What a Flat CDF Tells Us
The cumulative distribution function records how much probability has accumulated up to each point. When it stays constant as \(x\) moves through an interval, no probability is being added there. That simple observation is useful for locating gaps in a distribution’s support, but it requires care: the endpoints of a gap can still be in the support, and atoms at endpoints can produce jumps.
Recall that the support of a distribution on the real line consists of points whose every open neighborhood has positive probability. A point can belong to the support even when the probability of the point itself is zero. For example, a continuous distribution can have an endpoint in its support without placing any atom there.
For \(a<b\), the increase in the CDF from \(a\) to \(b\) is the probability in a half-open interval:
This identity follows by writing \(\{X\leq b\}\) as the disjoint union of \(\{X\leq a\}\) and \(\{a<X\leq b\}\). The half-open interval matters: the increment excludes a possible atom at \(a\), but includes a possible atom at \(b\).
Flatness and Zero Probability
The exact statement for an open interval is that the CDF is constant throughout that interval if and only if the interval has probability zero. A constant CDF cannot accumulate probability between any two of its arguments. Conversely, if the interval has probability zero, none of its subintervals can add probability.
Proof. Suppose first that \(F\) is constant on \((a,b)\). For any \(a<x<y<b\),
Choose sequences \(x_n\) decreasing to \(a\) and \(y_n\) increasing to \(b\), with \(a<x_n<y_n<b\). The intervals \((x_n,y_n]\) can be chosen increasing, and their union is \((a,b)\). By continuity of probability for increasing sequences of events,
For the reverse direction, suppose \(P(a<X<b)=0\). If \(a<x<y<b\), then \((x,y]\subset(a,b)\), so
Thus \(F(y)=F(x)\) for every pair \(x,y\) in \((a,b)\), and \(F\) is constant there. \(\square\)
The same reasoning shows why the interval endpoints must be treated separately. If \(P(a<X<b)=0\), then \(F(x)=F(a)\) for every \(x\in[a,b)\). But at \(b\),
So a CDF can be flat up to \(b\) and then jump at \(b\). A similar jump can occur at \(a\) when the CDF is viewed from the left. The open interval’s probability says nothing by itself about atoms at either endpoint.
Worked Example: A Discrete Distribution With a Flat Region
Suppose \(X\) takes the value \(-1\) with probability \(0.25\) and the value \(2\) with probability \(0.75\). Its CDF is
For every \(x\) between \(-1\) and \(2\), the CDF equals \(0.25\). The interval \((-1,2)\) contains neither possible value of \(X\), so
The support is \(\{-1,2\}\): every neighborhood of either point contains its positive point mass, and a point other than these two can be surrounded by a neighborhood with probability zero. Thus \((-1,2)\) is a gap in the support, while both endpoints belong to the support. The CDF jumps at \(-1\) and \(2\); flatness strictly between them does not remove either endpoint.
Zero-Probability Intervals and Support Gaps
Flat CDF regions identify intervals with zero probability. To connect that fact precisely to support, it helps to establish that a distribution on the real line assigns probability one to its support. This ensures that an interval containing no support points cannot carry hidden probability.
Proof. If \(x\) is outside the support, the definition of support gives an open neighborhood of \(x\) with probability zero. There are rational numbers \(r<s\) such that \(x\in(r,s)\) and \((r,s)\) lies inside that neighborhood. Therefore \(P(r<X<s)=0\).
There are only countably many intervals with rational endpoints. The complement of the support is covered by the collection of those rational-endpoint intervals that have probability zero. A countable union of probability-zero events has probability zero, so \(P(X\text{ is outside the support})=0\). The support therefore has probability one.
Now if \((a,b)\) is disjoint from the support, the event \(\{a<X<b\}\) is contained in the event that \(X\) is outside the support, which has probability zero. Conversely, if \(P(a<X<b)=0\), every point \(x\in(a,b)\) has a small open neighborhood contained in \((a,b)\), and that neighborhood has probability zero. Hence \(x\) is not in the support. This proves both directions. \(\square\)
Together, these results give a practical interpretation: a CDF is flat on an open interval exactly when that interval is a gap in the support. The statement applies to discrete, continuous, and mixed distributions. It concerns probability in an interval, not merely whether a density formula happens to be zero there.
