Defining Bivariate CDFs

Why a Bivariate CDF Is Useful

A pair of random variables can have many different kinds of joint laws: both variables may be discrete, both may be continuous, or one may be determined by the other. A single definition covers all of these cases. The bivariate cumulative distribution function records the probability that both variables lie below specified thresholds. Its elementary properties are not extra assumptions about the model; they follow from the probability axioms.

This tutorial develops three such properties: the function cannot decrease when either threshold increases, its lower-boundary limits are zero, and it is right-continuous. These facts help describe what a bivariate CDF must do without requiring a probability mass function or density.

Definition and Coordinatewise Monotonicity

Definition: Let $X$ and $Y$ be real-valued random variables on the same probability space. Their bivariate cumulative distribution function is $$ F(x,y)=P(X\le x,\;Y\le y),\qquad x,y\in\mathbb{R}. $$ Equivalently, $F(x,y)$ is the probability of the event $\{X\le x\}\cap\{Y\le y\}$.

The definition uses a joint event: both inequalities must hold. In particular, knowing the separate probabilities $P(X\le x)$ and $P(Y\le y)$ does not by itself determine $F(x,y)$; the relationship between $X$ and $Y$ matters.

Theorem: If $x_1\le x_2$ and $y_1\le y_2$, then $$ F(x_1,y_1)\le F(x_2,y_2). $$ Consequently, $F$ is nondecreasing in each coordinate when the other coordinate is held fixed.

To see why, if $X\le x_1$ and $Y\le y_1$, then $X\le x_2$ and $Y\le y_2$ as well. Thus $\{X\le x_1,Y\le y_1\}\subseteq\{X\le x_2,Y\le y_2\}$. Probability is monotone under event inclusion, so the claimed inequality follows. Holding $y$ fixed, or holding $x$ fixed, gives the one-coordinate statements.

This is a coordinatewise order property: increasing either threshold cannot shrink the event. It does not say that the value must increase strictly. A threshold may move through a region where the random variables have no probability, leaving $F$ unchanged.

Worked Example: A Pair of Bernoulli Random Variables

Let $X$ and $Y$ be independent Bernoulli random variables with $P(X=1)=P(Y=1)=1/2$. Each takes values 0 and 1. Define $$ G(t)= \begin{cases} 0,&t<0,\\ 1/2,&0\le t<1,\\ 1,&t\ge 1. \end{cases} $$ For either variable, $P(X\le t)=G(t)$ and $P(Y\le t)=G(t)$.

Independence gives the bivariate CDF: $$ F(x,y)=P(X\le x)P(Y\le y)=G(x)G(y). $$ For example, at $x=0.4$ and $y=0.8$, both thresholds include the value 0 but exclude 1, so $F(0.4,0.8)=(1/2)(1/2)=1/4$. At $x=0.4$ and $y=1$, the first threshold still includes only 0, while the second includes both possible values. Hence $F(0.4,1)=(1/2)(1)=1/2$.

Increasing either threshold cannot reduce $G$, and therefore cannot reduce the product $G(x)G(y)$. The steps at 0 and 1 also show why a CDF need not be continuous: it can jump at values that carry positive probability.

Continuity of Probability and Right-Continuity

The continuity property of a bivariate CDF follows from a basic consequence of countable additivity. We give the probability argument explicitly because it is the key step in the proof.

Probability continuity for decreasing events: If $A_1\supseteq A_2\supseteq\cdots$ are events and $P(A_1)<\infty$, then $$ P\left(\bigcap_{n=1}^{\infty} A_n\right)=\lim_{n\to\infty}P(A_n). $$

For a proof, set $B_n=A_n\setminus A_{n+1}$. The events $B_n$ are pairwise disjoint. Moreover, $A_1$ is the disjoint union of $\bigcap_{n=1}^{\infty}A_n$ and all the $B_n$. For each $m$, $A_m$ is the disjoint union of $\bigcap_{n=1}^{\infty}A_n$ and the events $B_n$ for $n\ge m$. Countable additivity therefore gives $$ P(A_m)=P\left(\bigcap_{n=1}^{\infty}A_n\right)+\sum_{n=m}^{\infty}P(B_n). $$ The series $\sum_{n=1}^{\infty}P(B_n)$ is finite because its sum is at most $P(A_1)$. Its tails tend to zero as $m\to\infty$, proving the result. $\square$

Theorem: If $x_n\downarrow x$ and $y_n\downarrow y$, then $$ \lim_{n\to\infty}F(x_n,y_n)=F(x,y). $$ In particular, $F$ is right-continuous in either coordinate when the other is fixed.

Let $A_n=\{X\le x_n,\;Y\le y_n\}$. Since both threshold sequences decrease, the events $A_n$ decrease. Their intersection is exactly $\{X\le x,\;Y\le y\}$: membership in every $A_n$ implies $X\le x_n$ for every $n$, hence $X\le\inf_n x_n=x$, and likewise $Y\le y$. Conversely, if $X\le x$ and $Y\le y$, then $X\le x_n$ and $Y\le y_n$ for every $n$. Applying continuity of probability for decreasing events gives $$ \lim_{n\to\infty}P(A_n) =P\left(\bigcap_{n=1}^{\infty}A_n\right) =P(X\le x,\;Y\le y)=F(x,y). $$ This proves the joint statement. Taking $y_n=y$ throughout, or $x_n=x$ throughout, proves right-continuity in one coordinate. $\square$

The hypotheses specify the direction of approach. The result concerns thresholds approaching from above, not an assertion that $F$ is continuous from both sides. The Bernoulli example has jumps, but those jumps do not contradict right-continuity: at a jump location, the value of the CDF includes the probability at that location.

