Reading Support and Quantiles from Distribution Plots

Why Distribution Plots Need Careful Reading

A distribution plot can make important features visible at a glance: where probability can occur, whether the law has atoms, and how a chosen quantile relates to the tails. But each plot records different information. A CDF displays cumulative probability, a density displays probability per unit length for an absolutely continuous component, and a survival function displays upper-tail probability. None should be read as if it automatically supplied all the others.

The main discipline is to translate a visual feature into a probability statement. A flat CDF means zero probability over the corresponding interval; a jump gives the probability at a point. A density curve can help locate where continuous probability lies, but it cannot show an atom. To establish support, it is not enough to know that a tail extends far: one must also check that neighborhoods of the points in question have positive probability.

Definitions for Reading the Plots

Definition: The CDF of a real-valued random variable \(X\) is \(F(x)=P(X\le x)\). Its generalized-inverse \(p\)-quantile, for \(0<p<1\), is \(Q(p)=\inf\{x:F(x)\ge p\}\). The support of the distribution is the set of points \(x\) such that every open neighborhood of \(x\) has positive probability: \(P(x-\varepsilon<X<x+\varepsilon)>0\) for every \(\varepsilon>0\). An atom at \(x\) is a point with \(P(X=x)>0\).

For a distribution with density \(f\), probabilities over intervals are found by integrating the density. A density plot describes this continuous part of the distribution; it does not by itself describe any separate point masses. The survival function under the convention used here is \(S(x)=P(X>x)=1-F(x)\). Thus it gives a strict upper-tail probability.

Theorem: For \(0<p<1\), the generalized-inverse quantile satisfies \(Q(p)\le x\) if and only if \(p\le F(x)\). Also, the mass at \(x\) is \(P(X=x)=F(x)-F(x-)\), where \(F(x-)=\lim_{t\uparrow x}F(t)\).

For the quantile statement, let \(A=\{y:F(y)\ge p\}\). The CDF limits at minus and plus infinity imply that \(A\) is nonempty and bounded below, so \(q=\inf A\) is finite. If \(p\le F(x)\), then \(x\in A\), and hence \(q\le x\). Conversely, for every \(\varepsilon>0\), the definition of the infimum gives some \(y\in A\) with \(y<q+\varepsilon\). Monotonicity gives \(F(q+\varepsilon)\ge F(y)\ge p\). Taking \(\varepsilon\) down to zero and using right-continuity of \(F\) gives \(F(q)\ge p\). Therefore, if \(q\le x\), monotonicity implies \(F(x)\ge F(q)\ge p\). This proves the equivalence. For the atom identity, the events \(\{X\le t\}\) increase to \(\{X<x\}\) as \(t\uparrow x\), so \(F(x-)=P(X<x)\). Subtracting from \(F(x)=P(X\le x)\) gives the stated mass. \(\square\)

What Each Plot Reveals

Plot	Visible feature	Probability interpretation
CDF	Jump at \(x\)	The jump size is \(P(X=x)\).
CDF	Flat on an interval	There is no probability in the corresponding open interval.
CDF	First crossing of height \(p\)	The leftmost point where \(F(x)\ge p\) is \(Q(p)\).
Density	Area over an interval	The area is the probability of that interval for the absolutely continuous law.
Survival function	Value \(S(x)\)	This is \(P(X>x)\), the probability strictly above \(x\).

The support is a local concept. For example, being able to find positive probability above every threshold proves that a distribution is unbounded above, but does not rule out gaps. To include a point in the support, check that every neighborhood around it has positive probability. For an absolutely continuous distribution, interval probabilities can be checked by integrating \(f\); from a CDF, they can be checked using CDF differences.

Worked Example: Finding Atoms, a Gap, and Quantiles from a CDF

Suppose \(X\) has probability \(1/4\) at \(-1\), and with probability \(3/4\) is uniform on \([1,3]\). Its CDF is

F(x)= \begin{cases} 0, & x<-1,\\ 1/4, & -1\le x<1,\\ 1/4+3(x-1)/8, & 1\le x<3,\\ 1, & x\ge 3. \end{cases}

At \(-1\), the CDF jumps from \(0\) to \(1/4\), so \(P(X=-1)=1/4\). It is flat from \(-1\) to \(1\), showing that \(P(-1<X<1)=0\). On \([1,3]\), it rises continuously to \(1\). The support is \(\{-1\}\cup[1,3]\): every neighborhood of \(-1\) contains its atom, every neighborhood of a point in \([1,3]\) contains positive uniform probability, and points strictly between \(-1\) and \(1\) have small neighborhoods with zero probability.

For the median, solve \(F(x)=1/2\) on the rising section:

\frac14+\frac{3(x-1)}8=\frac12 \quad\Longrightarrow\quad \frac{3(x-1)}8=\frac14 \quad\Longrightarrow\quad x=1+\frac23=\frac53.