Worked Example: A Gap Between Two Continuous Pieces
Let \(X\) have density
The total probability is \(0.4(1-0)+0.6(4-3)=0.4+0.6=1\), so this is a valid density. For \(1\leq x<3\), all probability on the first interval has accumulated and none on the second has accumulated. Therefore
In particular, \(P(1<X<3)=0\), so \((1,3)\) is a gap in the support. The support is \([0,1]\cup[3,4]\). To check the endpoints, every neighborhood of \(1\) contains points just below \(1\) with positive density, and every neighborhood of \(3\) contains points just above \(3\) with positive density. Thus \(1\) and \(3\) belong to the support even though \(P(X=1)=P(X=3)=0\). There are no jumps at those points; the flat CDF ends at \(3\) because probability begins accumulating again to its right.
Reading Endpoints and Tails
Flat intervals can mark internal gaps, while constant tails can indicate the outer boundaries of support. If \(F(x)=0\) for all \(x<c\), then \(P(X<c)=0\). If, in addition, the support is nonempty and bounded below with lower endpoint \(c\), that endpoint is in the support because the support is closed. It need not be an atom: \(F(c)\) can still equal zero. Likewise, if \(F(x)=1\) for all \(x>d\), then there is no probability above \(d\); the upper support endpoint \(d\) need not have positive point mass.
This distinction is useful when reading a graph. A jump represents an atom, while a flat stretch represents zero probability in the stretch’s interior. Support records where probability occurs arbitrarily nearby, so it includes limit points of positive-probability regions as well as points with positive mass.
Worked Example: A Mixed Law and Its Support Endpoints
Suppose \(X\) has probability \(0.2\) at \(-2\), probability \(0.5\) uniformly distributed over \([0,1]\), and probability \(0.3\) at \(3\). The weights sum to \(0.2+0.5+0.3=1\). Its CDF is
On \((-2,0)\), the CDF is \(0.2\), and \(P(-2<X<0)=0\). On \((1,3)\), it is \(0.7\), and \(P(1<X<3)=0\). The support is \(\{-2\}\cup[0,1]\cup\{3\}\). The points \(-2\) and \(3\) are support points because they have positive atoms. The points \(0\) and \(1\) are support points because every neighborhood contains part of the uniform component with positive probability, although neither endpoint has an atom. Thus the two flat regions correspond to gaps, but their boundary points can be in the support for different reasons.
Common Pitfalls
- Treating the endpoints as part of the zero-probability interval. From \(P(a<X<b)=0\), one cannot conclude that \(P(X=a)=P(X=b)=0\). Check endpoint atoms separately.
- Equating support membership with positive point mass. A point can have zero probability by itself and still be in the support, as at the endpoints of a uniform interval.
- Using a zero density as proof of a support gap. A density describes the absolutely continuous component. A distribution can also have singular probability, so a zero density on an interval alone does not establish that the interval has zero probability. The decisive condition is \(P(a<X<b)=0\), or equivalently a flat CDF there.
- Calling every flat stretch an outer endpoint. A flat stretch between regions of probability is an internal gap. A constant tail indicates absence of probability beyond a boundary; the CDF’s value and any endpoint jump still need to be checked.
Check Your Understanding
Use the interval-probability identity and the definition of support to answer these questions.
- If \(F\) is constant on \((a,b)\), what is \(P(a<X<b)\)? Why does this not determine \(P(X=b)\)?
- A CDF is constant on \([2,5)\) and jumps at \(5\). Which part of the interval can be inferred to have zero probability, and what can the jump tell you?
- For a uniform distribution on \([0,1]\), explain why \(0\) and \(1\) belong to the support although each has probability zero.
- Why is a density equal to zero on an interval not, by itself, enough to prove that the interval has zero probability?
- A distribution has support \(\{-1\}\cup[2,4]\). Name an open gap between support pieces and state what the CDF does across that gap.