Quantiles of Symmetric Distributions

Symmetry and Opposite Quantiles

For a distribution symmetric about $c$, values the same distance to the left and right of $c$ have matching probabilities. This balance connects lower-tail and upper-tail quantiles. It also explains why $c$ is a median, though it does not guarantee that $c$ is the unique median or that the usual lower quantile at probability $1/2$ equals $c$.

We use the lower generalized quantile $Q(p)=\inf\{x:F(x)\ge p\}$, for $0<p<1$. For distributions with atoms, reflecting a lower quantile naturally gives an upper generalized quantile, so both conventions are needed to state the identity without extra assumptions.

Definition: A random variable $X$ has a distribution symmetric about $c$ if $X-c$ and $-(X-c)$ have the same distribution. Equivalently, $X$ and $2c-X$ have the same distribution. The lower generalized quantile is $Q(p)=\inf\{x:F(x)\ge p\}$; the upper generalized quantile used here is $Q^+(p)=\inf\{x:F(x)>p\}$, for $0<p<1$.

The distinction between $Q$ and $Q^+$ matters at a jump or at a probability level where the CDF is flat. If $F$ is continuous and strictly increasing, they agree for probabilities in $(0,1)$. In that setting, the familiar statement is that opposite lower quantiles add to $2c$.

The Quantile Reflection Identity

Symmetry reflects locations across $c$ and exchanges the two tails. The exact identity pairs a lower quantile at probability $p$ with an upper generalized quantile at probability $1-p$.

Theorem: If $X$ is symmetric about $c$, then for every $0<p<1$, $$ Q(p)=2c-Q^+(1-p). $$ In particular, if $F$ is continuous and strictly increasing, then $Q^+(1-p)=Q(1-p)$, and $$ Q(p)+Q(1-p)=2c. $$

To prove the general identity, first center the distribution by writing $Z=X-c$, and let $G$ be the CDF of $Z$. Then $Z$ and $-Z$ have the same distribution. For every real $x$,

G(-x)=P(Z\le -x)=P(Z\ge x)=1-G(x-),

where $G(x-)=P(Z<x)$ is the left limit of the CDF. The equality involving $G(x-)$, rather than $G(x)$, accounts for any probability mass at $x$.

Set $q(p)=\inf\{x:G(x)\ge p\}$ and $q^+(u)=\inf\{x:G(x)>u\}$. Under the reflection $y=-x$, the condition $G(y)>1-p$ becomes

G(-x)>1-p \quad\Longleftrightarrow\quad 1-G(x-) > 1-p \quad\Longleftrightarrow\quad G(x-)<p.

The supremum of the $x$ satisfying $G(x-)<p$ is $q(p)$. Indeed, if $x<q(p)$, then $G(x)<p$, so $G(x-)<p$. If $x>q(p)$, the definition of the infimum gives some $a<x$ with $G(a)\ge p$, and therefore $G(x-)\ge G(a)\ge p$. Thus reflection sends the set defining $q^+(1-p)$ to a set whose supremum is $q(p)$, giving $q(p)=-q^+(1-p)$. Shifting back by $c$ yields $Q(p)=2c-Q^+(1-p)$. When $F$ is continuous and strictly increasing, the equations $F(Q(u))=u$ and $Q^+(u)=Q(u)$ hold, giving the stated lower-quantile identity. $\square$

Worked Example: A Symmetric Normal Distribution

Let $X\sim N(10,4)$, so the center is $c=10$ and the standard deviation is $2$. Its CDF is continuous and strictly increasing. Write $\Phi$ for the standard normal CDF. Standardization gives

Q(p)=10+2\Phi^{-1}(p).

The standard normal is symmetric about zero, so $\Phi^{-1}(1-p)=-\Phi^{-1}(p)$. Therefore

Q(p)+Q(1-p) =\bigl(10+2\Phi^{-1}(p)\bigr) +\bigl(10+2\Phi^{-1}(1-p)\bigr) =20.

For example, the 90th and 10th percentiles are $10+2\Phi^{-1}(0.9)$ and $10-2\Phi^{-1}(0.9)$, respectively. Their midpoint is $10$, the symmetry center. The median is $Q(0.5)=10+2\Phi^{-1}(0.5)=10$.

Worked Example: A Uniform Distribution on a Symmetric Interval

Let $X$ be uniform on $[2,8]$. The interval is symmetric about $c=5$, and for $2\le x\le8$, $F(x)=(x-2)/6$. Solving $F(x)=p$ gives

Q(p)=2+6p,\qquad 0<p<1.

