Joint Probability Mass Function¶

Definition¶

For discrete random variables $X$ and $Y$ , the joint PMF is

\begin{aligned} p_{X,Y}(x,y)=\mathbb{P}(X=x,Y=y). \end{aligned}

This is the Probability Mass Function of the pair: it assigns a mass to each possible ordered outcome, and summing those masses over an event gives its probability.

Interpretation¶

A joint PMF assigns probability to each pair of discrete values and describes how the variables behave together: the probability of any event about the pair $(X,Y)$ is the sum of the masses in that event.

Two dot diagrams on the four pairs of values 0 and 1, with dot sizes showing the joint masses: a general table with masses 0.1 to 0.4, and X = Y with mass 1/2 on each diagonal pair.

Each disc sits at a pair $(x,y)$ , and its size shows the mass $p_{X,Y}(x,y)$ . On the left the four masses $0.1$ , $0.2$ , $0.3$ and $0.4$ add up to $1$ . On the right $X=Y\sim\operatorname{Bern}(\tfrac{1}{2})$ $Bern (\frac{1}{2})$ : all mass lies on the diagonal, $\tfrac{1}{2}$ at $(0,0)$ and $\tfrac{1}{2}$ at $(1,1)$ , and the pairs $(0,1)$ and $(1,0)$ have mass $0$ .

The joint PMF holds more information than the two separate PMFs. In the right panel each of $X$ and $Y$ is $\operatorname{Bern}(\tfrac{1}{2})$ , exactly as for two independent fair coins, where each of the four pairs has mass $\tfrac{1}{4}$ ; only the joint PMF tells the two situations apart. The separate PMFs are recovered as marginal PMFs by summing over the other variable, and independence means that every joint mass is the product of the two marginal masses.

Several notions are built directly on the joint PMF: the conditional PMF rescales the masses with one fixed value of $Y$ , the joint CDF adds the masses in a lower-left corner, and two-dimensional LOTUS averages a function $g(X,Y)$ against them. For pairs of continuous variables, the joint density plays the same role, with sums replaced by integrals.

Key Results¶

$p_{X,Y}(x,y)\ge0$ and $\displaystyle \sum_x\sum_yp_{X,Y}(x,y)=1$ $1$ .
For a set $B$ of pairs, $\displaystyle \mathbb{P}((X,Y)\in B)=\sum_{(x,y)\in B}p_{X,Y}(x,y)$ $(x, y) \in B \sum p_{X, Y} (x, y)$ .

Examples¶

Two independent fair dice have $p_{X,Y}(i,j)=1/36$ for $i,j=1,\ldots,6$ $1, \dots, 6$ .
If $X=Y\sim\operatorname{Bern}(1/2)$ $Bern (1/2)$ , the masses at $(0,0)$ and $(1,1)$ are $1/2$ , and all other masses are zero.

Prerequisites¶

Probability Mass Function

Sources¶

Nicky van Foreest · Probability Distributions · Week 1

Aliases: joint PMF, joint mass function Symbols: element of, sum

Last updated: 20 October 2026 03:02