Joint CDFs from Discrete PMFs

From Joint Masses to a Cumulative Probability

A joint PMF records the probability at each possible pair of values. A bivariate CDF answers a cumulative question instead: how much probability lies at or below a specified value in both coordinates? For a discrete law, the calculation is a sum of joint masses over the lower-left region determined by the two thresholds.

The thresholds need not themselves be possible values. If a coordinate takes integer values, for example, raising its threshold from 1 to 1.8 does not add mass when there are no support values between 1 and 1.8. The cumulative probability changes only when a threshold passes a value with positive mass.

Definition: Let $(X,Y)$ be a random vector with a discrete joint PMF $f(u,v)=P(X=u,Y=v)$ on a finite or countable support $S\subseteq\mathbb{R}^2$. Its bivariate CDF is $$ F(x,y)=P(X\leq x,\;Y\leq y). $$ Equivalently, it is the total joint PMF mass at support points in the lower-left region: $$ F(x,y)=\sum_{\substack{(u,v)\in S\\u\leq x,\;v\leq y}} f(u,v). $$

Both inequalities matter. A support point contributes only if its first coordinate is at most $x$ and its second coordinate is at most $y$. The sum is over the joint PMF, so each included term is the probability of a pair, not a probability for just one coordinate.

How to Accumulate the Mass

For a finite support, identify all points that satisfy both threshold conditions and add their masses. For a countable support, use the same rule, with a countable series. Since the PMF masses are nonnegative and sum to 1, this series is well defined and finite. If either threshold is below every support value in its coordinate, the lower-left region contains no support points and the CDF is 0.

A useful way to organize a finite calculation is to move through the table in two directions: include all eligible rows up to $x$, and within those rows include only eligible columns up to $y$. This is a cumulative sum, not a sum over a rectangle centered at the origin, nor a sum over points lying exactly on the boundary.

Worked Example: A Finite Joint PMF Table

Suppose the joint PMF is given below. The row labels are possible values of $X$, and the column labels are possible values of $Y$.

$X\backslash Y$	0	1
0	0.10	0.05
1	0.20	0.15
2	0.25	0.25

The entries sum to $0.10+0.05+0.20+0.15+0.25+0.25=1$, as required for a PMF. To find $F(1,0)$, include points with $X\leq1$ and $Y\leq0$. These are $(0,0)$ and $(1,0)$, so

$$ F(1,0)=0.10+0.20=0.30. $$

For $F(1.4,0.6)$, the eligible $X$-values are still 0 and 1, while the only eligible $Y$-value is 0. Therefore $F(1.4,0.6)=0.10+0.20=0.30$. Neither noninteger threshold includes the next support value in its coordinate.

At $F(1,1)$, all four points in the first two rows qualify:

$$ F(1,1)=0.10+0.05+0.20+0.15=0.50. $$

Finally, $F(2,0)=0.10+0.20+0.25=0.55$, and $F(2,1)=1$, since the latter includes the entire support.

The Cumulative-Sum Formula

The lower-left sum is not merely a convenient recipe; it follows directly from how probability adds over disjoint outcomes. Each support point corresponds to the event that $(X,Y)$ equals that point. The event $X\leq x,\;Y\leq y$ is the union of exactly those point events whose coordinates satisfy the two inequalities.

Theorem: If $(X,Y)$ has a discrete joint PMF $f$ on a finite or countable support $S$, then for every $x,y\in\mathbb{R}$, $$ F(x,y)=\sum_{\substack{(u,v)\in S\\u\leq x,\;v\leq y}}f(u,v). $$ For any enumeration $S=\{(u_i,v_i):i\geq1\}$ of a countably infinite support, the sum means the limit of the partial sums over included points.

Proof. For each $(u,v)\in S$, let $A_{u,v}=\{X=u,\;Y=v\}$. Distinct support points give disjoint events. The event $\{X\leq x,\;Y\leq y\}$ consists of exactly the outcomes in $A_{u,v}$ for which $u\leq x$ and $v\leq y$. Thus

\{X\leq x,\;Y\leq y\} = \bigcup_{\substack{(u,v)\in S\\u\leq x,\;v\leq y}} A_{u,v},

where the union is disjoint. By finite or countable additivity of probability, the probability of this union is the sum of the probabilities of its constituent events. Since $P(A_{u,v})=f(u,v)$, the stated formula follows. $\square$

For an infinite support, the same reasoning also explains how to approximate the CDF. List the support points and add their eligible masses in that order. Every partial sum is at most the full CDF, and the partial sums increase toward it. A finite initial segment may be sufficient when the remaining PMF mass is small.

Truncation bound: For an enumeration $(u_i,v_i)$ of a countably infinite support, define $$ F_M(x,y)=\sum_{i=1}^{M} f(u_i,v_i)\mathbf{1}\{u_i\leq x,\;v_i\leq y\}. $$ Then $$ 0\leq F(x,y)-F_M(x,y)\leq \sum_{i>M} f(u_i,v_i). $$

To verify the bound, subtract the first $M$ terms from the full nonnegative series. The difference is the sum of the eligible terms with indices greater than $M$. It is nonnegative, and it cannot exceed the sum of all PMF masses with indices greater than $M$. This tail tends to 0 because the PMF series sums to 1.

