/4\), while \(F(1)=P(X=0)+P(X=1)=1/4+1/2=3/4\). At a threshold equal to a listed value, include that value’s probability.

A practical check is to verify that the grouped probabilities sum to 1 and that each cumulative total adds exactly the next probability. Then, for a requested threshold, mark which listed values satisfy \(x_j\leq x\) and add only their probabilities.