Marginals and Boundary Limits of a Joint CDF

Why Marginal and Boundary Limits Matter

A joint cumulative distribution function describes probabilities involving two thresholds at once. Its boundary behavior answers a natural question: what happens to the probability when one threshold is made so large that it no longer restricts the event? The answer recovers the distribution of the other random variable. The same limits also provide important checks when deciding whether a proposed function could be a joint CDF.

Let $X$ and $Y$ be real-valued random variables, finite with probability one. Their joint CDF is

F_{X,Y}(x,y)=P(X\le x,\;Y\le y),\qquad -\infty<x,y<\infty.

The marginal CDFs are $F_X(x)=P(X\le x)$ and $F_Y(y)=P(Y\le y)$. The key idea is to remove one restriction from the joint event by sending its threshold to positive infinity.

Theorem: For real-valued $X$ and $Y$, $$ F_X(x)=\lim_{y\to+\infty}F_{X,Y}(x,y), \qquad F_Y(y)=\lim_{x\to+\infty}F_{X,Y}(x,y). $$ The other boundary limits are $$ \lim_{y\to-\infty}F_{X,Y}(x,y)=0,\qquad \lim_{x\to-\infty}F_{X,Y}(x,y)=0, $$ $$ \lim_{y\to+\infty}F_{X,Y}(x,y)=F_X(x),\qquad \lim_{x\to+\infty}F_{X,Y}(x,y)=F_Y(y), \qquad \lim_{\substack{x\to+\infty\\y\to+\infty}}F_{X,Y}(x,y)=1. $$

Why Taking a Threshold to Infinity Works

Fix $x$, and let $y$ increase. The events $\{X\le x,\;Y\le y\}$ expand as $y$ increases. Their union is $\{X\le x\}$: whenever $X\le x$, the finite value of $Y$ is eventually below a sufficiently large threshold. By continuity of probability for an increasing sequence of events, the probabilities approach $P(X\le x)$. This gives the marginal CDF of $X$.

\bigcup_{n=1}^{\infty}\{X\le x,\;Y\le n\}=\{X\le x\}, \qquad \lim_{n\to\infty}P(X\le x,\;Y\le n)=P(X\le x).

The same reasoning, with $x$ increasing while $y$ is fixed, gives the marginal CDF of $Y$. For a threshold tending to negative infinity, the corresponding event shrinks to the empty event: a finite random variable cannot remain below thresholds that decrease without bound. Its probability therefore tends to zero.

Finally, when both thresholds increase without bound, the event $\{X\le x,\;Y\le y\}$ eventually includes any particular finite pair of values. The union of these events is the whole sample space, so the joint CDF tends to one. These arguments explain the boundary limits directly from the definition of probability; they do not depend on a density or on whether the variables are discrete or continuous.

Recovering Marginals from a Discrete Joint CDF

For a discrete joint distribution, taking a threshold to positive infinity includes every possible value of the other variable. In a probability table, this is equivalent to adding the relevant row or column probabilities. The CDF limits and the table’s marginal sums are two ways to express the same operation.

Worked Example: Marginals from a Joint Probability Table

Suppose a hypothetical pair $(X,Y)$ has the following joint probabilities. The rows correspond to $X$, and the columns correspond to $Y$.

	$Y=-1$	$Y=1$	Row total
$X=0$	0.10	0.20	0.30
$X=2$	0.30	0.40	0.70
Column total	0.40	0.60	1.00

Find the marginal CDFs, then calculate $F_{X,Y}(0,-1)$. Use the boundary-limit method to interpret the marginals.

Step 1: Recover $F_X$. For $x<0$, no possible value of $X$ is at most $x$, so $F_X(x)=0$. For $0\le x<2$, only $X=0$ is included. Its total probability is $0.10+0.20=0.30$. For $x\ge2$, both values are included, giving total probability $1$. Thus

F_X(x)= \begin{cases} 0,&x<0,\\ 0.30,&0\le x<2,\\ 1,&x\ge2. \end{cases}

This is also $\lim_{y\to+\infty}F_{X,Y}(x,y)$, since an arbitrarily large $y$ includes both possible $Y$ values.

