Quantiles at Jumps of a Discrete CDF

Why Discrete Quantiles Require Care

For a discrete random variable, the cumulative distribution function (CDF) is constant between its atom locations and increases in jumps at those locations. A requested probability level can lie anywhere inside one of these jumps. The quantile is found by asking where the CDF first reaches that level—not by interpolating within the jump.

For finite support, this gives a simple procedure: add the masses in increasing order of their support values, and stop when the cumulative total is at least the requested level. For infinite support, the same search works when a support value reaches the level. But it is important not to assume that every countable law has such a value: at an accumulation point, the CDF can reach a level without an atom there.

Definition: Let $X$ have CDF $F(x)=P(X\leq x)$. For $0<u<1$, its $u$-quantile is the generalized inverse $$ Q(u)=\inf\{x\in\mathbb{R}:F(x)\geq u\}. $$ If $X$ is discrete with atom probabilities $p(a)=P(X=a)$, then $$ F(x)=\sum_{a\leq x}p(a), $$ where the sum is over atom locations $a$. The quantile is the first location, in the sense of the infimum in this definition, where the CDF reaches or exceeds $u$.

Reading a CDF Jump

At an atom location $a$, the probability mass creates a jump from the left limit $F(a-)=\lim_{x\uparrow a}F(x)$ to $F(a)$. The levels strictly inside that jump, including its upper endpoint, have quantile $a$.

Theorem: If $a$ is an atom and $F(a-)<u\leq F(a)$, then $Q(u)=a$.

Proof. Since $F(a)\geq u$, the point $a$ belongs to the set $\{x:F(x)\geq u\}$. For every $x<a$, monotonicity gives $F(x)\leq F(a-)<u$, so no point below $a$ belongs to that set. Its infimum is therefore $a$, and $Q(u)=a$. $\square$

The strict inequality at the lower end matters. If $u=F(a-)$, the level is at the boundary before the jump. An earlier point may already have reached that level, so the quantile need not be $a$. For a finite ordered support, it is the preceding support value when that value exists. In more general countable laws, the boundary can also be associated with an accumulation point.

Worked Example: A Three-Point Distribution

Suppose $P(X=0)=0.20$, $P(X=2)=0.50$, and $P(X=5)=0.30$. Find $Q(0.40)$, $Q(0.70)$, and $Q(0.90)$.

First accumulate the masses in increasing order:

F(0)=0.20,\qquad F(2)=0.20+0.50=0.70,\qquad F(5)=0.20+0.50+0.30=1.

The level $0.40$ is above $F(0)=0.20$ and at most $F(2)=0.70$, so $Q(0.40)=2$. The level $0.70$ is exactly the upper endpoint of the jump at 2, and the definition uses $F(x)\geq u$, so $Q(0.70)=2$ as well. Finally, $0.70<0.90\leq1$, placing $0.90$ in the jump at 5, hence $Q(0.90)=5$.

The corresponding jump intervals are $0<u\leq0.20$ at 0, $0.20<u\leq0.70$ at 2, and $0.70<u\leq1$ at 5. The lower endpoint of each interval belongs to the preceding cumulative level, not to the new jump.

A Cumulative-Mass Search

When the support is finite, list its values in increasing order and form cumulative totals. When an infinite support is naturally listed as $x_0<x_1<\cdots$, use the same calculation, continuing until the cumulative total first reaches or exceeds $u$. The stopping condition is essential: for an infinite support the total need not equal 1 at any finite step.

Order the atom locations.
Write the support values in increasing order, when such a listing is available, and record each mass $p(x_j)$.

Accumulate probability.
Set $C_j=\sum_{k\leq j}p(x_k)$. For finite support, verify that the final total is 1. For infinite support, the totals should increase to 1; they need not equal 1 at a finite step.

Stop at the first reached level.
If $C_{j-1}<u\leq C_j$, then $Q(u)=x_j$. If no atom value reaches $u$, return to the generalized-inverse definition rather than assigning a nonexistent first hit.

The last qualification is not merely technical. Some countable supports have accumulation points, and the cumulative mass can approach or reach a requested level there. The infimum defining $Q(u)$ always remains the governing definition. The “first atom whose cumulative mass reaches $u$” shortcut is valid only when such an atom exists.

Worked Example: A Geometric Distribution on the Nonnegative Integers

Let $P(X=k)=0.6(0.4)^k$ for $k=0,1,2,\ldots$. Find $Q(0.50)$ and $Q(0.90)$. The geometric-series formula gives $$ F(k)=\sum_{j=0}^{k}0.6(0.4)^j=1-(0.4)^{k+1}. $$

For $u=0.50$, $F(0)=0.6\geq0.50$, and there is no support value below 0. Thus $Q(0.50)=0$.

For $u=0.90$, calculate successive totals: $$ F(0)=0.6,\qquad F(1)=1-(0.4)^2=0.84,\qquad F(2)=1-(0.4)^3=0.936. $$

Neither 0.6 nor 0.84 reaches 0.90, while 0.936 does. Therefore the first support value reaching the level is 2 and $Q(0.90)=2$. Notice that $F(k)<1$ for every finite $k$, even though $F(k)$ increases to 1 as $k$ tends to infinity. A search for a quantile level below 1 stops at a finite integer here; a search demanding a finite step with total exactly 1 would be the wrong procedure.

