Distribution-Feature Diagnostic Workshop

A Consistent Diagnostic Routine

A distribution can be presented as a probability mass function, a density, a cumulative distribution function, or a survival function. Each representation highlights different features, and moving between them is a useful check against mistakes. In particular, support is not the same as the set of points with positive probability: a distribution can assign zero probability to a point while every neighborhood of that point has positive probability.

For a random variable $X$, write $F(x)=P(X\le x)$ for its CDF and $S(x)=P(X>x)$ for its survival function. When $X$ has atoms, the distinction between $P(X\ge x)$ and $P(X>x)$ matters at those points. The diagnostics below keep that distinction explicit.

Definition: The support of the distribution of $X$ is the set of real numbers $x$ such that every open neighborhood of $x$ has positive probability: $$ \operatorname{supp}(X)=\{x\in\mathbb{R}:P(x-\varepsilon<X<x+\varepsilon)>0\text{ for every }\varepsilon>0\}. $$ An atom at $x$ is a point for which $P(X=x)>0$. The support contains every atom, but it can also contain points that are not atoms.

For a distribution with a density, the support is determined by probability in neighborhoods, not by the value of a particular density formula at isolated points. A density can be changed at individual points without changing the distribution. For discrete distributions, each point with positive mass is in the support, but limit points of those values may be in the support as well.

What the CDF and Survival Function Reveal

The CDF is nondecreasing and right-continuous, with limits 0 and 1 at negative and positive infinity. The survival function uses a strict exceedance: $S(x)=P(X>x)$. Therefore $S(x)=1-F(x)$, including at an atom. The jump of the CDF at $x$ gives the probability of that atom.

Result: For every real $x$, $$ S(x)=1-F(x), \qquad P(X=x)=F(x)-F(x-), $$ where $F(x-)=\lim_{t\uparrow x}F(t)$.

To verify the jump identity, take $A_n=\{X\le x-1/n\}$. These events increase to $\{X<x\}$, so continuity of probability for increasing events gives $F(x-)=P(X<x)$. Since $\{X\le x\}$ is the disjoint union of $\{X<x\}$ and $\{X=x\}$, subtracting probabilities yields $F(x)-F(x-)=P(X=x)$. The survival identity follows because $\{X>x\}$ is the complement of $\{X\le x\}$. $\square$

Worked Example: A Three-Point Probability Mass Function

Suppose $X$ has the following probability mass function:

P(X=-1)=0.2,\qquad P(X=1)=0.5,\qquad P(X=3)=0.3.

The probabilities sum to $0.2+0.5+0.3=1$, so this is a valid PMF. The atoms are $-1,1,3$, and the support is exactly these three points: each has positive mass, and any point outside this finite set has a sufficiently small neighborhood with probability zero.

The CDF is

F(x)= \begin{cases} 0,&x<-1,\\ 0.2,&-1\le x<1,\\ 0.7,&1\le x<3,\\ 1,&x\ge 3. \end{cases}

Its jumps are $0.2$ at $-1$, $0.5$ at $1$, and $0.3$ at $3$, matching the PMF. The survival function, using $P(X>x)$, is

S(x)= \begin{cases} 1,&x<-1,\\ 0.8,&-1\le x<1,\\ 0.3,&1\le x<3,\\ 0,&x\ge 3. \end{cases}

For example, $S(1)=0.3$, not $0.8$, because $S(1)$ excludes the mass at 1. The distribution has bounded support, and its upper tail is exactly zero for $x\ge3$.

For $0<p<1$, define the lower quantile by $Q(p)=\inf\{x:F(x)\ge p\}$. Thus $Q(p)=-1$ for $0<p\le0.2$, $Q(p)=1$ for $0.2<p\le0.7$, and $Q(p)=3$ for $0.7<p<1$. In particular, $Q(0.5)=1$. The unique median is 1: at 1, both $P(X\le1)=0.7$ and $P(X\ge1)=0.8$ are at least one half; any value below 1 has $P(X\le m)<0.5$, while any value above 1 has $P(X\ge m)<0.5$. The mode is 1 because it has the largest point mass, $0.5$.

