Quantiles Under Monotone Transformations

Why Quantiles Need a Convention

A quantile describes a location in a distribution, but for a discrete distribution there may be a jump across the probability level of interest. There can also be an interval of values on which the CDF is exactly at that level. Consequently, the phrase “the $u$-quantile” needs a convention. Once that convention is fixed, a monotone transformation carries quantiles in a predictable way: an increasing map preserves their order, while a decreasing map reverses it.

The distinction matters in practice. A decreasing transformation does not simply send the lower $u$-quantile to the lower $u$-quantile. It reverses the tail being described, and for distributions with atoms the lower and upper quantile conventions can give different answers.

Definition: Let $F_X(x)=P(X\le x)$ be the CDF of a real-valued random variable $X$. For $0<u<1$, define the lower quantile and upper quantile by $$ Q_X^-(u)=\inf\{x\in\mathbb R:F_X(x)\ge u\}, \qquad Q_X^+(u)=\inf\{x\in\mathbb R:F_X(x)>u\}. $$ Both infima are finite for $0<u<1$. The upper quantile is the infimum of the locations where the CDF exceeds $u$; the infimum need not itself be a location where the CDF is strictly greater than $u$.

The lower quantile uses the first threshold at which the CDF is at least $u$. The upper quantile uses the infimum of the locations at which it is strictly greater than $u$. For a continuous strictly increasing CDF, these definitions agree. At a jump, or across a flat portion of the CDF, they need not.

Increasing Bijections Preserve Quantile Order

Suppose $g:I\to J$ is a strictly increasing bijection between intervals, $P(X\in I)=1$, and $Y=g(X)$. For $y\in J$, the inverse $g^{-1}(y)$ exists and the event $Y\le y$ is the event $X\le g^{-1}(y)$. Thus

F_Y(y)=F_X\!\left(g^{-1}(y)\right), \qquad y\in J.

The quantile identities below use quantile levels $0<u<1$. In this setting the quantiles lie in the respective intervals $I$ and $J$; values outside $J$ must still be accounted for when identifying an infimum over all real $y$.

Theorem: Under the increasing-bijection assumptions above, for every $0<u<1$, $$ Q_Y^-(u)=g\!\left(Q_X^-(u)\right), \qquad Q_Y^+(u)=g\!\left(Q_X^+(u)\right). $$

For either choice of threshold, the sets inside the quantile definitions correspond under $g$ for $y\in J$: $F_Y(y)\ge u$ exactly when $F_X(g^{-1}(y))\ge u$, and the same equivalence holds with $>$ in place of $\ge$. Because $g$ is increasing, it preserves the order of those locations. If $y\notin J$, then $Y\in J$ almost surely. A point below the interval $J$ has $F_Y(y)=0$, and a point above $J$ has $F_Y(y)=1$. Neither kind of point changes the infimum of the threshold set for $0<u<1$. Therefore each quantile infimum is the image under $g$ of the corresponding quantile infimum for $X$. $\square$

Worked Example: An Increasing Affine Transformation

Let $X$ have an exponential distribution with rate $2$, so $F_X(x)=1-e^{-2x}$ for $x\ge0$, and let $Y=3X+2$. Find the $0.25$-quantile and the median of $Y$.

For $0<u<1$, solving $1-e^{-2x}=u$ gives $Q_X(u)=-\frac12\ln(1-u)$. This CDF is continuous and strictly increasing on its support, so its lower and upper quantiles agree. The map $g(x)=3x+2$ is strictly increasing, and hence the quantiles of $Y$ are obtained by applying $g$ to those of $X$.

At $u=0.25$, $$ Q_Y(0.25)=3\left(-\frac12\ln(0.75)\right)+2 =2-\frac32\ln(0.75) \approx 2.4315. $$

At $u=0.5$, $$ Q_Y(0.5)=3\left(-\frac12\ln(0.5)\right)+2 =2+\frac32\ln 2 \approx 3.0397. $$

As a check, $Y\le y$ is equivalent to $X\le (y-2)/3$ for $y\ge2$, so $F_Y(y)=1-\exp[-2(y-2)/3]$. Substituting either calculated quantile into this CDF gives its stated probability level.

Decreasing Bijections Reverse the Quantile Order

For a decreasing bijection, the event defining the transformed CDF reverses direction. Let $g:I\to J$ be a strictly decreasing bijection between intervals, let $P(X\in I)=1$, and put $Y=g(X)$. For $y\in J$,

F_Y(y)=P\!\left(X\ge g^{-1}(y)\right) =1-F_X\!\left(g^{-1}(y)-\right),

where $F_X(x-)=\lim_{t\uparrow x}F_X(t)=P(X<x)$ is the left limit of the CDF. The left limit is essential: an atom at $x$ belongs to the event $X\ge x$, not to $X<x$.

