Common Mistakes With Atoms and Densities

Why These Errors Matter

When a distribution has both point masses and a density, a probability calculation must account for two different kinds of contribution. An atom assigns positive probability to one exact value; a density assigns probability across sets of positive length through integration. Confusing these roles can produce a probability larger than one, miss an endpoint mass, or assign positive probability where none exists.

The reliable approach is to identify each contribution and its range before calculating. In particular, a density value is not itself a point probability, a CDF jump records an atom, and a density integral contributes only where the density is nonzero.

Definition: A random variable $X$ has an atom at $x$ if $P(X=x)>0$. A function $f$ is a density for a continuous component if that component assigns probability $\int_A f(x)\,dx$ to a measurable set $A$. For a law consisting of countably many atoms and a density component, a representation is $$ P(X\in A)=\sum_{j:x_j\in A}p_j+\int_A f(x)\,dx, $$ where $p_j=P(X=x_j)>0$, $f\geq 0$, and the atom masses and density component together have total probability one. The density component assigns probability zero to every singleton.

Mistake 1: Treating a Density Value as a Probability

The probability of a particular value is $P(X=x)$, not $f(x)$. For a law that has a density and no atom at $x$, $P(X=x)=0$, even if $f(x)$ is positive. Density values describe local probability per unit of the variable's scale; probabilities come from integrating over sets. For example, if $f(x)=1/2$ on $(0,2)$, the probability in an interval of length $0.2$ inside that range is $0.1$, not $1/2$.

A density can also be changed at a single point without changing any probability, because the integral over a singleton is zero. Thus, the value written for a density at one endpoint is not a way to specify an atom there. An atom must be given separately as a point probability.

Worked Example: A Positive Density Value Does Not Give a Point Probability

Suppose $X$ has density $f(x)=1/2$ for $0<x<2$, and $f(x)=0$ otherwise. Find $P(X=1)$ and $P(0.4<X\leq 1)$.

Because the law has a density and no atom at 1, a singleton has probability zero: $$ P(X=1)=\int_{\{1\}}f(x)\,dx=0. $$ The value $f(1)=1/2$ is not the probability of $X=1$.

For the interval, integrate the density over its length: $$ P(0.4<X\leq 1)=\int_{0.4}^{1}\frac12\,dx =\frac12(1-0.4)=0.3. $$ Changing whether the endpoint 1 is included does not change this result for this density-only law, since $P(X=1)=0$.

Mistake 2: Overlooking CDF Jumps

For a random variable with CDF $F(x)=P(X\leq x)$, the probability at a point is measured by the jump there. It is not generally correct to infer point probabilities from the density alone: a law may have an atom as well as a density, and the density does not describe that atom.

Theorem: For every real-valued random variable $X$ and every real $x$, $$ P(X=x)=F(x)-F(x-), $$ where $F(x-)=\lim_{t\uparrow x}F(t)=P(X<x)$.

To justify the left limit, take any sequence $t_n<x$ increasing to $x$. The events $\{X\leq t_n\}$ increase to $\{X<x\}$. Continuity of probability for an increasing sequence of events gives $F(t_n)\to P(X<x)$. Therefore $F(x-)=P(X<x)$. Since $\{X\leq x\}$ is the disjoint union of $\{X<x\}$ and $\{X=x\}$, subtracting their probabilities yields $F(x)-F(x-)=P(X=x)$. $\square$

Worked Example: Reading an Atom From a CDF Jump

Suppose $X$ has probability $0.25$ at 1 and, with probability $0.75$, is uniform on $(0,2)$. Find $P(X=1)$, $F(1-)$, and $F(1)$.

The uniform component has density $1/2$ on $(0,2)$, so its contribution to $P(X<1)$ is $$ 0.75\int_0^1\frac12\,dx=0.75(0.5)=0.375. $$ The atom at 1 does not contribute to $P(X<1)$. Hence $F(1-)=0.375$, while $P(X=1)=0.25$. The CDF at 1 includes both the continuous probability up to 1 and the atom: $$ F(1)=P(X\leq1)=0.375+0.25=0.625. $$ The jump is $0.625-0.375=0.25$, as required.

Mistake 3: Omitting or Double-Counting Endpoint Atoms

Interval notation matters when atoms are present. For example, $P(a<X\leq b)=F(b)-F(a)$, but $P(a\leq X\leq b)$ also includes any atom at $a$. A formula that uses an integral alone can omit atoms inside the interval; a formula that adds atoms to an already atom-inclusive CDF difference can count them twice.