Quantiles, Medians, and Modes in a Mixed Law

A distribution can combine point masses with a continuous component. Its CDF then has jumps at atoms and continuous increases between them. A quantile is still read from the CDF: $Q(p)=\inf\{x:F(x)\ge p\}$. A median $m$ is any value satisfying both $P(X\le m)\ge 1/2$ and $P(X\ge m)\ge1/2$. A lower median $Q(1/2)$ is a median, but there can be other medians as well.

Worked Example: An Atom Separated From a Uniform Component

Let $X$ have probability $1/3$ at 0, and with the remaining probability $2/3$ follow a uniform distribution on $(1,3)$. The continuous component has density $1/2$ on $(1,3)$, so its contribution to the full distribution has density $ (2/3)(1/2)=1/3$ there. The full law has both an atom and a density.

The support is $\{0\}\cup[1,3]$. The atom 0 is in the support; every neighborhood of 1 or 3 also contains points in $(1,3)$ with positive probability, so the endpoints belong to the support even though neither endpoint has positive probability.

The CDF is

F(x)= \begin{cases} 0,&x<0,\\ 1/3,&0\le x<1,\\ x/3,&1\le x<3,\\ 1,&x\ge3. \end{cases}

On $[1,3)$, the formula follows by adding the atom $1/3$ to the continuous probability $(x-1)/3$. At $x=3$, the CDF reaches 1. The only atom is at 0, where the CDF jumps from 0 to $1/3$. The survival function is $1$ for $x<0$, $2/3$ for $0\le x<1$, $(3-x)/3$ for $1\le x<3$, and 0 for $x\ge3$.

For $0<p\le1/3$, the first point where the CDF reaches $p$ is 0, so $Q(p)=0$. For $1/3<p<1$, solving $x/3=p$ gives $Q(p)=3p$. In particular, the median is $Q(1/2)=3/2$. It is the unique median: at $m=3/2$, the CDF and the probability $P(X\ge m)$ both equal $1/2$; values below it fail the first median inequality, and values above it fail the second.

A careless mode calculation is not appropriate here. The atom’s probability, $1/3$, and the density height, $1/3$, are different kinds of quantities and cannot be compared as if both were point probabilities. The density is constant on $(1,3)$, but the atom is a separate part of the law. State clearly whether a mode means a largest atom or a maximizer of a specified density; a mixed distribution need not have a single mode that summarizes both parts.

Support Can Include Non-Atoms

For a finite discrete distribution, support and the set of positive-mass points coincide. That shortcut fails for some countably infinite distributions. The definition by neighborhoods works in both cases and ensures support is a closed set.

Result: The support defined by positive probability in every neighborhood is closed and has probability one.

For closedness, if $x$ is outside the support, some open interval around $x$ has probability zero. Every point sufficiently close to $x$ also has a smaller neighborhood contained in that interval, so it too is outside the support. The complement of the support is therefore open.

For the probability-one claim, use the countable collection of open intervals with rational endpoints. If $x$ is outside the support, it lies in some zero-probability open interval, and there is a rational-endpoint interval containing $x$ inside that interval. Thus the complement of the support is covered by the collection of rational-endpoint intervals having probability zero. This collection is countable, so its union has probability zero by countable subadditivity. Therefore the support has probability one. $\square$

Worked Example: A Countable Distribution With a Support Limit Point

Suppose $P(X=1/n)=2^{-n}$ for positive integers $n$. These masses sum to 1 because $\sum_{n=1}^{\infty}2^{-n}=1$. Each $1/n$ is an atom. Also, 0 is in the support: for every $\varepsilon>0$, choose $n$ large enough that $1/n<\varepsilon$. Then the neighborhood $(-\varepsilon,\varepsilon)$ contains $1/n$, which has positive probability.

