Marginal CDFs from Boundary Limits

Why Boundary Limits Recover Marginals

A joint CDF records probabilities involving two coordinate restrictions at once. A marginal CDF asks about just one coordinate. The connection is simple to state but important to interpret: to retain the restriction on $X$ while removing the restriction on $Y$, let the upper bound on $Y$ grow without limit. Eventually every possible finite value of $Y$ is included. The remaining probability is exactly the probability that $X$ is below its specified bound.

This operation does not require independence, a density, or a particular form of dependence. It follows from the definition of a joint CDF and a basic continuity property of probability.

Definition: For real-valued random variables $X$ and $Y$, their joint CDF is $$ F(x,y)=P(X\leq x,\;Y\leq y), \qquad x,y\in\mathbb{R}. $$ Their marginal CDFs are $F_X(x)=P(X\leq x)$ and $F_Y(y)=P(Y\leq y)$.

Theorem: For every real $x$ and $y$, $$ F_X(x)=\lim_{b\to\infty}F(x,b), \qquad F_Y(y)=\lim_{a\to\infty}F(a,y). $$ The limits are taken with the other argument fixed.

In the first limit, $b$ increases while $x$ stays fixed. In the second, $a$ increases while $y$ stays fixed. The coordinate being sent to infinity is the restriction being removed; the coordinate held fixed is the one whose marginal distribution remains.

Why the Limits Hold

We prove the first identity; the proof of the second is the same with the roles of $X$ and $Y$ exchanged. Fix $x$, and for each positive integer $m$ define the event $A_m=\{X\leq x,\;Y\leq m\}$. These events are increasing: $A_m\subseteq A_{m+1}$. Their union is $\{X\leq x\}$, because whenever $X\leq x$, the real-valued random variable $Y$ has some finite realized value and therefore $Y\leq m$ for some sufficiently large integer $m$.

For completeness, countable additivity implies continuity of probability for increasing events. Set $D_1=A_1$ and $D_m=A_m\setminus A_{m-1}$ for $m\geq2$. The events $D_m$ are disjoint, $A_k=\bigcup_{m=1}^kD_m$, and $\bigcup_{m=1}^{\infty}A_m=\bigcup_{m=1}^{\infty}D_m$. Thus $P(A_k)=\sum_{m=1}^kP(D_m)$, and countable additivity gives $\lim_{k\to\infty}P(A_k)=\sum_{m=1}^{\infty}P(D_m)=P(\bigcup_m A_m)$. It follows that $\lim_{k\to\infty}F(x,k)=P(X\leq x)=F_X(x)$.

Finally, $F(x,b)$ is nondecreasing in $b$, since increasing $b$ enlarges the event in its definition. Its limit as a real $b$ tends to infinity therefore agrees with its limit along positive integers: for any $b$ between consecutive integers, the value is bounded by the values at those integers. This proves the first identity for the stated real-variable limit. The same reasoning proves the second. $\square$

Worked Examples

Worked Example: Independent Exponential and Uniform Variables

Suppose $X\sim\operatorname{Exp}(2)$ and $Y\sim\operatorname{Uniform}(0,3)$ are independent. Here the exponential parameter is the rate. For $x\geq0$, the CDF of $X$ is $1-e^{-2x}$; for $0\leq y\leq3$, the CDF of $Y$ is $y/3$. The joint CDF, using independence, is

F(x,y)= \begin{cases} (1-e^{-2x})\,y/3, & x\geq0,\ 0\leq y\leq3,\\ 1-e^{-2x}, & x\geq0,\ y>3,\\ 0, & x<0\text{ or }y<0. \end{cases}

Hold $x$ fixed and let $y$ increase. If $x<0$, the joint CDF remains zero. If $x\geq0$, then for all $y>3$, it equals $1-e^{-2x}$. Consequently $F_X(x)=0$ for $x<0$ and $F_X(x)=1-e^{-2x}$ for $x\geq0$.

To obtain $F_Y$, hold $y$ fixed and let $x$ increase. For $0\leq y\leq3$, the limit of $(1-e^{-2x})y/3$ as $x\to\infty$ is $y/3$, since $e^{-2x}\to0$. The limit is zero for $y<0$, and it is one for $y>3$. Thus the marginal CDFs agree with the stated exponential and uniform distributions.

For a numerical check, take $x=(\log 2)/2$ and $y=1.5$. Then $e^{-2x}=1/2$, so $F(x,y)=(1/2)(1.5/3)=1/4$. Sending $y$ above 3 gives $F_X(x)=1/2$, while sending $x$ to infinity gives $F_Y(1.5)=1.5/3=1/2$. The joint value at these finite bounds is not either marginal value.

Worked Example: Marginals in a Dependent Mixture

Let a component be selected with probability $1/2$. In component 1, $X$ and $Y$ are independent exponentials with rates 1 and 2; in component 2, they are independent exponentials with rates 3 and 1. The component selection is shared by both coordinates, so this construction can produce dependence. For nonnegative $x,y$, its joint CDF is

F(x,y)=\tfrac12(1-e^{-x})(1-e^{-2y}) +\tfrac12(1-e^{-3x})(1-e^{-y}).