Countable Supports and Thresholds

On integer-valued supports, floor functions make threshold calculations compact. For an integer $i$, the condition $i\leq x$ is equivalent to $i\leq\lfloor x\rfloor$. If the support is nonnegative, a negative threshold therefore contributes no values. With other supports, retain the inequalities directly rather than applying a floor formula that assumes integer coordinates.

Worked Example: A Countable Rectangular Support

Let $X,Y$ take nonnegative integer values with joint PMF

f(i,j)=\left(1-\frac12\right)\left(1-\frac13\right) \left(\frac12\right)^i\left(\frac13\right)^j, \qquad i,j=0,1,2,\ldots.

This is a valid PMF: summing over $i$ and $j$ gives $\bigl(\frac12\sum_{i=0}^{\infty}(1/2)^i\bigr) \bigl(\frac23\sum_{j=0}^{\infty}(1/3)^j\bigr)=1$. For $x,y\geq0$, only $i\leq\lfloor x\rfloor$ and $j\leq\lfloor y\rfloor$ enter the CDF. Put $m=\lfloor x\rfloor$ and $n=\lfloor y\rfloor$. Then

\begin{aligned} F(x,y) &=\sum_{i=0}^{m}\sum_{j=0}^{n} \frac13\left(\frac12\right)^i\left(\frac13\right)^j\\ &=\left[\frac12\sum_{i=0}^{m}\left(\frac12\right)^i\right] \left[\frac23\sum_{j=0}^{n}\left(\frac13\right)^j\right]\\ &=\left(1-\left(\frac12\right)^{m+1}\right) \left(1-\left(\frac13\right)^{n+1}\right). \end{aligned}

For example, at $x=1.2$ and $y=2$, $m=1$ and $n=2$, giving $F(1.2,2)=(1-\frac14)(1-\frac1{27})=\frac34\cdot\frac{26}{27}=\frac{13}{18}$. If either threshold is negative, $F(x,y)=0$ because the corresponding coordinate has no negative support values.

The support need not fill a grid. A joint PMF may put mass only on a diagonal or on another restricted set. In that case, the CDF sum still uses the same lower-left condition, but only support points with positive mass can contribute.

Worked Example: Mass on a Diagonal

Suppose the only possible pairs are $(k,k)$, $k=0,1,2,\ldots$, with $P(X=k,Y=k)=(1-r)r^k$, where $0<r<1$. The masses sum to 1 by the geometric series. A diagonal point qualifies for $F(x,y)$ precisely when $k\leq x$ and $k\leq y$, or equivalently $k\leq\min(x,y)$. If either threshold is negative, the CDF is 0. Otherwise, put $m=\lfloor\min(x,y)\rfloor$. Then

F(x,y)=\sum_{k=0}^{m}(1-r)r^k=1-r^{m+1}.

For $r=0.4$, $x=2.4$, and $y=1.1$, we have $m=1$, so $F(2.4,1.1)=1-0.4^2=0.84$. Although $X$ can be as large as 2 under the first threshold, the second threshold excludes the point $(2,2)$. At $x=y=2.4$, that point is included, and the value becomes $1-0.4^3=0.936$.

Why the Joint Support Matters

A CDF calculation depends on the locations of the joint support points as well as on their masses. In the diagonal example, the lower-left region contains only diagonal points. In a grid example, it may contain several rows and columns. The two thresholds act together, so inspecting only the values of one coordinate at a time is not enough to determine which joint masses enter the sum.

This construction is also useful for checking a proposed CDF value: the answer must lie between 0 and 1, and increasing either threshold cannot remove any previously included support points. A threshold that moves through a gap in the relevant coordinate values may leave the CDF unchanged; passing a support value can add mass, provided the other coordinate condition is also met.

Common Mistake

Pitfall: Summing every mass with $u\leq x$, even when its second coordinate exceeds $y$, uses the wrong region. The CDF is $P(X\leq x,\;Y\leq y)$, so a point $(u,v)$ contributes only if both $u\leq x$ and $v\leq y$. Also, a threshold between support values does not include the next value: at $x=1.4$, the condition $u\leq x$ does not include a support point with $u=2$.

A related error is to replace a joint PMF term with a product of separate probabilities without justification. The lower-left formula requires the actual joint masses $f(u,v)$. Use the values supplied by the joint PMF and sum exactly the terms whose pairs lie in the specified region.

Check Your Understanding

For each question, identify the eligible support points before summing any masses.

A joint PMF has mass 0.2 at $(0,1)$, 0.3 at $(1,0)$, and 0.5 at $(1,1)$. Find $F(1,0)$.
If $X$ and $Y$ take nonnegative integer values, what support values of each coordinate can contribute at $x=2.7,\;y=1$?
For diagonal support $(k,k)$, explain why $F(2.2,0.8)$ does not include the point $(1,1)$.
State a bound for the error when a countable-support CDF sum is truncated after $M$ terms.
Can the CDF change as $x$ increases while $y$ is held fixed? Describe when a newly eligible value of $X$ adds mass.

What You'll Learn