But $P(X=0)=0$. Thus the support is $\{0\}\cup\{1/n:n\ge1\}$, while the set of atoms is only $\{1/n:n\ge1\}$. The point 0 is a limit point of the atoms, not an atom itself. Saying that the support consists only of the points with positive mass would miss 0.

Density and Tail Diagnostics

A density describes probabilities through areas: if $X$ has density $f$, then $P(a<X\le b)=\int_a^b f(x)\,dx$. A positive density on an interval places that interval in the support, and an endpoint can belong to the support even when its point probability is zero. Tail behavior is often clearest from $S(x)$: it records the probability of exceeding the threshold $x$.

Worked Example: A Continuous Law With a Polynomial Upper Tail

Consider the density $f(x)=2x^{-3}$ for $x\ge1$, and $f(x)=0$ for $x<1$. It integrates to one:

\int_{-\infty}^{\infty}f(x)\,dx =\int_1^\infty 2x^{-3}\,dx =\left[-x^{-2}\right]_1^\infty =1.

Every interval inside $[1,\infty)$ that meets its interior has positive probability, while intervals below 1 have none. Hence the support is $[1,\infty)$, with no atoms. Integrating the density gives

F(x)= \begin{cases} 0,&x<1,\\ 1-x^{-2},&x\ge1, \end{cases} \qquad S(x)= \begin{cases} 1,&x<1,\\ x^{-2},&x\ge1. \end{cases}

For $0<p<1$, solve $1-x^{-2}=p$ on $x\ge1$. This gives $x=(1-p)^{-1/2}$, so $Q(p)=(1-p)^{-1/2}$. The median is $Q(1/2)=\sqrt{2}$. The density decreases on $[1,\infty)$, so its maximum occurs at 1; this is a density mode, not an atom. The upper tail decays polynomially, since $S(x)=x^{-2}$ for $x\ge1$, and it is never exactly zero at a finite threshold.

As a numerical check, $S(2)=1/4$, and $Q(0.75)=2$; these agree because $P(X>2)=0.25$ and $F(2)=0.75$.

Common Diagnostic Mistakes

Equating support with positive point masses. Positive masses are in the support, but accumulation points can be in the support without being atoms.
Forgetting the strict inequality in survival. $S(x)=P(X>x)=1-F(x)$. At an atom, this excludes the mass at $x$; $P(X\ge x)$ does not.
Reading a density value as a point probability. For a continuous law, $P(X=x)=0$ even if $f(x)>0$. Probabilities come from integrals over intervals.
Comparing an atom’s mass directly with a density height. These are not interchangeable measures of concentration. Specify which kind of mode is being reported, particularly for a mixed law.
Calling every upper tail unbounded because the formula is nonzero nearby. A law with maximum possible value $b$ has $S(x)=0$ for $x\ge b$; inspect the formula and its domain.

Key takeaway: Read a distribution by checking its neighborhood support, CDF jumps, quantile crossings, and survival probabilities. Keep atoms, density heights, and tail probabilities conceptually separate, then verify that each description agrees with the others.

Check Your Understanding

For each question, justify your answer from the definition or formula rather than relying on a label alone.

A distribution has atoms at $2^{-n}$, for every positive integer $n$, with positive mass at each. Is 0 necessarily an atom? Is it in the support?
If a CDF jumps from 0.35 to 0.60 at $x=4$, what is $P(X=4)$, and what is $S(4)$?
A continuous distribution has support $[a,b]$ and a positive density on $(a,b)$. Must either endpoint have positive probability? Can either endpoint be in the support?
For the density $2x^{-3}$ on $x\ge1$, calculate $S(3)$ and $Q(8/9)$. Check that the two values are consistent.
Why is it not generally valid to compare the size of an atom with the height of a density when finding a mode of a mixed distribution?