Deriving the CDF of an Atom-Continuous Mixture

From a Point Mass and a Density to a CDF

A continuous density contributes probability over intervals, while a point mass contributes probability at one exact value. To find the cumulative distribution function, $F(x)=P(X\leq x)$, combine those contributions while keeping track of whether the threshold $x$ has reached the point mass. This last detail matters: an atom affects the CDF at its location and to its right, but not to its left.

Let $g$ be a probability density, and let $G(x)=\int_{-\infty}^{x}g(t)\,dt$ be its CDF. Fix an atom location $a\in\mathbb{R}$ and a mass $q$ with $0<q<1$. Define the law of $X$ by assigning mass $q$ to $a$ and using the density $g$, with total weight $1-q$, for the remaining probability. Precisely, for each Borel set $A$,

P(X\in A)=q\,\mathbf{1}_{\{a\in A\}}+(1-q)\int_A g(t)\,dt.

Here $\mathbf{1}_{\{a\in A\}}$ is 1 when $a\in A$ and 0 otherwise. The density $g$ integrates to 1, so the two terms assign total probability $q+(1-q)=1$.

Definition: The CDF of this atom-and-density law is $$ F(x)=q\,\mathbf{1}_{\{a\leq x\}}+(1-q)G(x) =q\,\mathbf{1}_{\{a\leq x\}}+(1-q)\int_{-\infty}^{x}g(t)\,dt. $$ The indicator is 0 for $x<a$ and 1 for $x\geq a$. Thus the atom adds a jump of size $q$ when $x$ reaches $a$.

Why the Formula Has This Form

The event $\{X\leq x\}$ can receive probability from the atom only if $a\leq x$. The continuous component contributes its weighted probability below the threshold, namely $(1-q)\int_{-\infty}^{x}g(t)\,dt$. These contributions are disjoint, so their probabilities add.

\begin{aligned} F(x) &=P(X\leq x)\\ &=q\,\mathbf{1}_{\{a\leq x\}}+(1-q)\int_{-\infty}^{x}g(t)\,dt. \end{aligned}

This derivation also shows why a single formula can cover different positions of the atom: the integral automatically reflects the continuous component’s support, while the indicator reflects the atom’s location.

Result: If $G$ is continuous, then the left limit at the atom is $F(a-)= (1-q)G(a)$, while $F(a)=q+(1-q)G(a)$. Hence $$ F(a)-F(a-)=q. $$ At every $x\neq a$, $F$ is continuous.

To verify the result, take $x<a$ and let $x$ increase to $a$. The indicator remains zero, and continuity of $G$ gives $F(a-)=(1-q)G(a)$. At $x=a$, the indicator becomes one, so $F(a)=q+(1-q)G(a)$; subtracting gives the jump $q$. At any other point, the indicator is constant in a neighbourhood of that point and $G$ is continuous, so their weighted sum is continuous there. The continuity of $G$ follows because it is the integral CDF of a density.

When the Atom Is Outside the Continuous Support

If the atom lies outside the region where the density is positive, the CDF has a flat segment between that region and the atom. The formula remains unchanged; only its piecewise expression needs to reflect where the density begins and ends.

Worked Example: An Atom Above a Uniform Component

Let the continuous component be Uniform$(0,2)$, whose density is $g(t)=1/2$ for $0<t<2$ and zero elsewhere. Give the atom mass $q=1/4$ at $a=3$. Find the CDF.

The continuous CDF is 0 for $x<0$, $x/2$ for $0\leq x<2$, and 1 for $x\geq2$. The atom contributes nothing until $x=3$. Therefore,

F(x)= \begin{cases} 0, & x<0,\\ \frac{3x}{8}, & 0\leq x<2,\\ \frac{3}{4}, & 2\leq x<3,\\ 1, & x\geq3. \end{cases}

For $0\leq x<2$, the continuous contribution is $(1-q)G(x)=\frac34\cdot\frac{x}{2}=\frac{3x}{8}$. From 2 up to (but not including) 3, the continuous component has contributed all of its weight, $3/4$, and the atom has not yet been included. At $x=3$, the CDF becomes $3/4+1/4=1$. In particular, the jump at 3 is $1-3/4=1/4$.

Worked Example: An Atom Below the Continuous Support

Let the continuous density be $g(t)=(t-1)/2$ for $1<t<3$, and zero elsewhere. It is normalized because

\int_1^3 \frac{t-1}{2}\,dt =\left.\frac{(t-1)^2}{4}\right|_1^3 =1.