This CDF is continuous and strictly increasing across the support. For any $0<p<1$,

$$ Q(p)+Q(1-p) =(2+6p)+(2+6(1-p)) =10 =2c. $$

At $p=0.8$, for instance, $Q(0.8)=6.8$ and $Q(0.2)=3.2$. Their midpoint is $5$. At the central probability, $Q(0.5)=5$, so the median agrees with the symmetry center.

Why the Symmetry Center Is a Median

A median $m$ is a value for which both $P(X\le m)\ge 1/2$ and $P(X\ge m)\ge 1/2$. Symmetry gives the center this property even when there is positive probability exactly at the center.

Theorem: If $X$ is symmetric about $c$, then $c$ is a median: $P(X\le c)\ge1/2$ and $P(X\ge c)\ge1/2$.

Let $a=P(X<c)$, $b=P(X>c)$, and $r=P(X=c)$. Reflection about $c$ maps the event $X<c$ to $X>c$, so symmetry implies $a=b$. Since these events and $X=c$ partition the sample space, $2a+r=1$. Consequently,

P(X\le c)=a+r=\frac{1+r}{2}\ge\frac12, \qquad P(X\ge c)=b+r=\frac{1+r}{2}\ge\frac12.

Thus $c$ is a median. The argument also shows why a central atom does not invalidate the conclusion: it increases both probabilities. $\square$

Worked Example: A Symmetric Discrete Distribution

Let $P(X=-2)=P(X=2)=1/2$. This distribution is symmetric about $c=0$. Its lower quantile is

Q(p)= \begin{cases} -2,&0<p\le 1/2,\\ 2,&1/2<p<1. \end{cases}

At $p=1/2$, the lower quantile is $-2$, not $0$. In fact, every $m\in[-2,2]$ is a median: $P(X\le m)\ge1/2$ and $P(X\ge m)\ge1/2$ throughout this interval. In particular, the symmetry center $0$ is a median, but it is not the lower generalized quantile $Q(1/2)$.

The upper generalized quantile resolves the reflection identity at the boundary probability. Since $Q^+(1/2)=2$, we have $Q(1/2)=0-Q^+(1/2)=-2$, as required. For $p<1/2$, $Q(p)=-2$ and $Q^+(1-p)=2$; for $p>1/2$, $Q(p)=2$ and $Q^+(1-p)=-2$. The identity holds in all three cases.

Interpreting Symmetry in a Quantile Plot

A quantile plot places probability $p$ on the horizontal axis and $Q(p)$ on the vertical axis. For a continuous distribution with a strictly increasing CDF that is symmetric about $c$, the identity $Q(p)+Q(1-p)=2c$ says that paired points at probabilities $p$ and $1-p$ have vertical coordinates equally far from $c$. The graph is therefore reflected into itself by a half-turn about $(1/2,c)$.

For example, if $Q(0.9)=c+d$, then $Q(0.1)=c-d$. The two quantiles are equally distant from the center, even if their numerical distance from zero is not relevant. The plot’s symmetry is about the point $(1/2,c)$, not necessarily about the origin.

With atoms or flat regions of the CDF, a quantile plot can have jumps or intervals of probabilities associated with the same value. In that case, pair lower and upper generalized quantiles using $Q(p)=2c-Q^+(1-p)$, rather than assuming the lower-quantile curve must have exact point-for-point symmetry at every probability. The symmetry of the distribution remains; it is the inverse convention at a boundary probability that creates the apparent discrepancy.

Common Mistake

A common mistake is to conclude that symmetry always forces $Q(1/2)=c$. The reasoning confuses the fact that $c$ is a median with the stronger claim that the lower generalized quantile at $1/2$ must equal $c$. The symmetric two-point example disproves that claim: its center is $0$, while $Q(1/2)=-2$. The correction is to distinguish medians from a particular quantile convention. If the distribution has a unique median and the quantile is well behaved at $1/2$, the values coincide; in general, symmetry guarantees that $c$ is a median, not that it is the only median.

Key takeaway: Symmetry about $c$ pairs lower and upper generalized quantiles by $Q(p)=2c-Q^+(1-p)$. When the CDF is continuous and strictly increasing, this reduces to $Q(p)+Q(1-p)=2c$. The center is always a median, but the median need not be unique.

Check Your Understanding

Use the definitions and distinctions in this tutorial to answer the following.

A distribution is continuous and strictly increasing in its CDF, and symmetric about $c=3$. If its 75th percentile is $11$, what is its 25th percentile?
Why does the general quantile reflection identity use $Q^+$ rather than $Q$ at probability $1-p$?
For the equal-probability distribution on $-2$ and $2$, give $Q(1/2)$, the symmetry center, and the full set of medians.
State the two probability inequalities that verify a value is a median, and explain why symmetry makes them hold at $c$.
In a quantile plot for a continuous, strictly increasing symmetric distribution, what point is the center of the half-turn symmetry?