Point Masses and Degenerate Laws

Point Masses Describe Probability Concentrated at Locations

Some random variables can take many possible values, while others can take only a few. In the simplest case, a random variable is constant: it equals the same number with probability one. More generally, a finite atomic law assigns positive probability to a finite list of locations. In either case, the probability is attached to specific values, so an event’s probability can be calculated by adding the weights at the locations that belong to that event.

This representation is useful whenever probability is concentrated on isolated outcomes, such as a fixed measurement or a finite set of possible payouts. It also clarifies what “degenerate” means: the variable has no uncertainty in its value, even though its value is still a random variable in the formal sense.

Definition: For a real number \(a\), the point mass at \(a\), denoted \(\delta_a\), is the probability distribution that assigns probability one to the value \(a\). For any Borel set \(A\subseteq\mathbb{R}\), \(\delta_a(A)=1\) if \(a\in A\), and \(\delta_a(A)=0\) if \(a\notin A\). A random variable \(X\) has a degenerate law at \(a\) if \(P(X=a)=1\); its distribution is then \(\delta_a\).

A point mass is a probability distribution, not a density value. Its entire probability is located at one point. A finite atomic law extends this idea to several locations: it assigns a probability weight to each one, and the weights together account for all probability.

Definition: A finite atomic law with distinct locations \(x_1,\ldots,x_k\) and weights \(p_1,\ldots,p_k\) is written \[ P_X=\sum_{i=1}^{k}p_i\delta_{x_i}, \] where \(p_i>0\) for every \(i\) and \(\sum_{i=1}^{k}p_i=1\). The locations \(x_i\) form the support of this law, and \(p_i=P(X=x_i)\) is the mass at \(x_i\). A location with probability zero is not included in the support.

The locations and weights provide everything needed to calculate probabilities of events involving \(X\). For example, an interval event includes just those support locations lying in the interval. Whether an interval endpoint is included matters only if that endpoint itself is one of the locations with positive mass.

Calculating Probabilities From the Masses

The basic calculation follows from the fact that the possible values \(x_1,\ldots,x_k\) are distinct and the events \(\{X=x_i\}\) are disjoint. For any Borel set \(A\), the event \(\{X\in A\}\) occurs precisely when \(X=x_i\) for one of the locations \(x_i\) that lies in \(A\). Thus its probability is the sum of the corresponding weights.

Theorem: If \(P_X=\sum_{i=1}^{k}p_i\delta_{x_i}\), then for every Borel set \(A\subseteq\mathbb{R}\), \[ P(X\in A)=\sum_{i=1}^{k}p_i\,\mathbf{1}\{x_i\in A\}, \] where \(\mathbf{1}\{x_i\in A\}\) equals 1 when \(x_i\in A\) and 0 otherwise.

Proof. The event \(\{X\in A\}\) is the disjoint union of the events \(\{X=x_i\}\) for which \(x_i\in A\). By finite additivity of probability, \[ P(X\in A)=\sum_{i:\,x_i\in A}P(X=x_i) =\sum_{i=1}^{k}p_i\,\mathbf{1}\{x_i\in A\}. \] The second sum simply includes a weight when its location belongs to \(A\), and contributes zero otherwise. \(\square\)

Taking \(A=\mathbb{R}\) in the formula gives \(1=\sum_{i=1}^{k}p_i\), as required for a probability distribution. Taking \(A=\{x_j\}\) gives \(P(X=x_j)=p_j\), because the locations are distinct. If a requested value is not among the support locations, its point probability is zero.

Worked Example: A Constant Measurement

A calibration check returns a fixed reference value of 12.5 units every time it is performed. Model the recorded value \(X\) as constant and find the probabilities of \(X=12.5\), \(X\in[12,13]\), and \(X>12.5\).

Since the measurement is fixed, \(P(X=12.5)=1\). The law is the single point mass \(P_X=\delta_{12.5}\), with one support location and weight 1.

The interval \([12,13]\) contains the support location 12.5, so \[ P(X\in[12,13])=1. \] The event \(X>12.5\) does not contain that location, so \[ P(X>12.5)=0. \] The same reasoning gives \(P(X=12.4)=0\): the event asks for a value different from the constant value.

Worked Example: Adding Weights Inside an Interval

Suppose \(X\) has support locations 0, 2, and 5, with respective probabilities \(1/4\), \(1/2\), and \(1/4\). Find \(P(1<X\leq5)\), \(P(X=2)\), and \(P(X\notin\{0,5\})\).

First check the weights: \(1/4+1/2+1/4=1\). For \(1<X\leq5\), the eligible locations are 2 and 5. Therefore, \[ P(1<X\leq5)=P(X=2)+P(X=5)=\frac12+\frac14=\frac34. \]

The event \(X=2\) contains just the location 2, so \(P(X=2)=1/2\). The event \(X\notin\{0,5\}\) also contains just the support location 2. Hence \[ P(X\notin\{0,5\})=P(X=2)=\frac12. \] In this example, the endpoint 5 is included in the first interval and contributes its full weight \(1/4\).

