From Outcomes to a Probability Distribution
A random variable turns outcomes of a chance process into numerical values. Once the mapping from outcomes to numbers is specified, we need to describe how probability is distributed across those values. Two ideas do this: the support lists the values that can occur with positive probability, and the probability mass function assigns a probability to each such value.
The distinction between an outcome and a random-variable value matters. Several different outcomes may produce the same value, so the probability of that value is the sum of the probabilities of all outcomes that map to it. This aggregation is the main step when constructing a distribution from an outcome table.
A discrete random variable has a finite or countable set of possible values. In this tutorial, the examples use finite outcome tables, so the support and all nonzero pmf values can be listed explicitly. The notation \(p_X(x)\) emphasizes which random variable the pmf describes; when the random variable is clear, \(p(x)\) is also commonly used.
What Makes a Valid Probability Mass Function?
A pmf must satisfy two conditions. First, each probability must be nonnegative. Second, the probabilities across all possible values must add to one. The second condition says that the random variable must take one of its support values with total probability one.
These checks are necessary whether the pmf is given as a formula or as a table. A negative entry cannot be a probability, and a total below or above one means probability has been omitted, double-counted, or otherwise assigned incorrectly. A value of zero is allowed for a value being considered in a larger list, but it is not included in the support as defined here.
Worked Example: Support and PMF for a Die Roll
Suppose a fair six-sided die is rolled once, and \(X\) is the number of dots shown. Find the support and pmf of \(X\).
The outcomes are \(1,2,3,4,5,6\), each with probability \(1/6\), because the die is fair. Here each outcome maps to a different value of \(X\). Therefore,
Every listed probability is nonnegative. The total probability is
Thus the listed values are the support, and this is a valid pmf. The units here are dots, although the values are often simply reported as counts.
Constructing a PMF from a Finite Outcome Table
A finite outcome table gives a direct construction method. Start with mutually exclusive outcomes that cover every possibility in the chance process. Each outcome has an assigned probability, and the random variable has a value for that outcome. To find \(P(X=x)\), add the probabilities of all outcomes whose assigned value is \(x\). Then list the values with positive total probability as the support.
The outcome probabilities must themselves be valid: they must be nonnegative and sum to one. If outcomes are equally likely, each gets the same probability, but equal likelihood is not required. When probabilities differ, the same aggregation rule applies.
Confirm that the listed outcomes are mutually exclusive and exhaustive, and that their probabilities are nonnegative and sum to one.
For each distinct value \(x\), collect every outcome that maps to \(x\).
Set \(p_X(x)\) equal to the sum of the probabilities of the outcomes in that group.
Keep values with positive probability, check that no pmf entry is negative, and confirm the entries sum to one.
Worked Example: Counting Heads in Two Coin Tosses
Suppose a fair coin is tossed twice. Let \(X\) be the number of heads. Find the support and pmf using the outcome table.
The four ordered outcomes are \(HH, HT, TH,\) and \(TT\). Because the coin is fair and the two tosses are independent, each ordered outcome has probability \(1/4\). The random variable maps outcomes to counts as follows.
| Outcome | Probability | Value of \(X\): number of heads |
|---|---|---|
| HH | \(1/4\) | 2 |
| HT | \(1/4\) | 1 |
| TH | \(1/4\) | 1 |
| TT | \(1/4\) | 0 |
The value \(X=1\) comes from two outcomes, \(HT\) and \(TH\). Its probability is their combined probability, not the probability of just one of them:
The other probabilities are \(P(X=0)=P(TT)=1/4\) and \(P(X=2)=P(HH)=1/4\). Thus the support and pmf are
The entries are nonnegative, and
The table's outcome probabilities also sum to one: \(4(1/4)=1\). This confirms that the four listed outcomes account for the entire two-toss experiment.
Why the Aggregation Rule Works
The rule for combining probabilities follows from a basic property of probability: probabilities of disjoint events add. For a fixed value \(x\), the event \(X=x\) consists of precisely those outcomes that the mapping sends to \(x\). The groups of outcomes for different values of \(X\) do not overlap, because one outcome is assigned only one value. Together, they cover the whole outcome space.
