Quantile Functions as Generalized Inverses

Why a Generalized Inverse Is Needed

For a strictly increasing continuous function, finding an inverse means solving an equation such as $F(x)=p$. A distribution function need not be strictly increasing or continuous, however. It can be flat over an interval, jump at an atom, or combine continuous and discrete probability. In those cases, the equation $F(x)=p$ may have many solutions or none at all.

A quantile should still identify a threshold at which the accumulated probability reaches a specified level. The generalized inverse does this by choosing the smallest threshold whose CDF is at least that level. The word “smallest” is important: it gives a precise answer even when the CDF is flat, and the “at least” condition accommodates jumps.

Definition: Let $X$ be a real-valued random variable with distribution function $F(x)=P(X\le x)$. For $0<p<1$, the $p$-quantile (or generalized inverse of $F$ at $p$) is $$ Q(p)=\inf\{x\in\mathbb{R}:F(x)\ge p\}. $$ The set inside the infimum is the CDF threshold set for $p$.

This definition uses only the distribution function, so it applies whether the law is continuous, discrete, mixed, or singular. We restrict attention here to $0<p<1$; endpoint conventions for $p=0$ or $p=1$ can require extended-real values and depend on the distribution’s range.

Why the Infimum Is a Real Threshold

An infimum need not ordinarily belong to the set whose infimum is taken. For a CDF and $0<p<1$, it does. This fact is what makes the definition a usable threshold rather than merely a limiting location.

Theorem: For any real-valued random variable $X$ and any $p\in(0,1)$, the set $\{x:F(x)\ge p\}$ is nonempty and bounded below. Its infimum is finite and satisfies $F(Q(p))\ge p$.

Proof. The defining limits of a CDF are $F(x)\to 0$ as $x\to-\infty$ and $F(x)\to 1$ as $x\to+\infty$. Since $p>0$, there is a real number $b$ such that $F(b)<p$. Monotonicity of $F$ gives $F(x)\le F(b)<p$ for every $x\le b$, so the threshold set is bounded below. Since $p<1$, there is a real number $c$ such that $F(c)>p$, making the set nonempty. Its infimum $q$ is therefore finite.

For each positive integer $n$, the definition of infimum allows a point $x_n$ in the threshold set with $q\le x_n<q+1/n$. Thus $x_n\to q$, with each $x_n\ge q$. Right-continuity of $F$ implies $F(x_n)\to F(q)$. Because $F(x_n)\ge p$ for every $n$, it follows that $F(q)\ge p$. Hence the infimum is attained and equals $Q(p)$. $\square$

The argument relies on right-continuity, a defining property of distribution functions. It does not require a density, a continuous CDF, or a unique solution to $F(x)=p$.

The Threshold Characterization

The generalized inverse has a simple interpretation that will be useful whenever a quantile is used as a cutoff.

Theorem: For $0<p<1$ and any real $x$, $$ Q(p)\le x \quad\Longleftrightarrow\quad p\le F(x). $$

Proof. If $Q(p)\le x$, the preceding result gives $F(Q(p))\ge p$. Since $F$ is nondecreasing, $F(x)\ge F(Q(p))\ge p$. Conversely, if $F(x)\ge p$, then $x$ belongs to the threshold set $\{y:F(y)\ge p\}$. Its infimum cannot exceed one of its elements, so $Q(p)\le x$. This proves both directions. $\square$

In probability language, the equivalence says that $Q(p)$ is the boundary for thresholds $x$ at which $P(X\le x)$ has reached at least $p$. It does not assert that $F(Q(p))=p$. At an atom, the CDF may jump from below $p$ to above it. A useful way to record the possible difference is $$ F(Q(p)-)\le p\le F(Q(p)), $$ where $F(q-)=\lim_{x\uparrow q}F(x)=P(X<q)$. The left inequality follows because every $x<Q(p)$ has $F(x)<p$; taking the left limit gives $F(Q(p)-)\le p$. The right inequality follows from the theorem that the infimum is attained.

Worked Examples Across Different Laws

Worked Example: A Continuous Uniform Distribution

Let $X\sim\operatorname{Uniform}(0,1)$. Its CDF is $F(x)=0$ for $x<0$, $F(x)=x$ for $0\le x\le1$, and $F(x)=1$ for $x>1$. Find $Q(0.3)$.

For $x<0$, $F(x)=0<0.3$. On $[0,1]$, the threshold condition is $F(x)=x\ge0.3$, which holds exactly when $x\ge0.3$. For $x>1$, the condition also holds, but those values are larger. The smallest threshold is therefore $$ Q(0.3)=\inf\{x:F(x)\ge0.3\}=0.3. $$

Here $F(Q(0.3))=0.3$, so solving $F(x)=p$ gives the same answer. That convenient equality is not guaranteed for other types of laws.

