A Point Mass Plus Uniform Area Component

Why Separate the Point and Area Components?

A joint distribution need not be entirely discrete or entirely continuous. One part of its probability may be concentrated at a single point, while the rest is spread uniformly over a two-dimensional region. To calculate probabilities correctly, keep these contributions distinct: the point contributes its full mass whenever it belongs to the event, while the uniform component contributes in proportion to the area of the event inside its rectangle.

This mixture also illustrates why a joint CDF can have a jump even when most of the distribution has a density. The density describes the area component, but it cannot account for probability concentrated at one point. A complete representation must include both.

Definition of the Mixture

Definition: Let $0\leq p\leq 1$, and let $R=[a,b]\times[c,d]$ be a rectangle with $a<b$ and $c<d$. Fix a point $(x_0,y_0)\in\mathbb{R}^2$. A point-mass-plus-uniform-area distribution assigns probability $p$ to the point $(x_0,y_0)$ and distributes the remaining probability $1-p$ uniformly over $R$. For every measurable set $A\subseteq\mathbb{R}^2$, $$ P\big((X,Y)\in A\big) = p\,\mathbf{1}_{\{(x_0,y_0)\in A\}} + (1-p)\frac{\operatorname{area}(A\cap R)}{(b-a)(d-c)}. $$ Here $\mathbf{1}_{\{(x_0,y_0)\in A\}}$ is 1 if the point is in $A$, and 0 otherwise.

The formula also covers $p=0$, a purely uniform distribution over $R$, and $p=1$, a purely concentrated point mass. The point $(x_0,y_0)$ need not lie inside $R$. If it does lie inside $R$, its probability $p$ is still separate from the uniform component’s probability in any region around it. In particular, the point mass is not represented by an ordinary two-dimensional density.

The total probability is 1: taking $A=\mathbb{R}^2$, the point indicator is 1 and $\operatorname{area}(R)=(b-a)(d-c)$, so the two terms sum to $p+(1-p)=1$.

Worked Example: Writing the Joint CDF

Suppose $p=0.3$, the point mass is at $(1,1)$, and the uniform component is on $R=[0,2]\times[0,3]$. Derive the joint CDF $F(x,y)=P(X\leq x,Y\leq y)$.

The atom belongs to the event $\{X\leq x,Y\leq y\}$ exactly when $x\geq1$ and $y\geq1$. Thus its contribution is $0.3\,\mathbf{1}_{\{x\geq1,\ y\geq1\}}$.

For the uniform component, the intersection of $(-\infty,x]\times(-\infty,y]$ with $R$ has width $\min(\max(x,0),2)$ and height $\min(\max(y,0),3)$. Its area divided by the rectangle’s area, $2\cdot3=6$, is the uniform component’s CDF. Including its mixture weight gives $$ F(x,y) = 0.3\,\mathbf{1}_{\{x\geq1,\ y\geq1\}} + 0.7\frac{\min(\max(x,0),2)\min(\max(y,0),3)}{6}. $$

For example, at $(x,y)=(1,1.5)$, the point is included. The clipped width and height are 1 and 1.5, so $$ F(1,1.5)=0.3+0.7\frac{1\cdot1.5}{6} =0.3+0.175=0.475. $$

Joint CDF and Marginal Distributions

For general $a,b,c,d$, define the clipped fractions $$ G_X(x)=\frac{\min(\max(x-a,0),b-a)}{b-a}, \qquad G_Y(y)=\frac{\min(\max(y-c,0),d-c)}{d-c}. $$ Each fraction is 0 below its interval, increases linearly from 0 to 1 within the interval, and equals 1 above it. These are the one-dimensional CDFs of the uniform component’s two coordinates. The uniform component occupies a rectangle, so the area below $(x,y)$ is the product of its clipped width and clipped height.

