Why the Shape of a CDF Matters
A cumulative distribution function (CDF) records how probability accumulates as the value of a random variable increases. Its graph therefore gives more than a visual summary: jumps identify point probabilities, smooth rises indicate probability spread over intervals, and flat portions indicate ranges receiving no probability. These features help classify a distribution and interpret probabilities without confusing a value of the CDF with probability at one point.
The key is to read the graph in two ways. First, compare the value immediately before a location with the value at that location. Then inspect how the graph behaves between locations. A graph consisting only of jumps behaves differently from one that rises smoothly, while a graph with both features represents a combination.
Reading Jumps, Smooth Rises, and Flat Portions
At a value \(a\), the jump in a CDF is the amount by which the graph rises at that value. The probability at the single point \(a\) is exactly that jump. Between jumps, a smooth increasing portion shows probability accumulating across a range of values. If the CDF is differentiable and absolutely continuous on such a range, its derivative there is the density. A flat portion shows no accumulation across the interval.
A graph may have a jump and also rise smoothly elsewhere. It is then mixed: some probability is concentrated at individual values, and some is spread over intervals. A purely discrete CDF has only jumps and is flat between them. A purely continuous CDF has no jumps and, under the continuous-distribution convention above, increases through its density.
Here \(F_X(a-)\) means the left-hand limit of the CDF at \(a\), the height just before any jump at \(a\). The interval formula follows by subtracting the event \(\{X\leq a\}\) from \(\{X\leq b\}\). The result includes \(b\) and excludes \(a\), which is why it works for discrete, continuous, and mixed distributions alike.
Each jump gives a point mass: its location is the value and its height is the probability at that value.
Flat portions carry no probability across their interiors. Smooth increasing portions indicate a continuous component when the CDF is absolutely continuous there.
Only jumps indicate a discrete distribution; smooth accumulation without jumps indicates a continuous distribution; jumps together with smooth accumulation indicate a mixed distribution.
Use the locations with probability nearby for support, and use CDF differences for interval probabilities, paying attention to open and closed endpoints.
What the Graph Reveals About Support
The support is the set of values for which every neighborhood has positive probability. For a discrete distribution, the support consists of the locations with positive jumps. For a continuous distribution, it is typically the range over which the density is positive, including boundary points when every neighborhood on the relevant side has positive probability. A mixed distribution can have both isolated support points and intervals.
Flat portions help reveal gaps in the support, but endpoint details matter. If the CDF is flat on an open interval, that interval receives no probability. An endpoint of the flat portion can still be in the support if there is an atom at the endpoint or probability accumulating immediately on its other side. The CDF graph should be read together with its jumps, not by treating every flat stretch and its endpoints as one undifferentiated gap.
Worked Example: Classifying a Step-Shaped CDF
A hypothetical discrete measurement \(X\) has a CDF that is zero below \(-1\), jumps to \(0.20\) at \(-1\), stays flat until it jumps to \(0.70\) at \(2\), stays flat again until it jumps to \(1\) at \(5\), and remains at \(1\) thereafter. Classify the distribution, find its support and point masses, and calculate \(P(X\geq 2)\).
Method. Read each jump as a point mass, then use the CDF immediately before \(2\) to account for the inclusive event \(X\geq 2\).
The graph has jumps and is flat between them, with no smooth increase. It describes a discrete distribution. The support is the set of jump locations:
At \(-1\), the jump rises from \(0\) to \(0.20\), so \(P(X=-1)=0.20\). At \(2\), it rises from \(0.20\) to \(0.70\), so \(P(X=2)=0.70-0.20=0.50\). At \(5\), it rises from \(0.70\) to \(1\), so \(P(X=5)=1-0.70=0.30\).
For \(P(X\geq 2)\), add the masses at \(2\) and \(5\): \(0.50+0.30=0.80\). Equivalently, \(P(X\geq 2)=1-F_X(2-)=1-0.20=0.80\). The masses check because \(0.20+0.50+0.30=1.00\).
Common mistake avoided: The CDF height at \(2\) is \(0.70\), but that is \(P(X\leq 2)\), not \(P(X=2)\). The point mass is the jump height, \(0.50\).
Worked Example: Classifying a Smooth CDF
A hypothetical continuous measurement has CDF \(F_X(x)=0\) for \(x\leq 0\), \(F_X(x)=x/4\) for \(0<x<4\), and \(F_X(x)=1\) for \(x\geq 4\). Classify the distribution, identify its support and density, and find \(P(1<X\leq 3)\).
Method. Check for jumps at the endpoints, use the derivative on the increasing portion to identify the density, and subtract CDF values for the interval probability.
At \(0\), the CDF is \(0\) and approaches \(0\) from the right, so there is no jump. At \(4\), it approaches \(1\) from the left and equals \(1\) at \(4\), so there is no jump there either. The CDF increases smoothly between these endpoints, so the distribution is continuous.
The CDF rises only from \(0\) to \(4\), giving support \([0,4]\). On the increasing interval, its derivative is:
For the requested event, \(F_X(3)=3/4\) and \(F_X(1)=1/4\). Thus
The result is consistent with the graph: the CDF rises by \(1/2\) between \(1\) and \(3\). There is no endpoint atom to add or subtract.
Why the Jump Formula Holds
The CDF at \(a\) is \(P(X\leq a)\). Its left limit is \(P(X<a)\): as values approach \(a\) from below, the accumulated event includes all outcomes with \(X<a\) but not outcomes with \(X=a\). The difference between those two events is precisely \(\{X=a\}\).