Result: Suppose the law has the atom-and-density representation above, with no additional component. For finite $a<b$, $$ P(a<X\leq b)=\sum_{j:a<x_j\leq b}p_j+\int_{(a,b]}f(x)\,dx. $$ For a density, the integral over an endpoint does not depend on whether that endpoint is included. The atom sum, however, must follow the interval's endpoint convention.

This follows directly by applying the representation to $A=(a,b]$: each atom contributes its mass exactly when its location lies in that set, and the density component contributes its integral over that set. The integral is unchanged by including or excluding either endpoint, since a finite set has length zero. This same decomposition gives, for example, $P(a\leq X\leq b)$ by including atoms at both $a$ and $b$, while retaining the same density integral. $\square$

Worked Example: Count an Interior Atom Exactly Once

Let $X$ have mass $0.25$ at 1 and, with probability $0.75$, a uniform distribution on $(0,2)$. Find $P(1\leq X\leq1.5)$.

The interval contains the atom at 1, contributing $0.25$. The continuous component contributes its mixture weight times the uniform probability over the interval: $$ 0.75\int_1^{1.5}\frac12\,dx =0.75\left(\frac12\right)(1.5-1) =0.75(0.25)=0.1875. $$ Thus $$ P(1\leq X\leq1.5)=0.25+0.1875=0.4375. $$ The atom is included once in the atom sum. Equivalently, the answer is $F(1.5)-F(1-)$; adding the atom again to that CDF difference would double-count it.

Mistake 4: Integrating Beyond the Density's Support

A density is zero outside its specified region of support, except that its pointwise values on sets of length zero do not affect probabilities. Do not extend a convenient formula beyond the range where it is defined and then integrate it as if it still applied. For an interval that extends beyond the density region, intersect the interval with that region before integrating. Separately check whether any atoms lie in the original interval.

Worked Example: Truncate the Integral to the Density Region

Suppose $X$ has mass $0.2$ at 3 and has density $f(x)=1.6x$ for $0<x<1$, with $f(x)=0$ elsewhere. (The density component has total mass $0.8$, since $\int_0^1 1.6x\,dx=0.8$.) Find $P(-1<X\leq0.5)$ and $P(X>1)$.

The first interval includes no atom: the atom at 3 lies outside it. Its density contribution comes only from the intersection $(0,0.5]$: $$ P(-1<X\leq0.5)=\int_0^{0.5}1.6x\,dx =0.8(0.5)^2=0.2. $$ For $X>1$, the density is zero throughout the relevant region, but the atom at 3 is included. Therefore $$ P(X>1)=0.2+\int_1^\infty f(x)\,dx=0.2+0=0.2. $$ Integrating $1.6x$ beyond 1 would be an error because that formula is not the density there.

A Reliable Probability-Calculation Checklist

For a mixed law, calculate from the representation of the law rather than trying to make one formula do two jobs. First identify atoms and their masses; then identify the region where the density formula applies. For the event of interest, include each atom precisely when its location belongs to the event, and integrate the density only over the event's intersection with its region of definition.

Read the event's endpoints.
Decide exactly which values are included; this determines whether endpoint atoms count.

List included atoms.
Add each point mass whose location lies in the event, once and only once.

Restrict the integral.
Integrate the density over the event intersected with the region where that density is defined.

Check the total.
Confirm that the atom contributions and density contribution are nonnegative and do not exceed one.

A useful diagnostic is to compare a proposed point probability with the CDF jump. If the calculation gives $P(X=x)>0$ but the CDF has no jump at $x$, something is inconsistent. Conversely, a visible jump cannot be accounted for by integrating an ordinary density over a singleton: that integral is zero.

Common pitfall: Writing $P(a<X\leq b)=\int_a^b f(x)\,dx$ for a mixed law silently assumes that the law has no atoms in the interval. The correction is to add the masses of atoms in $(a,b]$, and to integrate only the density component over $(a,b]$. If using $F(b)-F(a)$, those atom contributions are already included; do not add them again.

Check Your Understanding

For each question, pay attention to whether the law has an atom, a density, or both.

A density has value 2 at $x=4$. Does that fact alone determine $P(X=4)$? Explain.
A CDF has left limit 0.31 and value 0.46 at $x=2$. What is $P(X=2)$?
A mixed law has an atom at the left endpoint $a$. Which CDF difference gives $P(a<X\leq b)$, and why does it exclude that atom?
A density formula is specified only on $(0,1)$. How should you calculate its contribution to an event extending from $-2$ to $0.6$?
When calculating an interval probability from $F(b)-F(a)$, why should you not add the atoms in $(a,b]$ a second time?