Theorem: For the mixture defined above, the joint CDF and marginal CDFs are $$ F(x,y) = p\,\mathbf{1}_{\{x_0\leq x,\ y_0\leq y\}} + (1-p)G_X(x)G_Y(y), $$ $$ F_X(x)=P(X\leq x) = p\,\mathbf{1}_{\{x_0\leq x\}}+(1-p)G_X(x), \qquad F_Y(y)=P(Y\leq y) = p\,\mathbf{1}_{\{y_0\leq y\}}+(1-p)G_Y(y). $$

Proof of the joint CDF. Apply the defining probability formula to $A=(-\infty,x]\times(-\infty,y]$. The point $(x_0,y_0)$ belongs to $A$ if and only if $x_0\leq x$ and $y_0\leq y$, giving the indicator term. The intersection $A\cap R$ is a rectangle, possibly empty or truncated at one or more sides. Its width is $\min(\max(x-a,0),b-a)$, and its height is $\min(\max(y-c,0),d-c)$. Dividing their product by $(b-a)(d-c)$ gives $G_X(x)G_Y(y)$. Multiplying by $1-p$ and adding the point contribution proves the joint formula.

Proof of the first marginal. Apply the defining formula to $A=(-\infty,x]\times\mathbb{R}$. The point is included exactly when $x_0\leq x$. The uniform part of this set intersects $R$ in a vertical strip whose width is $\min(\max(x-a,0),b-a)$ and whose height is $d-c$. Dividing the strip area by the area of $R$ gives $G_X(x)$. This proves the formula for $F_X$. Applying the same argument to $A=\mathbb{R}\times(-\infty,y]$ gives the stated formula for $F_Y$. $\square$

The marginal law of $X$ therefore has an atom of mass $p$ at $x_0$, together with a uniform component on $[a,b]$ of total mass $1-p$. Likewise, $Y$ has an atom of mass $p$ at $y_0$ and a uniform component on $[c,d]$. These descriptions remain valid when an atom location is outside the corresponding interval. For example, if $x_0\notin[a,b]$, the marginal still has mass $p$ at $x_0$; that atom does not move into the interval.

Worked Example: Marginal CDFs and Interval Probabilities

Use the model with $p=0.3$, atom at $(1,1)$, and uniform component on $[0,2]\times[0,3]$. Find $F_X(1)$, $P(1<X\leq1.5)$, and $P(Y\leq1)$.

For $X$, the uniform component’s CDF at 1 is $1/2$, since 1 is halfway across $[0,2]$. The atom at $x_0=1$ is included in $F_X(1)$, so $$ F_X(1)=0.3+0.7\left(\frac{1}{2}\right)=0.65. $$

The interval $(1,1.5]$ excludes the atom at $x_0=1$. Its length within $[0,2]$ is $1.5-1=0.5$, giving uniform-component probability $0.5/2=0.25$. Therefore $$ P(1<X\leq1.5)=0.3(0)+0.7(0.25)=0.175. $$

For $Y\leq1$, the atom at $y_0=1$ is included. The uniform component contributes the fraction $1/3$ of its vertical interval, so $$ P(Y\leq1)=0.3+0.7\left(\frac{1}{3}\right) =\frac{3}{10}+\frac{7}{30} =\frac{8}{15}. $$

The inclusion of the atom at an interval endpoint depends on the event’s inequality. Here $Y\leq1$ includes it, while $1< X$ excludes the atom at $X=1$.

Probabilities for Rectangles and General Regions

For any measurable region, the same two-part calculation applies: check whether the atom belongs to the region, then calculate the area of its intersection with $R$. This works whether the region is rectangular, triangular, curved, or a union of pieces. When the region has a boundary, membership of the atom must be checked using the exact inequalities defining the event. For the area term, changing the inclusion or exclusion of a boundary curve does not change its area, provided the boundary has area zero.

