Why a Valid PMF Sums to One

Why the PMF Conditions Matter

A probability mass function assigns a probability to each possible value of a discrete random variable. Those assignments must be consistent with the probability model: each assigned probability must be possible, and the probabilities across all possible values must account for the entire outcome space. These two requirements become essential when checking a proposed distribution, especially when its support is infinite.

The support is the set of values that can occur with positive probability. A random variable may also be described on a larger set that includes impossible values; those values have probability zero. This tutorial focuses on the two basic checks for a finite or countably infinite discrete support. It does not address probability densities, which use a different normalization condition.

Definition: For a discrete random variable $X$, its probability mass function (PMF) is $p_X(x)=P(X=x)$. A candidate function $p_X$ is a valid PMF on a finite or countable set of possible values if (1) $p_X(x)\geq 0$ for every value and (2) the sum of $p_X(x)$ over all possible values is 1. Values outside the support are assigned probability 0.

Where the Two Conditions Come From

For each value $x$, the event $\{X=x\}$ has a probability. Probability cannot be negative, so $P(X=x)\geq 0$. Also, the events $\{X=x\}$, as $x$ ranges over all possible values, are mutually exclusive: $X$ cannot equal two different values at once. Together, they cover every outcome in the model, because $X$ must take one of its possible values.

For a finite support, the events form a finite partition of the sample space, so their probabilities add to the probability of the whole sample space, which is 1. For a countably infinite support, countable additivity gives the corresponding infinite sum. Thus the conditions are not arbitrary tests: they follow directly from the basic rules of probability.

p_X(x)=P(X=x)\geq 0, \qquad \sum_{x\in \mathcal{S}}p_X(x)=1,

Here $\mathcal{S}$ is the set of possible values of $X$. If a formula is stated for a larger set, include only its possible values in the sum, or equivalently include all values and assign zero probability outside the support. The infinite sum condition means that the sequence of partial sums approaches 1; it is not enough for a finite initial portion to have a convenient total.

Justification: The events $\{X=x\}$, for $x\in\mathcal{S}$, are disjoint and their union is the whole sample space. Therefore, by finite or countable additivity, $$ \sum_{x\in\mathcal{S}}P(X=x) = P\left(\bigcup_{x\in\mathcal{S}}\{X=x\}\right) = P(\text{whole sample space}) = 1. $$ Each summand is a probability, so each is nonnegative as well.

Both conditions are necessary, and together they are sufficient for a probability assignment on a finite or countable set: one can construct a probability model whose outcomes are those values and whose probabilities are the assigned masses. In practice, checking both conditions is the first step before using a candidate PMF to calculate event probabilities.

Checking a Finite Candidate PMF

For a finite support, the check is direct. First list the proposed values, including any values assigned probability zero. Then inspect every mass for nonnegativity and add the masses. If either check fails, the assignment is not a PMF, even if the other check passes.

Worked Example: Validating a Finite Warning-Count Model

A hypothetical monitoring system records $X$, the number of alerts generated during one scheduled test. A proposed model assigns $p_X(k)=(k+1)/10$ for $k=0,1,2,3$, and assigns zero probability to all other values. Check whether this is a valid PMF, then find the probability that at least two alerts occur.

Method. Check nonnegativity at every listed value, add all masses, and only then calculate the requested event probability.

For the four possible values, the masses are $$ p_X(0)=\frac{1}{10},\quad p_X(1)=\frac{2}{10},\quad p_X(2)=\frac{3}{10},\quad p_X(3)=\frac{4}{10}. $$

Each numerator is positive, so all four masses are nonnegative. Their total is $$ \frac{1}{10}+\frac{2}{10}+\frac{3}{10}+\frac{4}{10} = \frac{1+2+3+4}{10} = \frac{10}{10} = 1. $$ The zero probabilities outside $\{0,1,2,3\}$ are nonnegative and do not change the sum. Both conditions hold, so the proposal is a valid PMF.

The event of at least two alerts is $X\geq 2$, which includes $X=2$ and $X=3$. Therefore, $$ P(X\geq 2)=p_X(2)+p_X(3)=\frac{3}{10}+\frac{4}{10}=\frac{7}{10}=0.7. $$

Check. The event probability $0.7$ is between 0 and 1, and the remaining probability is $1/10+2/10=0.3$; the two parts add to 1. The common mistake avoided here is checking only the displayed formula at one value rather than checking every mass and the total.

Why Nonnegativity Cannot Be Skipped

A total of 1 does not make an assignment valid if one of its entries is negative. Adding a negative number can offset an excessively large positive mass, producing a total that looks correct while assigning an impossible probability to an event.

Worked Example: A Total of One With an Impossible Mass

A proposed model for a hypothetical count of service requests assigns $p_X(0)=0.6$, $p_X(1)=-0.1$, and $p_X(2)=0.5$, with no other possible values. Determine whether it is a PMF. If it is not, give one valid alternative assignment obtained by replacing the negative mass with zero and renormalizing the remaining masses.

Method. Test nonnegativity and total mass separately. If constructing the requested alternative, first remove the negative entry, then divide the remaining nonnegative weights by their total.

The proposed total is $$ 0.6+(-0.1)+0.5=1.0. $$ But $p_X(1)=-0.1<0$, so the nonnegativity condition fails. This assignment is not a PMF, despite summing to 1.