If $x<0$ or $y<0$, the joint CDF is zero. For fixed $x\geq0$, let $y\to\infty$. Both exponential terms involving $y$ tend to zero, so

F_X(x)=\tfrac12(1-e^{-x})+\tfrac12(1-e^{-3x}),\qquad x\geq0,

and $F_X(x)=0$ for $x<0$. In the other direction, for fixed $y\geq0$, let $x\to\infty$. This gives

F_Y(y)=\tfrac12(1-e^{-2y})+\tfrac12(1-e^{-y}),\qquad y\geq0,

with $F_Y(y)=0$ for $y<0$. Each marginal is a mixture of exponential distributions; neither marginal requires identifying the component from the observed value of the other coordinate.

The joint value at $x=y=\log 2$ illustrates the distinction between joint and marginal probabilities. Since $e^{-x}=1/2$, $e^{-2y}=1/4$, and $e^{-3x}=1/8$,

F(\log2,\log2) =\tfrac12\left(\tfrac12\right)\left(\tfrac34\right) +\tfrac12\left(\tfrac78\right)\left(\tfrac12\right) =\tfrac{3}{16}+\tfrac{7}{32} =\tfrac{13}{32}.

The marginal values are $F_X(\log2)=\tfrac12(\tfrac12)+\tfrac12(\tfrac78)=\tfrac{11}{16}$ and $F_Y(\log2)=\tfrac12(\tfrac34)+\tfrac12(\tfrac12)=\tfrac58$. Their product is $55/128$, which differs from the joint value $13/32=52/128$. In particular, the joint CDF is not generally the product of its marginal CDFs.

Worked Example: Marginals of a Correlated Normal Pair

Let $Z_1,Z_2$ be independent standard normal random variables, and set $X=Z_1$ and $Y=\rho Z_1+\sqrt{1-\rho^2}Z_2$, where $-1<\rho<1$. This is a standard construction of a bivariate normal pair. Its joint CDF is $F(x,y)=P(X\leq x,Y\leq y)$; a closed-form expression for this joint probability is not needed to recover the marginals.

By the boundary-limit theorem, $F_X(x)=\lim_{y\to\infty}F(x,y)$. Since $X=Z_1$, this marginal CDF is the standard normal CDF $\Phi(x)$. Also $Y$ is a linear combination of independent standard normals with mean zero and variance $\rho^2+(1-\rho^2)=1$, so $Y$ is standard normal and $F_Y(y)=\Phi(y)$. For example, if $\rho=0.6$, the joint CDF at $(0,0)$ is a probability involving both restrictions; the two marginal CDF values are each $\Phi(0)=1/2$, regardless of $\rho$. This shows how the boundary limits identify the coordinate distributions without requiring a formula for the interior of the joint CDF.

Boundary Checks and a Common Pitfall

For any real-valued pair, a joint CDF also has useful zero-boundary limits. For fixed $x$, as $y\to-\infty$, the events $\{X\leq x,Y\leq y\}$ decrease to the empty event: no finite realized value of $Y$ can be below every real bound. Continuity of probability for decreasing events gives $\lim_{y\to-\infty}F(x,y)=0$. The decreasing-event property follows from countable additivity by applying continuity for increasing events to complements. Likewise, $\lim_{x\to-\infty}F(x,y)=0$ for fixed $y$. At the positive boundaries, the limits are the marginals, not necessarily 1, because the other coordinate's bound remains fixed.

A common mistake is to try to obtain $F_X(x)$ by evaluating $F(x,0)$, or by setting the two arguments equal and using $F(x,x)$. Neither operation removes the condition on $Y$. For example, in the independent exponential-uniform example, $F(x,0)=0$, while $F_X(x)=1-e^{-2x}$ for $x\geq0$. The correction is to take a limit in the other coordinate: $F_X(x)=\lim_{y\to\infty}F(x,y)$. Similarly, recover $F_Y(y)$ by taking $x\to\infty$, not by choosing an arbitrary finite $x$.

Key takeaway: To obtain the marginal CDF of one coordinate, keep its threshold fixed and send the other coordinate's threshold to positive infinity. The result holds for every joint distribution of real-valued $X$ and $Y$, whether the coordinates are independent or dependent.

Check Your Understanding

Use the boundary-limit identities and distinguish each marginal event from a joint event.

Write $F_Y(y)$ as a limit of the joint CDF, specifying which argument is held fixed.
Why does $\bigcup_{m=1}^{\infty}\{X\leq x,Y\leq m\}=\{X\leq x\}$ hold for real-valued $Y$?
In the independent exponential-uniform example, what is $F_X(x)$ for $x\geq0$, and why does taking $y=1.5$ not remove the restriction on $Y$?
In the mixture example, give the expression for $F_Y(y)$ when $y\geq0$.
For fixed $x$, what is $\lim_{y\to-\infty}F(x,y)$, and what event interpretation explains the answer?

Marginal CDFs from Boundary Limits

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