For a finite outcome space \(\Omega\), write \(q(\omega)=P(\text{outcome } \omega)\) for each outcome \(\omega\). If \(X(\omega)\) is the value assigned to that outcome, then
is the set of outcomes that produce \(x\). Since these outcomes are mutually exclusive,
Now let \(S\) be the set of values with positive probability. Every outcome belongs to exactly one value group, so summing the group probabilities counts each outcome probability exactly once:
The last equality holds because the outcome probabilities describe the full experiment. Also, each \(p_X(x)\) is a sum of nonnegative outcome probabilities, so \(p_X(x)\geq 0\). This justifies both validity conditions for a pmf constructed from a complete, valid outcome table.
Worked Example: Grouping Unequally Likely Outcomes
Consider a hypothetical chance process with four mutually exclusive and exhaustive outcomes, labelled A, B, C, and D. Their probabilities are \(0.10, 0.25, 0.35,\) and \(0.30\), respectively. A discrete random variable \(X\) assigns the value \(0\) to A, the value \(1\) to both B and C, and the value \(2\) to D. Construct and verify its pmf.
First check the outcome probabilities:
They are all nonnegative and total one, so they specify a valid probability distribution over the four outcomes. To find the pmf, combine probabilities whenever outcomes share a value of \(X\). Outcome A is the only outcome giving \(X=0\), and D is the only outcome giving \(X=2\). Both B and C give \(X=1\):
All three values have positive probability, so \(S=\{0,1,2\}\). The pmf entries are nonnegative and sum to one:
The result is a valid pmf. Notice that it would be incorrect to assign probability \(0.25\) to \(X=1\) just because outcome B has that probability. Outcome C also produces \(X=1\), and both outcome probabilities must be included.
Reading and Reporting a PMF
A pmf can be presented as a rule, a table, or a list of value-probability pairs. In each form, its meaning is the same: \(p_X(x)\) is the probability that the random variable equals \(x\), not the probability of a particular underlying outcome unless that outcome is the only one that produces \(x\).
A table should state what \(X\) measures and include the units or interpretation of its values. For example, a table of counts is clearer when it says “number of heads” rather than giving \(0,1,2\) without explanation. Where helpful, include zero-probability values in an intermediate calculation, but distinguish them from the support. Under the definition used here, a value belongs to the support only when its probability is strictly positive.
For a finite outcome table, a quick check is to compare totals in two ways. The probabilities of all outcomes should sum to one, and after grouping outcomes by the value of \(X\), the pmf entries should also sum to one. If the first total is one but the second is not, the grouping or arithmetic is wrong. If the first total is not one, the problem lies in the outcome probabilities or in whether the table covers the whole process.
Common Mistakes
- Listing outcomes instead of values of \(X\). In two coin tosses the outcomes are \(HH, HT, TH,\) and \(TT\), but the support for the number of heads is \(\{0,1,2\}\). The support lists values of the random variable.
- Failing to combine probabilities. If two or more outcomes produce the same \(x\), add their probabilities to find \(P(X=x)\). Do not copy only one outcome's probability into the pmf.
- Assuming outcomes are equally likely. Equal outcome probabilities are appropriate only when the model says the outcomes are equally likely. With unequal probabilities, use the assigned probabilities in the table.
- Checking only that the total is one. A list with a negative entry might still sum to one, but it cannot be a pmf. Check nonnegativity as well as the total.
- Including a zero-probability value in the support. The support is the set of values with positive probability. A value with probability zero is outside the support under this definition.
Check Your Understanding
For each question, identify the values of the random variable rather than merely listing the underlying outcomes.
- A fair coin is tossed three times, and \(X\) is the number of tails. What is the support of \(X\)?
- Three mutually exclusive outcomes have probabilities \(0.20, 0.30,\) and \(0.50\). The first two both produce \(X=4\), and the third produces \(X=7\). Find the pmf and check its total.
- Can a proposed pmf have entries \(0.4, 0.8,\) and \(-0.2\)? Explain which validity conditions it does or does not satisfy.
- A complete outcome table has probabilities that sum to one. Explain why the pmf probabilities also sum to one after grouping outcomes by the value of \(X\).
- A listed value \(x=5\) has \(P(X=5)=0\). Is \(5\) in the support as defined in this tutorial? Explain.