For the specified alternative, replace the mass at 1 by zero. The remaining weights are $0.6,0,0.5$, whose total is $$ 0.6+0+0.5=1.1. $$ Divide each weight by $1.1$. Since $1.1=11/10$, the resulting masses are $$ p_X(0)=\frac{0.6}{1.1}=\frac{6}{11},\quad p_X(1)=0,\quad p_X(2)=\frac{0.5}{1.1}=\frac{5}{11}. $$

Both masses are nonnegative, and their total is $6/11+0+5/11=1$, so this alternative is a valid PMF. It is one possible replacement model, not a way to preserve the original negative probability.

Check. The proposed assignment fails because of its negative entry; the alternative passes both checks. The common mistake avoided is treating a sum of 1 as sufficient evidence of validity.

Normalizing Nonnegative Weights

A useful finite-support procedure starts with nonnegative weights that are not yet probabilities. If their total is positive, dividing every weight by that total produces masses that sum to 1 and remain nonnegative. This procedure is valid because it rescales the entire assignment by the same positive constant. It cannot repair negative weights: multiplying a negative weight by a positive constant leaves it negative.

w_j\geq 0,\quad 0<\sum_{j=1}^{r}w_j<\infty \quad\Longrightarrow\quad p_j=\frac{w_j}{\sum_{k=1}^{r}w_k}, \qquad \sum_{j=1}^{r}p_j=1.

For instance, if three possible outcomes have weights $2,3,5$, their total is 10 and the normalized masses are $2/10,3/10,5/10$. Each is nonnegative and their sum is $10/10=1$. The important order is to verify that the weights are nonnegative and have a finite positive total before normalizing.

Checking a Countably Infinite PMF

When the support has infinitely many values, one cannot add all the masses by listing them. Instead, evaluate the corresponding infinite series, usually by a known series identity or by expressing its partial sums in a form whose limit can be found. A candidate with nonnegative terms still fails if its total mass is not 1.

Worked Example: A PMF on the Positive Integers

A hypothetical model for the number $N$ of inspection rounds until a recorded category is reached assigns $$ p_N(n)=\frac{c}{n(n+1)},\qquad n=1,2,3,\ldots, $$ where $c$ is a constant. Find the value of $c$ that makes this a valid PMF, and then find $P(N\geq 4)$.

Method. Require nonnegative masses and use a telescoping series to determine their total. Then sum the relevant tail using the same identity.

Since $n(n+1)>0$ for every positive integer $n$, the masses are nonnegative if $c\geq 0$. The identity $$ \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1} $$ makes the partial sum through $m$ terms equal to $$ \sum_{n=1}^{m}\frac{c}{n(n+1)} = c\sum_{n=1}^{m}\left(\frac{1}{n}-\frac{1}{n+1}\right) = c\left(1-\frac{1}{m+1}\right). $$

As $m$ tends to infinity, $1/(m+1)$ tends to zero, so the total mass is $c$. To make that total equal to 1, choose $c=1$. This value is nonnegative, and the infinite sum is exactly 1; hence the assignment is a valid PMF.

For the tail probability, use the partial-sum identity or telescope directly: $$ P(N\geq 4) = \sum_{n=4}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right) = \lim_{m\to\infty}\left(\frac{1}{4}-\frac{1}{m+1}\right) = \frac{1}{4}. $$

Check. The probability of $N\leq 3$ is $1/2+1/6+1/12=3/4$, so $P(N\leq3)+P(N\geq4)=3/4+1/4=1$. The common mistake avoided is assuming that infinitely many positive terms necessarily sum to 1; the series must actually be evaluated.

A Reliable Check for Exam Problems

For a finite candidate, write down every mass and add them. For an infinite candidate, establish nonnegativity term by term and evaluate the full series, including its limiting behavior. In either case, keep the two checks distinct: a correct total cannot excuse a negative term, and nonnegative terms do not guarantee that the total is 1.

Identify the support.
State which values are possible and note any values outside it that have probability zero.

Check every mass.
Verify that each assigned value is at least zero. For an infinite support, give an argument that applies to every index.

Check the total.
Add the finite list or evaluate the infinite series, and show that the result is 1.

Use the PMF only after it passes.
Then an event probability can be found by summing the masses of the values in that event.

This check is aligned with the SOA VEE Mathematical Statistics syllabus because it connects the definition of a discrete distribution to the basic probability rules used throughout statistical modeling. It also provides a simple defense against two frequent errors: overlooking a negative entry and assuming that a plausible-looking formula automatically has total mass one.

Check Your Understanding

For each problem, verify the relevant PMF conditions before using the assigned masses.

A candidate model assigns $p_X(k)=a(k+1)$ for $k=0,1,2,3$, and zero elsewhere. Find the value of $a$ that makes it a PMF, and calculate $P(X\geq2)$. (Answer: $a=0.1$; $P(X\geq2)=0.7$.)
A device-count model assigns masses $0.15,0.25,0.45,0.15$ at $x=0,1,2,3$, respectively. Check whether this is a PMF and find the probability of at least two devices. (Answer: It is valid; $P(X\geq2)=0.60$.)
A proposed count distribution assigns $p_X(0)=0.2$, $p_X(1)=0.5$, and $p_X(2)=0.4$, with no other possible values. Which PMF condition fails? (Answer: Nonnegativity holds, but the total is $1.1$, so the total-mass condition fails.)
For $n=0,1,2,\ldots$, let $p_N(n)=c(1/3)^n$. Find $c$, then calculate $P(N\geq2)$. (Answer: $c=2/3$; $P(N\geq2)=1/9$.)
For $n=1,2,3,\ldots$, let $p_N(n)=c/[n(n+1)]$. Find $c$ and $P(N\leq3)$. (Answer: $c=1$; $P(N\leq3)=3/4$.)

What You'll Learn