Region probability rule: For every measurable $A\subseteq\mathbb{R}^2$, $$ P\big((X,Y)\in A\big) = \underbrace{p\,\mathbf{1}_{\{(x_0,y_0)\in A\}}}_{\text{point contribution}} + \underbrace{(1-p)\frac{\operatorname{area}(A\cap R)}{\operatorname{area}(R)}}_{\text{uniform area contribution}}. $$

This rule is not limited to events of the form $\{X\leq x,Y\leq y\}$. It directly gives probabilities for arbitrary regions and avoids trying to infer a two-dimensional density for the entire mixture. The area contribution does have density $(1-p)/\operatorname{area}(R)$ on $R$, but that density integrates to $1-p$, not 1. The remaining mass is the atom.

Worked Example: A Region Containing the Atom

Again take $p=0.3$, the atom at $(1,1)$, and $R=[0,2]\times[0,3]$. Find $$ P(0.5<X\leq1.5,\ 0<Y\leq2). $$

The atom $(1,1)$ satisfies all four inequalities, so it contributes $0.3$. The event’s intersection with $R$ is a rectangle of width $1.5-0.5=1$ and height $2-0=2$. Its area is $2$, while the full rectangle’s area is $2\cdot3=6$. Hence the uniform component assigns probability $2/6=1/3$ to the event before its mixture weight is applied. Thus $$ P(0.5<X\leq1.5,\ 0<Y\leq2) = 0.3+0.7\left(\frac{1}{3}\right) = \frac{3}{10}+\frac{7}{30} = \frac{8}{15}. $$

If the event instead required $X>1$, the atom would be excluded, and the probability would be only the weighted area contribution. The rectangle’s area would then be $(1.5-1)\cdot2=1$, yielding $0.7(1/6)=7/60$. The point contribution changes discontinuously when the event boundary passes the atom; the area contribution changes according to area.

A Reliable Calculation Procedure

For a new probability question, first identify the event and test whether it contains $(x_0,y_0)$. Next, intersect the event with the uniform rectangle and calculate the resulting area. Finally, add the atom’s mass, if included, to the uniform component’s weighted area fraction. Keeping those calculations separate prevents both omission of the atom and accidental assignment of the atom’s mass to the density.

Check the point.
Substitute $(x_0,y_0)$ into the event’s inequalities. Add $p$ if and only if the point belongs to the event.

Restrict to the support rectangle.
Find the intersection of the event with $R$; portions outside $R$ contribute no uniform-component probability.

Weight the area.
Divide the intersection’s area by $(b-a)(d-c)$, multiply by $1-p$, and add the point contribution from Step 1.

Common Mistakes and Why the Mixture Matters

Using only an area density. The area density accounts for total probability $1-p$, not the atom. Include the separate point contribution whenever the event contains $(x_0,y_0)$.
Counting the atom as part of the uniform area. The atom’s probability is $p$, regardless of how small a region around it is. Do not add it to the area before calculating the uniform fraction.
Ignoring the event’s endpoint convention. The events $X\leq x_0$ and $X<x_0$ have different probabilities when there is an atom at $x_0$. Check membership directly.
Calculating area outside the rectangle. Only $A\cap R$ contributes to the uniform part, even if the event extends far beyond $R$.

The mixture representation is useful whenever a joint law combines a special outcome with a spread-out component—for example, a process with a chance of producing one fixed pair of measurements and otherwise producing values uniformly over a range. The central lesson is structural: the CDF and region probabilities must preserve the concentrated mass as well as the continuous area component.

Check Your Understanding

For each question, keep the point contribution and the area contribution separate.

In the general mixture, what is the formula for $F(x,y)$, and which condition determines whether the point mass contributes?
If the atom lies outside $R$, does it disappear from the marginal distributions? Explain.
For $p=0.2$, a uniform component on $[0,4]\times[0,2]$, and an atom at $(1,1)$, calculate the probability of $0<X\leq2,\ 0<Y\leq1$.
In that same model, does the event $X<1,\ Y\leq1$ contain the atom? What is its atom contribution?
Why does the density of the area component integrate to $1-p$, rather than to 1?