Step 2: Recover $F_Y$. For $y<-1$, no value of $Y$ is included. For $-1\le y<1$, only $Y=-1$ is included, with total probability $0.10+0.30=0.40$. For $y\ge1$, both columns are included, giving $1$. Therefore

F_Y(y)= \begin{cases} 0,&y<-1,\\ 0.40,&-1\le y<1,\\ 1,&y\ge1. \end{cases}

Step 3: Calculate the joint CDF value. At $(x,y)=(0,-1)$, the event includes only the cell $X=0,\;Y=-1$. Hence $F_{X,Y}(0,-1)=0.10$.

Check. The row totals are $0.30$ and $0.70$, and the column totals are $0.40$ and $0.60$. Each set of totals sums to $1$, as required. The method avoids a common mistake: using a joint CDF value at a finite second threshold as though it were already a marginal probability.

Boundary Limits in a Continuous Example

A continuous distribution illustrates that the limits apply just as directly when probabilities are not arranged in table cells. Consider a hypothetical model in which $U$ is uniform on $[0,1]$, and define $X=U$ and $Y=U+1$. Both variables are finite with probability one. Define the uniform CDF $H$ by

H(t)= \begin{cases} 0,&t<0,\\ t,&0\le t\le1,\\ 1,&t>1. \end{cases}

Worked Example: Marginals and a Joint Probability from Boundary Limits

For this model, derive the joint CDF, recover both marginals, and calculate $F_{X,Y}(0.4,1.7)$. The distributional assumption is $U\sim\operatorname{Uniform}(0,1)$.

Step 1: Express the joint event in terms of $U$. The two inequalities are $U\le x$ and $U+1\le y$, the latter being $U\le y-1$. Together, they require $U\le\min(x,y-1)$. Therefore

F_{X,Y}(x,y)=H\bigl(\min(x,y-1)\bigr).

Step 2: Send $y$ to positive infinity. For fixed $x$, the smaller of $x$ and $y-1$ becomes $x$ once $y$ is sufficiently large. Thus $F_X(x)=H(x)$, which has the uniform distribution on $[0,1]$.

Step 3: Send $x$ to positive infinity. For fixed $y$, the smaller threshold becomes $y-1$ once $x$ is sufficiently large. Hence $F_Y(y)=H(y-1)$. This is the distribution of $U+1$, supported on $[1,2]$.

Step 4: Evaluate the requested joint CDF. At $x=0.4$ and $y=1.7$, the two bounds on $U$ are $U\le0.4$ and $U\le1.7-1=0.7$. The more restrictive bound is $0.4$, so

F_{X,Y}(0.4,1.7)=H(\min(0.4,0.7))=H(0.4)=0.4.

Check. The corresponding marginals are $F_X(0.4)=0.4$ and $F_Y(1.7)=H(0.7)=0.7$. The joint probability is no greater than either marginal probability: $0.4\le0.4$ and $0.4\le0.7$, consistent with the joint event being contained in each marginal event.

Boundary Conditions Are Necessary, Not Sufficient

A proposed joint CDF must have the boundary limits stated above. However, passing those boundary checks does not by itself establish validity. For example, probabilities assigned to rectangles must be nonnegative. The next example shows how a function can produce plausible marginal CDFs and still fail as a joint CDF.

Worked Example: A Boundary-Consistent Function That Is Not a Joint CDF

Consider a function built from signed masses at the four points $(0,0),(0,1),(1,0),(1,1)$, respectively:

-0.10,\qquad 0.50,\qquad 0.50,\qquad 0.10.

The four numbers sum to $1$. The row totals for $X=0,1$ are $0.40,0.60$, and the column totals for $Y=0,1$ are also $0.40,0.60$. These totals would give valid marginal CDFs: each jumps from $0$ to $0.40$ at zero, then from $0.40$ to $1$ at one. Consequently, the function has the expected limits $F_X(x)$ as $y\to+\infty$, $F_Y(y)$ as $x\to+\infty$, and $1$ as both thresholds tend to positive infinity. Its lower boundary limits are zero.

