Conditional Quantiles from Conditional CDFs

Why Conditional Quantiles Need a Careful Definition

A conditional CDF describes the distribution of one random variable after information about another has been observed. A conditional quantile answers a related question: given that information, what value marks a specified fraction of the conditional distribution? For example, a conditional median divides the conditional probability around one half.

For a continuous distribution with a strictly increasing CDF, the $u$-quantile can often be found by solving $F(q)=u$. That equation is not a general definition. A discrete CDF jumps over probability levels, and a mixed distribution can have both jumps and continuous portions. The definition must work in all these cases.

Definition: Fix a conditioning value $y$ for which a conditional distribution of $X$ given $Y=y$ is specified. Write its conditional CDF as $F_{X\mid Y}(x\mid y)=P(X\leq x\mid Y=y)$. For $0<u<1$, the lower conditional $u$-quantile is $$ Q_X(u\mid y)=\inf\{x\in\mathbb{R}:F_{X\mid Y}(x\mid y)\geq u\}. $$ The infimum convention selects the leftmost value at which the conditional CDF reaches or exceeds $u$.

When conditioning on an event of positive probability, the conditional CDF is defined by the usual conditional probability. When conditioning on a continuous random variable taking the exact value $y$, a conditional distribution is specified through a regular conditional law. Such laws are determined only up to sets of conditioning values having probability zero. Thus, at continuous conditioning values, a statement about $Q_X(u\mid y)$ refers to the chosen conditional-distribution version and is meaningful almost surely in $Y$; a particular version can be specified to give a pointwise formula.

General Properties of the Definition

The generalized-inverse definition always gives a finite real number for $0<u<1$, because a CDF tends to zero at $-\infty$ and to one at $+\infty$. Its central property is that the CDF at the quantile is at least $u$, while the probability strictly below the quantile is at most $u$.

Theorem: Let $F$ be any CDF and let $0<u<1$. Define $q=\inf\{x:F(x)\geq u\}$. Then $q$ is finite and $$ F(q-)\leq u\leq F(q), $$ where $F(q-)=\lim_{x\uparrow q}F(x)$. In probability notation for a random variable with CDF $F$, this is $P(X<q)\leq u\leq P(X\leq q)$.

Proof. Since $F(x)\to 1$ as $x\to+\infty$, the set $A=\{x:F(x)\geq u\}$ is nonempty. Since $F(x)\to 0$ as $x\to-\infty$ and $u>0$, there is a finite lower bound below which $F(x)<u$. Thus $q=\inf A$ is finite.

For every positive integer $k$, the definition of the infimum allows a choice $x_k\in A$ with $q\leq x_k<q+1/k$. Therefore $x_k\to q$, with $x_k$ approaching from the right or equalling $q$. Right-continuity of a CDF gives $F(x_k)\to F(q)$. Because $F(x_k)\geq u$ for every $k$, taking the limit yields $F(q)\geq u$.

If $x<q$, then $x\notin A$; otherwise $x$ would be a member of $A$ smaller than its infimum. Hence $F(x)<u$ for every $x<q$. Taking the limit as $x\uparrow q$ gives $F(q-)\leq u$. This proves both inequalities. $\square$

The inequalities allow for a jump at $q$: the target probability $u$ can lie anywhere between the probability strictly below $q$ and the probability at or below $q$. For a continuous, strictly increasing CDF, there is no jump and the quantile is the unique solution of $F(q)=u$. Indeed, continuity and the two inequalities give $F(q)=u$, while strict increase rules out two distinct solutions.

Worked Example: A Discrete Conditional Median

Suppose the only joint probabilities with $Y=A$ for a discrete $X$ and a two-valued variable $Y$ are $P(X=0,Y=A)=0.10$, $P(X=1,Y=A)=0.20$, and $P(X=2,Y=A)=0.10$. Find the conditional median and the conditional $0.80$-quantile of $X$ given $Y=A$.

First, $P(Y=A)=0.10+0.20+0.10=0.40$. Dividing each joint probability by $0.40$ gives $$ P(X=0\mid Y=A)=0.25,\qquad P(X=1\mid Y=A)=0.50,\qquad P(X=2\mid Y=A)=0.25. $$

The conditional CDF is zero for $x<0$, equals $0.25$ for $0\leq x<1$, equals $0.75$ for $1\leq x<2$, and equals $1$ for $x\geq2$. At $u=0.50$, the first value where the CDF is at least $0.50$ is $1$, so $Q_X(0.50\mid A)=1$. At $u=0.80$, the first such value is $2$, so $Q_X(0.80\mid A)=2$.

At the median, $F_{X\mid Y}(1-\mid A)=0.25$ and $F_{X\mid Y}(1\mid A)=0.75$. Thus $0.25\leq0.50\leq0.75$, as the theorem requires. The CDF does not take the value $0.50$; the quantile is still well-defined.