Worked Example: Independent Uniform Random Variables

Let $X$ and $Y$ be independent and each uniform on $[0,1]$. For one such variable define $$ H(t)= \begin{cases} 0,&t\le 0,\\ t,&0<t<1,\\ 1,&t\ge 1. \end{cases} $$ This is its CDF: for $0<t<1$, the interval $[0,t]$ has length $t$ within an interval of total length 1; below 0 the probability is zero, and at or above 1 it is one.

By independence, $$ F(x,y)=H(x)H(y). $$ For instance, $F(0.25,0.8)=0.25(0.8)=0.2$. If $x_n\downarrow x$ and $y_n\downarrow y$, continuity of $H$ from the right gives $H(x_n)\to H(x)$ and $H(y_n)\to H(y)$, so their product tends to $H(x)H(y)=F(x,y)$. The formula also makes coordinatewise monotonicity clear because $H$ is nondecreasing and takes values in $[0,1]$.

Boundary Limits

A real-valued random variable takes a finite real value, so thresholds moving far enough toward negative infinity eventually exclude any fixed outcome. This fact, together with continuity of probability, determines the lower boundary behavior of $F$.

Theorem: For each fixed real $y$ and $x$, $$ \lim_{a\to-\infty}F(a,y)=0, \qquad \lim_{b\to-\infty}F(x,b)=0. $$ Also, when both thresholds tend to positive infinity, $$ \lim_{a\to+\infty,\;b\to+\infty}F(a,b)=1. $$

For the first limit, take any sequence $a_n\downarrow-\infty$. The events $A_n=\{X\le a_n,\;Y\le y\}$ decrease, and their intersection is empty: no finite value of $X$ can be at most every $a_n$. Continuity for decreasing events yields $F(a_n,y)\to0$. Since $F(a,y)$ is nondecreasing in $a$, this establishes the limit as $a\to-\infty$. The proof for $b\to-\infty$ is the same, using the fact that $Y$ is finite.

For the upper-corner limit, take sequences $a_n\uparrow+\infty$ and $b_n\uparrow+\infty$. The events $C_n=\{X\le a_n,\;Y\le b_n\}$ increase, and their union is the entire probability space: for every outcome, both real values are finite and are eventually below the two thresholds. Continuity for increasing events follows by applying the decreasing-event result to complements, and gives $P(C_n)\to1$. Monotonicity then yields the stated joint limit.

Worked Example: Equal Variables With a Uniform Distribution

Let $U$ be uniform on $[0,1]$, and set $X=U$ and $Y=U$. The variables are dependent, but the definition of a bivariate CDF still applies. The two inequalities $U\le x$ and $U\le y$ hold together exactly when $U\le\min(x,y)$. Therefore $$ F(x,y)=H(\min(x,y)), $$ where $H$ is the uniform CDF defined above.

For $x=0.3$ and $y=0.7$, the smaller threshold is 0.3, so $F(0.3,0.7)=H(0.3)=0.3$. For $x=0.3$ and $y=0.2$, the smaller threshold is 0.2, giving $F(0.3,0.2)=0.2$. Increasing either threshold cannot increase the minimum in the decreasing direction: the minimum cannot become smaller, so $H(\min(x,y))$ cannot decrease. If either threshold tends to negative infinity while the other stays fixed, the minimum tends to negative infinity and $F$ tends to zero. If both tend to positive infinity, the minimum eventually exceeds 1 and $F$ equals 1.

Common Mistake: Treating Monotonicity as Strict or Continuity as Two-Sided

A frequent mistake is to assume that a bivariate CDF must increase whenever either coordinate increases. Monotonicity only gives a weak inequality. In the Bernoulli example, increasing $x$ from 1 to 2 changes no event at all, so $F(1,y)=F(2,y)$. Another mistake is to conclude that right-continuity makes the CDF continuous everywhere. A discrete distribution can have jumps; right-continuity says how the CDF behaves when thresholds approach from above, not from below.

The distinction matters in practice. These properties are consequences that every bivariate CDF must satisfy, whatever the joint law. They allow us to check the behavior of a proposed function at basic boundaries and threshold changes, while leaving its particular probability structure to be studied separately.

Key takeaway: A bivariate CDF is the probability of a lower-left threshold event. Event inclusion gives coordinatewise monotonicity, continuity of probability gives right-continuity, and the fact that real-valued random variables are finite gives the lower-boundary and joint upper-corner limits.

Check Your Understanding

Use the definition and probability arguments above to answer these questions.

Write the event whose probability is $F(x,y)$. Why does increasing $x$ while holding $y$ fixed not decrease this probability?
If $x_n\downarrow x$ and $y_n\downarrow y$, identify the intersection of the events $\{X\le x_n,Y\le y_n\}$.
Why does $F(a,y)\to0$ as $a\to-\infty$ for fixed $y$? State the property of probability used in the argument.
For $X=Y=U$ with $U$ uniform on $[0,1]$, calculate $F(0.4,0.6)$ and $F(0.4,0.2)$.
Can a bivariate CDF have a jump and still be right-continuous? Explain using a discrete example or the Bernoulli example above.

What You'll Learn