Worked Example: A Discrete Distribution

Suppose $X$ takes the values $-1,2,5$ with probabilities $0.2,0.5,0.3$, respectively. The CDF is $0$ for $x<-1$, $0.2$ for $-1\le x<2$, $0.7$ for $2\le x<5$, and $1$ for $x\ge5$. Find $Q(0.2)$ and $Q(0.6)$.

For $p=0.2$, the CDF is below $0.2$ at every $x<-1$, while $F(-1)=0.2$. Thus the first threshold is $-1$, and $Q(0.2)=-1$.

For $p=0.6$, $F(x)=0.2<0.6$ throughout $[-1,2)$, but $F(2)=0.7\ge0.6$. Therefore the first threshold is $2$, and $Q(0.6)=2$. In particular, $F(Q(0.6))=0.7$, not $0.6$: the CDF jumps over the requested probability level at the atom $X=2$.

Worked Example: A Mixed Distribution

Let $X$ have probability $0.4$ at $0$, and with probability $0.6$ have a uniform distribution on $[0,2]$. Its CDF is $$ F(x)= \begin{cases} 0, & x<0,\\ 0.4+0.3x, & 0\le x\le2,\\ 1, & x>2. \end{cases} $$ The formula on $[0,2]$ includes the atom at zero as well as the continuous component. Find $Q(0.4)$ and $Q(0.7)$.

For $p=0.4$, every $x<0$ has $F(x)=0<0.4$, whereas $F(0)=0.4$. Hence $Q(0.4)=0$.

For $p=0.7$, the threshold within $[0,2]$ satisfies $0.4+0.3x\ge0.7$. Subtracting $0.4$ and dividing by $0.3$ gives $x\ge1$. Values below $0$ also fail the threshold condition, so the smallest eligible value is $1$. Thus $Q(0.7)=1$, and direct substitution checks that $F(1)=0.4+0.3=0.7$.

The same definition handled both the atom and the continuous part. For levels up to $0.4$, the atom at zero supplies the first threshold; higher levels are reached within the continuous component.

Flat CDF Intervals and the Choice of Boundary

A CDF can be constant over an interval when the distribution assigns no probability there. On such a plateau, an equation $F(x)=p$ may have multiple solutions. The infimum definition selects the leftmost solution when the plateau’s CDF value is $p$, rather than leaving the quantile ambiguous.

Worked Example: A Quantile at a Flat Region

Let $X$ be an equal mixture of a uniform distribution on $[-2,-1]$ and a uniform distribution on $[1,2]$. The first interval carries probability $0.5$, and the second interval carries probability $0.5$. The CDF reaches $0.5$ at $-1$ and remains $0.5$ for $-1\le x<1$. Find $Q(0.5)$.

At every $x<-1$, $F(x)<0.5$, because the first uniform component has not yet accumulated all its probability. At $x=-1$, $F(-1)=0.5$. The CDF then remains at $0.5$ across the gap before increasing again. Consequently, several values satisfy $F(x)=0.5$, but the first is $-1$: $$ Q(0.5)=\inf\{x:F(x)\ge0.5\}=-1. $$

This illustrates why the generalized inverse is more precise than saying “solve $F(x)=p$.” It specifies which boundary to use even when that equation does not have a unique solution.

Common Mistake and Practical Meaning

A frequent mistake is to assume that a $p$-quantile must satisfy $F(Q(p))=p$. That is true in many continuous examples, but it fails at jumps: in the discrete example, $Q(0.6)=2$ and $F(2)=0.7$. It can also fail when the CDF is flat at the relevant level, as in the final example, where $F(x)=0.5$ over an entire interval. The definition instead guarantees $F(Q(p))\ge p$ and $F(x)<p$ for every $x<Q(p)$.

Quantiles are useful as probability cutoffs precisely because this threshold interpretation survives changes in the shape of a distribution. A continuous model, a distribution with point masses, and a law with separated regions of support all use the same definition. When using a quantile, check the CDF on both sides of the proposed value: the probability must be below $p$ before the threshold and at least $p$ at the threshold.

Key takeaway: For $0<p<1$, $Q(p)$ is the leftmost real threshold at which the CDF reaches or exceeds $p$. The “at least” condition handles jumps, and the infimum selects a boundary even when a CDF is flat.

Check Your Understanding

Use the generalized-inverse definition to answer each question.

For a CDF that is continuous and strictly increasing at level $p$, what equation does its $p$-quantile satisfy?
A CDF jumps from $0.3$ to $0.8$ at $x=4$. What is $Q(0.6)$, assuming the CDF is below $0.6$ for every $x<4$?
Why does right-continuity matter in showing that the infimum threshold is attained?
If $F(x)=p$ throughout an interval $[a,b]$, and $F(x)<p$ for every $x<a$, what is $Q(p)$?
Can $F(Q(p))$ be greater than $p$? Explain using a discrete distribution.