Calculating Quantiles from Conditional CDFs

For a continuous conditional law, first write the conditional CDF on each part of its support. Then set it equal to $u$ and solve when the CDF is continuous and strictly increasing there. Check that the result lies in the support and that the conditional CDF at the result is $u$.

Worked Example: A Uniform Conditional Distribution

Suppose $Y$ is uniform on $(1,3)$, and, given $Y=y$, $X$ is uniform on $(0,y)$. Find the conditional $u$-quantile for $0<u<1$, then calculate the conditional $0.80$-quantile when $y=2.5$.

For a fixed $y\in(1,3)$, the conditional density of $X$ is $1/y$ on $0<x<y$. Integrating from zero to $x$, and including the values outside the support, gives $$ F_{X\mid Y}(x\mid y)= \begin{cases} 0, & x\leq 0,\\ x/y, & 0<x<y,\\ 1, & x\geq y. \end{cases} $$

For $0<u<1$, the solution of $x/y=u$ is $x=uy$, which lies strictly between $0$ and $y$. Therefore $Q_X(u\mid y)=uy$. At $y=2.5$ and $u=0.80$, this gives $Q_X(0.80\mid 2.5)=0.80(2.5)=2.0$. Substitution checks the result: $F_{X\mid Y}(2.0\mid2.5)=2.0/2.5=0.80$.

For a mixed conditional law, a quantile may fall at an atom or in a continuous part. The generalized-inverse definition handles both cases without changing conventions.

Worked Example: A Conditional Law with an Atom

For a fixed conditioning value $y$, suppose $X$ has conditional probability $0.40$ at zero and, with probability $0.60$, has an exponential distribution with rate $2$ on the positive half-line. Find the conditional $0.30$-quantile and conditional median.

For $x<0$, the CDF is zero. For $x\geq0$, the atom and the positive continuous component contribute $$ F_{X\mid Y}(x\mid y)=0.40+0.60(1-e^{-2x}). $$

At zero, the CDF jumps from zero to $0.40$. Since $0.30\leq0.40$, the first point where the CDF is at least $0.30$ is zero: $Q_X(0.30\mid y)=0$.

For the median, $0.50$ exceeds the jump size $0.40$, so solve in the continuous part: $$ 0.40+0.60(1-e^{-2x})=0.50. $$ Subtracting $0.40$ gives $0.60(1-e^{-2x})=0.10$. Dividing by $0.60$ gives $1-e^{-2x}=1/6$, so $e^{-2x}=5/6$. Taking logarithms yields $x=\tfrac12\log(6/5)$, approximately $0.0912$. This is positive, as required for the continuous part, and substitution gives CDF value $0.50$. Thus $Q_X(0.50\mid y)=\tfrac12\log(6/5)$.

Conditional Medians and a Common Mistake

The conditional median under the lower-quantile convention is $Q_X(0.50\mid y)$. The theorem implies that at this value $q$, the conditional probability strictly below $q$ is at most one half, and the conditional probability at or below $q$ is at least one half. Equivalently, each of the events $X\leq q$ and $X\geq q$ has conditional probability at least one half. There may be more than one value satisfying these median inequalities; the lower-quantile convention selects a particular one.

Mistake: requiring $F_{X\mid Y}(q\mid y)=u$. This can fail for a discrete law because a CDF can jump from below $u$ to above it. The correction is to use the smallest $q$ with $F_{X\mid Y}(q\mid y)\geq u$, and check the two inequalities involving $F(q-)$ and $F(q)$.
Mistake: using a pointwise conditional formula without specifying its status. If $Y$ is continuous, conditioning on $Y=y$ is described through a conditional-distribution version, and that version is fixed only up to a set of $Y$-values with probability zero. State the conditional law being used and interpret its conclusions almost surely unless a version is specified pointwise.

Key takeaway: A conditional quantile is the lower generalized inverse of the conditional CDF. Its defining check is $F_{X\mid Y}(q-\mid y)\leq u\leq F_{X\mid Y}(q\mid y)$; equality with $u$ is a special feature of suitable continuous conditional laws, not a requirement of the definition.

Check Your Understanding

Use the lower generalized-inverse convention in each question.

A conditional CDF is $0.20$ for $0\leq x<1$, $0.65$ for $1\leq x<3$, and $1$ for $x\geq3$, with CDF zero for $x<0$. Find its $0.50$- and $0.80$-quantiles.
For a conditional uniform law on $(0,y)$, with $y>0$, derive the conditional $u$-quantile and identify the conditional median.
Why can a discrete conditional CDF have no value $x$ for which $F_{X\mid Y}(x\mid y)=u$, even though its $u$-quantile exists?
If $q=Q_X(0.25\mid y)$, state the inequalities involving $F_{X\mid Y}(q-\mid y)$ and $F_{X\mid Y}(q\mid y)$.
Why should a formula for a conditional quantile given a continuous $Y=y$ be understood relative to a specified conditional-distribution version?

Conditional Quantiles from Conditional CDFs

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