Aggregating Outcomes That Give the Same Value

A random variable can assign the same numerical value to several underlying outcomes. The law of the random variable is formed from its possible numerical values, so the probabilities of outcomes that produce the same value must be added. In the resulting representation, each location appears only once.

Worked Example: A Fair Die and a Payout

A fair six-sided die is rolled. The payout \(X\), in dollars, is \(-2\) for a roll of 1 or 2, \(1\) for a roll of 3, 4, or 5, and \(7\) for a roll of 6. Find the finite atomic law of \(X\), then calculate \(P(X\geq1)\), \(P(-2\leq X<7)\), and \(P(X=0)\).

Each die outcome has probability \(1/6\). Two outcomes give payout \(-2\), three give payout 1, and one gives payout 7. Therefore the masses at the distinct payout locations are \[ P(X=-2)=\frac{2}{6}=\frac13,\qquad P(X=1)=\frac{3}{6}=\frac12,\qquad P(X=7)=\frac{1}{6}. \] The weights sum to \(1/3+1/2+1/6=2/6+3/6+1/6=1\), so the law is \[ P_X=\frac13\delta_{-2}+\frac12\delta_1+\frac16\delta_7. \]

The event \(X\geq1\) includes locations 1 and 7, giving \[ P(X\geq1)=\frac12+\frac16=\frac23. \] The event \(-2\leq X<7\) includes \(-2\) and 1 but excludes 7, so \[ P(-2\leq X<7)=\frac13+\frac12=\frac56. \] Finally, 0 is not a support location, so \(P(X=0)=0\).

When Is a Finite Atomic Law Degenerate?

A finite atomic law is degenerate exactly when it has only one support location. The following result makes the distinction precise.

Theorem: Let \(X\) have a finite atomic law with distinct support locations \(x_1,\ldots,x_k\) and positive weights \(p_1,\ldots,p_k\). Then \(X\) is constant with probability one if and only if \(k=1\).

Proof. If \(k=1\), then the weights sum to 1, so the sole weight is \(p_1=1\). Consequently, \(P(X=x_1)=1\), and \(X\) is constant with probability one. Conversely, suppose \(X=c\) with probability one. Every support location must then equal \(c\): a distinct location \(x_i\ne c\) with positive weight would imply \(P(X=x_i)>0\), contradicting \(P(X=c)=1\). Since the locations are distinct, there can be only one support location. Thus \(k=1\). \(\square\)

The same finite representation gives expectations and variances directly when those quantities are useful. For any real-valued function \(g\) on the support, \[ E[g(X)]=\sum_{i=1}^{k}p_i g(x_i). \] This follows because \(g(X)\) takes the value \(g(x_i)\) on the event \(\{X=x_i\}\), whose probability is \(p_i\). In particular, \(E[X]=\sum_i p_i x_i\), and \[ \operatorname{Var}(X)=\sum_{i=1}^{k}p_i(x_i-E[X])^2. \] Each term in this variance sum is nonnegative. If there is one support point, the variance is zero. If there are at least two distinct support points with positive weights, at most one of them can equal \(E[X]\); at least one other term is strictly positive. Thus the variance is zero exactly for a degenerate finite atomic law.

Common Mistakes and Why Point Masses Matter

Counting support locations instead of summing their weights. An event containing two locations does not automatically have probability \(2/k\). Add the actual weights at those locations.
Including a location outside the event. For \(1<X\leq5\), the location 1 is excluded while 5 is included. Check each location against both inequalities.
Leaving repeated outcomes as separate locations. If several underlying outcomes yield the same value of \(X\), add their probabilities to obtain the mass at that value.
Assigning probability to a value that is not in the support. If no support location equals \(a\), then \(P(X=a)=0\), even if \(a\) lies between support locations.
Confusing a point mass with density height. A point mass is the probability assigned to a location; it is not a density evaluated at that location.

Point-mass representations make finite laws compact and calculations transparent. Once the support and weights are known, an event probability is a finite sum, and a constant-valued variable is the special case with one location and weight 1. This perspective is also useful for checking a proposed law: every listed weight must be nonnegative, the weights must sum to one, and repeated locations must be combined.

Check Your Understanding

Use the locations and weights to answer each question.

If \(P_X=\delta_4\), what are \(P(X=4)\) and \(P(X\in(4,6])\)? Explain each answer.
A law has masses \(0.2\), \(0.5\), and \(0.3\) at locations \(-1\), 2, and 6. Find \(P(-1\leq X<6)\).
A fair coin produces \(X=0\) on heads and \(X=3\) on tails. Write its law using point masses and state whether it is degenerate.
A random experiment has four equally likely outcomes, and three of them give the same value of \(X\). What probability weight should be assigned to that value?
For a finite atomic law with two distinct locations of positive mass, explain why the law cannot be degenerate.