Point Masses from Double CDF Increments

Why a Joint CDF Needs a Double Increment

A joint CDF records probability accumulated below a pair of thresholds. Its value at \((x,y)\), \(F(x,y)\), includes all outcomes with \(X\leq x\) and \(Y\leq y\), not just the outcome \((X,Y)=(x,y)\). To isolate the probability at that single point, we must remove the probability contributed by outcomes lying to the left or below it. The required calculation is a double increment: a difference in each coordinate.

This distinction matters for distributions with atoms, but also for mixed distributions. Probability may accumulate along a curve or over a region near \((x,y)\) without assigning positive probability to the point itself. The four CDF values in the double increment cancel those contributions and leave precisely the probability at the point.

Definition of the Four CDF Values

For random variables \(X\) and \(Y\), their joint CDF is \(F(x,y)=P(X\leq x,Y\leq y)\). At a fixed point \((x,y)\), define the left limits by taking the thresholds up to \(x\) or \(y\) from below:

\begin{aligned} F(x^-,y) &= \lim_{u\uparrow x} F(u,y),\\ F(x,y^-) &= \lim_{v\uparrow y} F(x,v),\\ F(x^-,y^-) &= \lim_{\substack{u\uparrow x\\v\uparrow y}} F(u,v). \end{aligned}

The two-coordinate limit in the last line can be taken along any sequences \(u_n\uparrow x\) and \(v_n\uparrow y\). These limits exist because a CDF is nondecreasing in each coordinate. The superscript minus indicates a limit from below; it does not mean subtraction.

Definition: The double increment of the joint CDF at \((x,y)\) is \(F(x,y)-F(x^-,y)-F(x,y^-)+F(x^-,y^-)\). It equals the point probability \(P(X=x,Y=y)\).

Why the Formula Gives a Point Probability

First, the one-sided limits have direct event interpretations:

\begin{aligned} F(x^-,y) &= P(X<x,Y\leq y),\\ F(x,y^-) &= P(X\leq x,Y<y),\\ F(x^-,y^-) &= P(X<x,Y<y). \end{aligned}

For example, take any sequence \(u_n\uparrow x\). The events \(\{X\leq u_n,Y\leq y\}\) increase to \(\{X<x,Y\leq y\}\): every value \(X<x\) is eventually at most \(u_n\), while no value \(X=x\) is. Continuity of probability for increasing sequences of events therefore gives the first identity. Applying the same argument in the other coordinate gives the second; applying it in both coordinates gives the third.

Theorem: For any real-valued random variables \(X,Y\), with joint CDF \(F\), and any real \(x,y\),

$$ P(X=x,Y=y) = F(x,y)-F(x^-,y)-F(x,y^-)+F(x^-,y^-). $$

Proof. Let \(A=\{X\leq x,Y\leq y\}\), \(B=\{X<x,Y\leq y\}\), \(C=\{X\leq x,Y<y\}\), and \(D=\{X<x,Y<y\}\). By the event interpretations above, the double increment is \(P(A)-P(B)-P(C)+P(D)\). The indicator \(1_A-1_B-1_C+1_D\) is zero unless both \(X=x\) and \(Y=y\). On that event it is one: the outcome belongs to \(A\), but not to \(B\), \(C\), or \(D\). Taking expectations proves the formula. \(\square\)

Equivalently, the double increment is the limit of probabilities of shrinking rectangles. For \(u<x\) and \(v<y\), the rectangle increment gives \(P(u<X\leq x,v<Y\leq y)\). Taking \(u\uparrow x\) and \(v\uparrow y\), these rectangles decrease to \(\{X=x,Y=y\}\). Continuity of probability for decreasing sequences applies because probabilities are finite, and yields the same result.

Worked Calculations

Worked Example: A Point Mass in a Finite Distribution

Suppose \((X,Y)\) takes the values \((0,0),(0,1),(1,0),(1,1)\) with respective probabilities \(0.2,0.3,0.1,0.4\). Find \(P(X=1,Y=0)\) using the joint CDF.

At \((1,0)\), \(F(1,0)\) includes outcomes with \(X\leq1\) and \(Y\leq0\), namely \((0,0)\) and \((1,0)\). Thus \(F(1,0)=0.2+0.1=0.3\).

For the first left limit, only outcomes with \(X<1\) and \(Y\leq0\) count, so \(F(1^-,0)=0.2\). There are no outcomes with \(Y<0\), giving \(F(1,0^-)=0\) and \(F(1^-,0^-)=0\). Therefore

$$ P(X=1,Y=0)=0.3-0.2-0+0=0.1. $$

This agrees with the assigned probability at \((1,0)\). The calculation shows why \(F(1,0)\) itself is not the point mass: it also counts the probability at \((0,0)\).