When the Quantile Is Not an Atom

For a countable atomic law, the support values may accumulate at a point that is not itself an atom. The quantile can be that accumulation point. In this case, no individual support value need have cumulative mass equal to or greater than the requested level before reaching the infimum.

Worked Example: An Accumulation Point Quantile

Put mass $2^{-n}$ at $x_n=1-\frac{1}{n}$ for each integer $n\geq2$, and mass $0.5$ at 2. The masses at the $x_n$ sum to $$ \sum_{n=2}^{\infty}2^{-n}=\frac{1/4}{1-1/2}=\frac12, $$ so the total probability is $0.5+0.5=1$. Find $Q(0.50)$.

The locations $x_n$ increase toward 1. At an individual such location, the cumulative mass is $$ F(x_n)=\sum_{k=2}^{n}2^{-k} =\frac12-2^{-n}<\frac12. $$

For every $x$ with $1\leq x<2$, all masses at $x_n$ have been accumulated and the atom at 2 has not, so $F(x)=0.5$. In particular, $F(1)=0.5$, although there is no atom at 1. For $x<1$, the cumulative mass is less than 0.5: below 1, at most finitely many of the locations $x_n$ have been included, and each such finite cumulative sum is less than 0.5. Hence the set $\{x:F(x)\geq0.50\}$ begins at 1, and $$ Q(0.50)=\inf\{x:F(x)\geq0.50\}=1. $$

There is no probability mass at 1, and none of the atom locations $x_n$ reaches cumulative probability 0.50. This example shows why the cumulative search is a convenient rule for common ordered supports, not a replacement for the generalized inverse in every countable case.

Checking the Definition Directly

A general property helps verify a proposed answer. For $0<u<1$, the set $\{x:F(x)\geq u\}$ is nonempty because $F(x)$ tends to 1 as $x$ tends to infinity. It is bounded below because $F(x)$ tends to 0 as $x$ tends to minus infinity. Its infimum is therefore finite. In addition, the CDF's right-continuity ensures that the infimum itself satisfies the probability threshold.

Theorem: For every CDF $F$ and $0<u<1$, $Q(u)$ is finite, $F(Q(u))\geq u$, and $F(x)<u$ for every $x<Q(u)$.

Proof. Write $A=\{x:F(x)\geq u\}$. Since $F(x)\to1$ as $x\to\infty$, $A$ is nonempty. Since $F(x)\to0$ as $x\to-\infty$, there is a lower bound for $A$. Thus $q=\inf A$ is finite. By the definition of infimum, for each positive integer $n$, there is $z_n\in A$ such that $q\leq z_n<q+1/n$. Consequently $z_n\to q$ from the right. Right-continuity gives $F(z_n)\to F(q)$, and $F(z_n)\geq u$ for all $n$, so $F(q)\geq u$. Finally, if $x<q$, then $x\notin A$; otherwise $x$ would be a member of $A$ below its infimum. Therefore $F(x)<u$. Since $q=Q(u)$, all claims follow. $\square$

This result justifies the meaning of “first” in the definition: the quantile itself reaches the requested level, and every smaller argument falls short. But it does not say that $Q(u)$ must be an atom location. The accumulation-point example satisfies the theorem exactly.

Common Mistakes

Using a strict threshold. The quantile condition is $F(x)\geq u$, not $F(x)>u$. Thus the upper endpoint of an atom's jump belongs to that atom, as in $Q(0.70)=2$ in the three-point example.
Assigning the lower jump endpoint to the new atom. For finite ordered support, if $u=C_{j-1}$, the preceding support value has already reached the level. The jump formula for $x_j$ uses $C_{j-1}<u\leq C_j$.
Requiring an infinite cumulative sum to equal 1 at a finite step. Infinite-support cumulative totals generally approach 1. Check that they increase to 1, and stop when a total first reaches the requested $u<1$, if a finite first hit exists.
Assuming every discrete quantile is an atom. A countable set of atoms can accumulate at a point where the CDF reaches the requested level. In general, use $Q(u)=\inf\{x:F(x)\geq u\}$.

Key takeaway: For a finite ordered support, accumulate the masses and select the first support value whose cumulative total is at least $u$. For an infinite support, the same rule applies when that first hit exists. In full generality, the quantile is the generalized inverse, and it may be an accumulation point rather than an atom.

Check Your Understanding

For each question, pay particular attention to whether the probability level is inside a jump or at its boundary.

A random variable has masses 0.25 at 1, 0.50 at 3, and 0.25 at 4. Find $Q(0.25)$, $Q(0.60)$, and $Q(0.75)$.
For an ordered finite support with cumulative totals $C_1=0.2$, $C_2=0.55$, and $C_3=1$, which support value is $Q(0.55)$? Explain why the inequality at the jump endpoint matters.
For an infinite support, must the cumulative probability equal 1 at some finite support value? Explain.
State the generalized-inverse definition of $Q(u)$, and explain what it guarantees about points smaller than $Q(u)$.
How can a discrete distribution have a quantile at a point with no probability mass? Describe the role of an accumulation point.