This argument explains why the height of a jump, rather than the CDF height itself, gives a point probability. It also explains why a continuous CDF has no point masses: if the CDF has no jump at \(a\), then \(F_X(a)=F_X(a-)\) and \(P(X=a)=0\).
On a smooth portion, a related interpretation applies when the CDF is absolutely continuous. The increase in the CDF from \(a\) to \(b\) is the integral of its density over that interval:
Thus the graph’s vertical rise over a range is the probability accumulated there. A flat segment has zero rise and therefore zero probability across its interior. A purely continuous but irregular CDF need not be differentiable everywhere; absence of jumps alone does not establish that a density exists. For the smooth CDF graphs considered here, the derivative identifies the density on the smooth portions.
Worked Example: Identifying a Mixed Distribution
A hypothetical device’s measurement \(X\) has CDF \(0\) for \(x<0\), \(0.30+0.35x\) for \(0\leq x<2\), and \(1\) for \(x\geq 2\). Classify the distribution, find the point mass and continuous probability, state the support, and calculate \(P(0<X\leq 1)\).
Method. Compare the CDF just before and at \(0\) to identify an atom, then inspect the linear rise from \(0\) to \(2\) for the continuous component.
At \(0\), the left limit is \(F_X(0-)=0\), while \(F_X(0)=0.30+0.35(0)=0.30\). The jump is \(0.30\), so \(P(X=0)=0.30\). For \(0<x<2\), the CDF rises linearly with slope \(0.35\); this is a continuous component with density \(0.35\) on that interval.
At the upper end, the linear expression approaches \(0.30+0.35(2)=1.00\). The continuous component therefore has total probability \(1.00-0.30=0.70\). The jump and continuous probabilities add to \(0.30+0.70=1.00\). Since the graph has both a jump and a smooth rise, the distribution is mixed.
The support is \([0,2]\): there is an atom at \(0\), and every neighborhood of each point in the interval, including \(2\), receives positive probability from the continuous component. For the requested interval, \(F_X(1)=0.30+0.35(1)=0.65\) and \(F_X(0)=0.30\). Therefore,
The lower endpoint is excluded, so its atom is not included. This agrees with the continuous rise of \(0.35\) between \(0\) and \(1\).
Common Exam Traps
- Reading the CDF height as a point mass. If \(F_X(a)=0.70\), that value means \(P(X\leq a)=0.70\). To find \(P(X=a)\), subtract \(F_X(a-)\).
- Ignoring whether a segment is flat or rising. A flat segment has no probability accumulating through its interior. A rising segment may represent continuous probability, even if the graph also contains jumps elsewhere.
- Calling every distribution with a jump discrete. A jump proves there is a point mass, but a smooth rise elsewhere means the distribution is mixed, not purely discrete.
- Forgetting the interval endpoints. \(F_X(b)-F_X(a)\) gives \(P(a<X\leq b)\). If the problem asks for \(P(a\leq X\leq b)\), a possible mass at \(a\) must also be included.
- Equating a flat interval with unsupported endpoints. A flat stretch rules out probability in its interior, but an endpoint can still carry an atom or be approached by probability from the other side.
Check Your Understanding
For each graph description, classify the distribution and use its jumps and CDF values carefully.
- A CDF jumps from \(0\) to \(0.25\) at \(1\), from \(0.25\) to \(0.75\) at \(4\), and from \(0.75\) to \(1\) at \(7\), staying flat between jumps. State the distribution type, support, \(P(X=4)\), and \(P(1<X\leq 7)\). (Answer: discrete; support \(\{1,4,7\}\); \(P(X=4)=0.50\); \(P(1<X\leq 7)=0.75\).)
- A CDF is \(0\) below \(-1\), is \((x+1)/6\) from \(-1\) to \(5\), and is \(1\) above \(5\), with no jumps at either endpoint. State the type and support, and find \(P(-0.5<X\leq 2)\). (Answer: continuous; support \([-1,5]\); \(P(-0.5<X\leq 2)=5/12\approx0.4167\).)
- A CDF is \(0\) below \(-1\), jumps to \(0.20\) at \(-1\), then rises linearly as \(0.20+0.20(x+1)\) for \(-1\leq x<3\), reaching \(1\) at \(3\). Find the type, atom at \(-1\), continuous probability, and \(P(-1<X\leq1)\). (Answer: mixed; \(P(X=-1)=0.20\); continuous probability \(0.80\); \(P(-1<X\leq1)=0.40\).)
- A CDF jumps from \(0\) to \(0.40\) at \(0\), stays flat at \(0.40\) until \(1\), rises linearly to \(0.70\) at \(2\), stays flat until \(5\), then rises linearly to \(1\) at \(7\). Identify its type and support, and find \(P(1<X\leq6)\). (Answer: mixed; support \(\{0\}\cup[1,2]\cup[5,7]\); \(P(1<X\leq6)=0.45\).)
- A CDF has no jumps and rises smoothly from \(0\) to \(0.60\) on \([2,4]\), remains flat until \(6\), then rises smoothly from \(0.60\) to \(1\) on \([6,8]\). State the type and support, calculate \(P(3<X\leq7)\) if each rise is linear, and explain whether \(X=4\) has positive probability. (Answer: continuous; support \([2,4]\cup[6,8]\); \(P(3<X\leq7)=0.50\); \(P(X=4)=0\).)