Worked Example: An Atom Alongside a Continuous Area Component

With probability \(p=0.2\), let \((X,Y)=(0,0)\). With probability \(0.8\), let \(X\) and \(Y\) be independent uniform random variables on \([0,1]\). Calculate the double increment at \((0,0)\) and at \((0.5,0.5)\).

At \((0,0)\), the CDF is \(F(0,0)=0.2\): the atom is included, while the continuous component has probability zero of being exactly \((0,0)\). If either threshold is negative, neither component contributes. Hence \(F(0^-,0)=F(0,0^-)=F(0^-,0^-)=0\), and

$$ P(X=0,Y=0)=0.2-0-0+0=0.2. $$

For thresholds \(s,t\) strictly between 0 and 1, the atom is included and the uniform component contributes probability \(st\). Thus \(F(s,t)=0.2+0.8st\). In particular, \(F(0.5,0.5)=0.2+0.8(0.25)=0.4\). The three left-limit values at \((0.5,0.5)\) are also \(0.4\), since this CDF expression has no jump there. Therefore

$$ P(X=0.5,Y=0.5)=0.4-0.4-0.4+0.4=0. $$

The continuous component assigns probability to every small area around \((0.5,0.5)\), but it assigns zero probability to the single point. The atom at the origin, by contrast, creates a positive double increment.

Worked Example: A Singular Continuous Component and an Atom

Let \(a=0.4\). With probability \(p=0.3\), set \((X,Y)=(a,a)\). With probability \(0.7\), let \(U\) be uniform on \([0,1]\) and set \(X=Y=U\). Find the point mass at \((a,a)\).

The uniform component lies on the diagonal, so its probability is concentrated along a curve rather than spread over a two-dimensional area. Nevertheless, it has probability zero of taking the specific value \(U=a\). For the full mixture, the four CDF values are

\begin{aligned} F(a,a) &= 0.3+0.7(0.4)=0.58,\\ F(a^-,a) &= 0.7(0.4)=0.28,\\ F(a,a^-) &= 0.7(0.4)=0.28,\\ F(a^-,a^-) &= 0.7(0.4)=0.28. \end{aligned}

The atom is excluded from each left-limit value; the diagonal component contributes \(0.7P(U<a)=0.7(0.4)\) to each. Substitution gives

$$ P(X=a,Y=a)=0.58-0.28-0.28+0.28=0.30. $$

Thus the diagonal component contributes no point mass, despite placing probability on the line through \((a,a)\). The double increment isolates only the atom.

Interpretation and Common Pitfalls

The formula is an inclusion-exclusion calculation: subtract the probabilities that extend left of \(x\) or below \(y\), then add back the region that was subtracted twice. What remains is the probability at the corner itself. Its value is always nonnegative because it equals a probability, even when individual CDF values are close and the arithmetic involves cancellation.

Using \(F(x,y)\) alone. This counts the whole lower-left region, not just its upper-right corner. Subtract the two one-coordinate left limits and add back the double left limit.
Taking just one difference. The quantity \(F(x,y)-F(x^-,y)\) includes all probability with \(X=x\) and \(Y\leq y\). It may include points below \(y\), so it is not generally \(P(X=x,Y=y)\).
Assuming a jump in one coordinate proves a point mass. A jump in \(F(x,y)\) as \(x\) is approached from below can reflect probability on the whole event \(X=x,Y\leq y\). The second difference is needed to isolate the specified \(y\)-value.
Confusing nearby probability with point probability. A continuous area component or a singular component on a curve can put positive probability in neighborhoods of a point while assigning zero probability to that point.

This point-mass calculation is useful whenever a joint distribution may contain atoms, whether or not the rest of the distribution is discrete. It also provides a precise diagnostic: evaluate all four corner limits before deciding whether the distribution assigns positive probability to a specified pair of values.

Check Your Understanding

Use the double-increment formula and the event interpretations to answer these questions.

Write the four joint-CDF terms whose combination gives \(P(X=x,Y=y)\).
In words, what event does \(F(x^-,y)\) represent?
Why can \(F(x,y)-F(x^-,y)\) exceed \(P(X=x,Y=y)\)?
A distribution has a continuous density near \((x,y)\), with no atom there. What is its double increment at that point, and why?
If a random vector lies on a curve, does that alone imply positive probability at each point of the curve? Explain.