But at $(x,y)=(0,0)$, the cumulative value includes only the mass at $(0,0)$:

F_{X,Y}(0,0)=-0.10.

A probability cannot be negative, so this function cannot be a joint CDF. Equivalently, the probability assigned to the rectangle $(-\tfrac12,0]\times(-\tfrac12,0]$ would be $-0.10$, violating nonnegativity. The boundary limits are necessary checks, but a full validity check must also rule out negative probabilities for regions such as rectangles.

Common Exam Trap: Taking the Wrong Limit

To recover $F_X(x)$, keep $x$ fixed and send the other threshold, $y$, to positive infinity. Sending $x$ to positive infinity instead removes the restriction on $X$, leaving $F_Y(y)$. Confusing these two limits swaps the marginal distributions.

For example, in the continuous model above, the incorrect calculation for $F_X(0.4)$ would be

\lim_{x\to+\infty}F_{X,Y}(x,0.4)=F_Y(0.4)=H(-0.6)=0.

That limit is a valid calculation, but it is the marginal CDF of $Y$ at $0.4$, not $F_X(0.4)$. The correct limit is

F_X(0.4)=\lim_{y\to+\infty}F_{X,Y}(0.4,y)=H(0.4)=0.4.

A reliable check is to identify which event remains after taking the limit: making $Y\le y$ unrestricted leaves the event $X\le x$; making $X\le x$ unrestricted leaves $Y\le y$.

Key takeaway: A joint CDF recovers $F_X(x)$ by taking $y\to+\infty$, and recovers $F_Y(y)$ by taking $x\to+\infty$. Its lower boundary limits are zero and its upper corner limit is one. These conditions must hold for a valid joint CDF, but they do not alone guarantee validity.

Check Your Understanding

For each problem, identify the relevant marginal or boundary limit and check that the result agrees with the stated model.

A pair $(X,Y)$ has probabilities $0.15,0.25,0.20,0.40$ at $(1,0),(1,2),(3,0),(3,2)$, respectively. Find $F_X(1)$, $F_Y(0)$, and $F_{X,Y}(1,0)$. (Answer: $F_X(1)=0.40,\;F_Y(0)=0.35,\;F_{X,Y}(1,0)=0.15$.)
Let $U\sim\operatorname{Uniform}(0,2)$, $X=U$, and $Y=3-U$. Find $F_X(1.2)$, $F_Y(2.2)$, and $F_{X,Y}(1.2,2.2)$. (Answer: $F_X(1.2)=0.60,\;F_Y(2.2)=0.60,\;F_{X,Y}(1.2,2.2)=0.20$.)
For any real-valued pair $(X,Y)$, state which limit gives $F_Y(1)$. Then state the value of $\lim_{y\to-\infty}F_{X,Y}(1,y)$. (Answer: $F_Y(1)=\lim_{x\to+\infty}F_{X,Y}(x,1)$; the lower limit is $0$.)
A proposed discrete joint law assigns masses $0.20,0.30,0.10,0.40$ at $(0,0),(0,1),(1,0),(1,1)$. Find $F_X(0)$, $F_Y(0)$, and the limit of the joint CDF as both thresholds tend to positive infinity. (Answer: $F_X(0)=0.50,\;F_Y(0)=0.30$, and the corner limit is $1$.)
A candidate function has the correct marginal CDF limits and upper corner limit, but assigns probability $-0.05$ to one rectangle. Is it a valid joint CDF? Give the decisive reason. (Answer: No; a probability assigned to any rectangle cannot be negative.)

	\(Y=-1\)	\(Y=1\)	Row total
\(X=0\)	0.10	0.20	0.30
\(X=2\)	0.30	0.40	0.70
Column total	0.40	0.60	1.00

What You'll Learn