Put an atom of mass $q=1/3$ at $a=0$. For $1\leq x<3$, the continuous CDF is

G(x)=\int_1^x\frac{t-1}{2}\,dt=\frac{(x-1)^2}{4}.

Consequently, the full CDF is

F(x)= \begin{cases} 0, & x<0,\\ \frac13, & 0\leq x<1,\\ \frac13+\frac{(x-1)^2}{6}, & 1\leq x<3,\\ 1, & x\geq3. \end{cases}

At $x=0$ the CDF jumps from 0 to $1/3$. It stays flat at $1/3$ until the continuous support begins at 1. At $x=3$, the expression from below approaches $1/3+4/6=1$, as required. This example illustrates that the flat portion can occur after the atom as well as before it.

When the Atom Lies Inside the Continuous Support

If $a$ is in a region where $g$ is positive, the continuous contribution is already accumulating when the atom is reached. The CDF still has exactly one jump, of size $q$, at $a$; the density component does not create a point mass there.

Worked Example: An Atom Within an Exponential Component

Let the continuous component have the Exponential$(1)$ density $g(t)=e^{-t}$ for $t\geq0$, with CDF $G(x)=1-e^{-x}$ for $x\geq0$. Put mass $q=1/3$ at $a=1$, which lies inside the continuous support. Then

F(x)= \begin{cases} 0, & x<0,\\ \frac23(1-e^{-x}), & 0\leq x<1,\\ \frac13+\frac23(1-e^{-x}), & x\geq1. \end{cases}

Immediately to the left of 1, the CDF approaches $\frac23(1-e^{-1})$. At 1, it is $\frac13+\frac23(1-e^{-1})$, so the jump is exactly $1/3$. The continuous part has no jump at 1.

For a further check, calculate $P(1/2<X\leq2)=F(2)-F(1/2)$. Since $1/2<1\leq2$, this interval includes the atom:

\begin{aligned} P(1/2<X\leq2) &=\left[\frac13+\frac23(1-e^{-2})\right] -\frac23(1-e^{-1/2})\\ &=\frac13+\frac23(e^{-1/2}-e^{-2})\\ &\approx 0.6475. \end{aligned}

The result includes the atom’s $1/3$ probability. By contrast, $P(1<X\leq2)$ excludes the atom because the interval is open at 1; subtracting $F(1)$, rather than $F(1-)$, removes that point mass.

Interval Probabilities and a Common Pitfall

For real numbers $u<v$, the CDF identity $P(u<X\leq v)=F(v)-F(u)$ can be made explicit for this law:

P(u<X\leq v) =q\,\mathbf{1}_{\{u<a\leq v\}}+(1-q)\int_u^v g(t)\,dt.

Indeed, subtracting the two CDF values cancels the continuous integral below $u$, leaving the integral from $u$ to $v$. The difference of the two atom indicators is 1 exactly when $u<a\leq v$, and is 0 otherwise. This endpoint convention is important: the interval includes its right endpoint and excludes its left endpoint.

Wrong: Add $q$ to the continuous CDF for every $x$. That would make the CDF positive even below the atom.
Correction: Add $q$ only when $x\geq a$, using $q\mathbf{1}_{\{a\leq x\}}$.
Wrong: Treat the density’s value $g(a)$ as the probability at $a$.
Correction: A density contributes probability by integration; a singleton has integral zero. The point probability here is $P(X=a)=q$.

The formula is useful because it keeps two distinct features visible: the density determines the gradual accumulation of probability, and the indicator determines where the jump occurs. Checking the left limit and value at $a$, along with the limits at the extremes, is an efficient way to catch a misplaced jump or an incorrect weight.

Check Your Understanding

Use the indicator-and-integral formula to answer the following.

A law has mass $1/5$ at $a=4$ and a continuous component Uniform$(0,1)$. Give the CDF on the intervals separated by 0, 1, and 4.
For an atom at $a$ with mass $q$, what are $F(a-)$, $F(a)$, and the jump $F(a)-F(a-)$ in terms of the continuous CDF $G$?
In the exponential example, does $P(1\leq X\leq2)$ include the atom? Explain using the CDF endpoint convention.
Why does a density component assign probability zero to the singleton $\{a\}$, even if its density is positive at $a$?
If the atom lies below the continuous support, which part of the CDF can be flat, and why?