Theorem: Under the decreasing-bijection assumptions above, for every $0<u<1$, $$ Q_Y^-(u)=g\!\left(Q_X^+(1-u)\right), \qquad Q_Y^+(u)=g\!\left(Q_X^-(1-u)\right). $$

Here is a set-based proof that also checks the points outside the range $J$. For any CDF $F_X$ and any $0<v<1$, $$ \sup\{x:F_X(x-)\le v\}=Q_X^+(v), \qquad \sup\{x:F_X(x-)<v\}=Q_X^-(v). $$ To verify the first identity, write $q=Q_X^+(v)$. If $x<q$, then $F_X(t)\le v$ for every $t<x$, so $F_X(x-)\le v$. If $x>q$, the definition of the infimum gives some $t<x$ with $F_X(t)>v$; hence $F_X(x-)\ge F_X(t)>v$. The supremum of the set is therefore $q$. For the second identity, write $r=Q_X^-(v)$. If $x<r$, then $F_X(x)<v$, since otherwise $x$ would belong to the set whose infimum is $r$. Monotonicity gives $F_X(x-)\le F_X(x)<v$. If $x>r$, some $t<x$ has $F_X(t)\ge v$, so $F_X(x-)\ge v$. Thus this set has supremum $r$. $\square$

Set $v=1-u$. For $y\in J$, write $y=g(x)$. The transformed-CDF formula shows that $F_Y(y)\ge u$ exactly when $F_X(x-)\le v$; it shows that $F_Y(y)>u$ exactly when $F_X(x-)<v$. Since $g$ is decreasing, the infimum of the corresponding $y$-locations is $g$ applied to the supremum of the $x$-locations. If $y\notin J$ and $y<g(q)$, where $q$ is the relevant quantile in $I$, then $y$ lies below $J$: because $J$ is an interval and $g(q)\in J$, a point outside $J$ below $g(q)$ cannot lie above $J$. As $Y\in J$ almost surely, $F_Y(y)=0<u$. Such points do not lower the infimum. Points above $J$ have $F_Y(y)=1$ and cannot lower it either. This proves both identities. $\square$

Worked Example: A Decreasing Transformation of an Exponential Variable

Let $X$ have an exponential distribution with rate $1$, and set $Y=1/X$. The map $g(x)=1/x$ is a strictly decreasing bijection from $(0,\infty)$ to $(0,\infty)$. Find the lower $0.8$-quantile of $Y$.

The exponential CDF is continuous and strictly increasing on its support, so its two quantile conventions agree. Solving $1-e^{-x}=u$ gives $Q_X(u)=-\ln(1-u)$. The decreasing-transformation identity uses the complementary level $1-u=0.2$: $$ Q_Y^-(0.8)=\frac{1}{Q_X(0.2)} =\frac{1}{-\ln(0.8)} \approx \frac{1}{0.22314355} \approx 4.4814. $$

Check directly: for $y>0$, $Y\le y$ is equivalent to $X\ge1/y$. Therefore $F_Y(y)=P(X\ge1/y)=e^{-1/y}$, and $$ F_Y(4.4814)\approx e^{-0.22314}=0.8. $$ The high quantile of $Y$ corresponds to a low quantile of $X$, because taking a reciprocal reverses order.

Worked Example: Why the Two Conventions Matter for Atoms

Suppose $X$ equals $0$ and $1$, each with probability $1/2$, and let $Y=-X$. Compute both $0.5$-quantiles under each convention.

For $X$, the CDF is $0$ below $0$, $1/2$ on $[0,1)$, and $1$ on $[1,\infty)$. Thus $$ Q_X^-(0.5)=0,\qquad Q_X^+(0.5)=1. $$ For $Y$, the values are $-1$ and $0$, each with probability $1/2$. Its CDF is $0$ below $-1$, $1/2$ on $[-1,0)$, and $1$ on $[0,\infty)$, giving $$ Q_Y^-(0.5)=-1,\qquad Q_Y^+(0.5)=0. $$

The decreasing identities agree with these calculations: $$ Q_Y^-(0.5)=-Q_X^+(0.5)=-1, \qquad Q_Y^+(0.5)=-Q_X^-(0.5)=0. $$ The two answers differ because the CDF has a flat stretch at height $1/2$; using only the phrase “the median” would conceal which convention is intended.

Common Mistakes and Practical Use

Using the same quantile convention after a decreasing map. The lower quantile of $Y$ is obtained from the upper quantile of $X$ at level $1-u$, not generally from the lower $u$-quantile of $X$.
Replacing $>$ by $\ge$ without checking for atoms. In the two-point example, this changes the $0.5$-quantile from $1$ to $0$ for $X$.
Assuming the infimum has to belong to the set. For the upper quantile, the infimum of locations where $F_X(x)>u$ need not itself satisfy $F_X(x)>u$. A continuous strictly increasing CDF has $F_X(Q_X^+(u))=u$.
Applying an inverse outside its domain. In a proof using $g^{-1}(y)$, first restrict to $y\in J$. Points outside the image interval must be handled using $Y\in J$ almost surely and the CDF values below or above $J$.

Quantile transformations are useful for reporting percentiles on a changed measurement scale. For an increasing conversion, such as converting a positive measurement to a larger unit, the quantile level is unchanged. For a decreasing score or reciprocal, the level is complemented and the lower and upper conventions switch. These identities can also provide a direct check on a transformed-distribution calculation.

Key takeaway: Increasing bijections preserve both quantile conventions at the same probability level. Decreasing bijections reverse the order and exchange the conventions: the lower $u$-quantile becomes the image of the upper $1-u$-quantile, and vice versa.

Check Your Understanding

Use the stated lower and upper quantile definitions, especially when the distribution has atoms.

For a continuous strictly increasing CDF, why do the lower and upper quantiles agree?
If $Y=5-X$, express $Q_Y^-(u)$ and $Q_Y^+(u)$ in terms of the quantiles of $X$.
A random variable takes the values $2$ and $6$, each with probability $1/2$. Find its lower and upper $0.5$-quantiles.
For $X$ exponential with rate $1$ and $Y=1/X$, write the lower $0.9$-quantile of $Y$ in terms of a quantile of $X$.
Why must a proof using $g^{-1}(y)$ distinguish $y$ inside the image interval from $y$ outside it?

Quantiles Under Monotone Transformations

What You'll Learn