Thus \(Q(1/2)=5/3\). For \(p=1/4\), the leftmost point where \(F(x)\ge1/4\) is \(-1\), so \(Q(1/4)=-1\), even though the CDF remains at \(1/4\) until \(x=1\). This illustrates why the quantile is the leftmost crossing, not an arbitrary point on a horizontal part of the graph.

Worked Example: Reading Support and a Median from a Density

Let \(X\) have density \(f(x)=2x\) for \(0\le x\le1\), and \(f(x)=0\) otherwise. The total area is

\int_0^1 2x\,dx=[x^2]_0^1=1.

The density is positive on \((0,1]\), and each neighborhood of \(0\) contains an interval \((0,\delta)\) with positive area \(\int_0^\delta 2x\,dx=\delta^2>0\). Each neighborhood of \(1\) also contains points in \((0,1)\) with positive density. Points outside \([0,1]\) have a sufficiently small neighborhood with zero probability. Therefore the support is \([0,1]\), including both endpoints even though \(f(0)=0\).

For \(0\le x\le1\), integration gives \(F(x)=\int_0^x2t\,dt=x^2\). The median solves \(F(x)=1/2\), so

Q(1/2)=\sqrt{1/2}=\frac{1}{\sqrt{2}}\approx0.7071.

The density plot shows a rising curve, but its height at a point is not the probability of that point. In this continuous example, every single point has probability zero; probabilities come from areas over intervals.

Reading Tail Direction from a Survival Plot

A survival plot is especially direct for upper-tail questions. Its height at \(x\) is \(P(X>x)\); a value that remains positive for arbitrarily large \(x\) signals an unbounded upper tail. It does not, by itself, prove that every point to the right belongs to the support. To establish support there, inspect local interval probabilities.

Worked Example: Support and Upper Quantiles from a Survival Function

Suppose

S(x)= \begin{cases} 1, & x<2,\\ \exp\bigl(-(x-2)/2\bigr), & x\ge2. \end{cases}

For \(x<2\), \(P(X>x)=1\), so there is no probability at or below such an \(x\); in particular, the distribution is bounded below by \(2\). For \(2\le a<b\), the probability in the interval \((a,b]\) is

P(a<X\le b)=S(a)-S(b) =\exp\bigl(-(a-2)/2\bigr)-\exp\bigl(-(b-2)/2\bigr)>0.

The strict inequality holds because \(a<b\) and the exponential function decreases as its argument increases. Every neighborhood of any \(x>2\) therefore contains an interval of positive probability. Every neighborhood of \(2\) contains some interval \((2,b]\) with \(b>2\), which also has positive probability. Points below \(2\) have neighborhoods with zero probability. Hence the support is exactly \([2,\infty)\). Moreover, \(S\) is continuous at \(2\) and everywhere else, so the law has no atoms.

For an upper-tail threshold with exceedance probability \(0.1\), solve \(S(x)=0.1\) for \(x\ge2\):

\exp\bigl(-(x-2)/2\bigr)=0.1 \quad\Longrightarrow\quad -\frac{x-2}{2}=\log(0.1)=-\log(10) \quad\Longrightarrow\quad x=2+2\log(10)\approx6.605.

At this threshold \(P(X>x)=0.1\). Equivalently, \(F(x)=1-S(x)=0.9\), so the same threshold is the 0.9-quantile.

Common Mistakes and Practical Checks

Equating a positive tail with a full support interval. Positive probability above every threshold establishes that the distribution is unbounded above, not that there are no gaps. Check positive probability in neighborhoods or intervals.
Reading a density value as a point probability. For a continuous law, \(f(x)\) is not \(P(X=x)\); interval probabilities are areas under the density.
Ignoring atoms when looking only at a density. A mixed distribution can have both a density and point masses. The density alone omits those masses; jumps in the CDF reveal them.
Choosing the wrong point on a CDF plateau. The generalized-inverse quantile is the leftmost point where the CDF reaches or exceeds the probability level.
Confusing \(P(X>x)\) and \(P(X\ge x)\). The survival convention here is strict. They differ when there is an atom at \(x\).

A reliable reading proceeds in order: identify the type of plot, translate visible jumps, flat regions, or areas into probabilities, and then check local probability before asserting a support boundary. For quantiles, use the leftmost crossing rule; for upper-tail thresholds, read the survival height directly and verify whether a point mass affects strict versus non-strict exceedance.

Check Your Understanding

Use the definitions and plot-reading rules above to answer these questions.

A CDF jumps from \(0.3\) to \(0.5\) at \(x=4\). What is \(P(X=4)\)?
A CDF is constant on \([2,5]\). Which open interval has probability zero?
For a continuous density that is zero at the left endpoint but positive immediately to its right, can the endpoint belong to the support? Explain using neighborhoods.
A survival function is positive at every finite \(x\). What can you conclude about the upper bound of the distribution, and what additional check is needed to establish support throughout a proposed interval?
If \(S(7)=0.08\), what is \(P(X>7)\), and what is \(F(7)\)?