Why CDF Jumps Matter
A cumulative distribution function records the probability that a random variable is no greater than a specified value. Its behavior at a single point can therefore answer a precise question: how much probability is assigned to that exact value? The answer is the size of the CDF's jump there.
This connection is useful whenever a distribution has values with positive probability. It applies to discrete distributions and also to mixed distributions, where some probability is assigned to individual points while other probability is spread across ranges. A CDF can have both jumps and increases between them, so the jump—not the CDF's total value—is what measures a point mass.
Deriving the Point-Mass Identity
The event \(\{X\le a\}\) consists of two disjoint events: \(\{X<a\}\) and \(\{X=a\}\). The left limit \(F(a-)\) gives the probability of the first event. Subtracting that probability from \(F(a)\), which includes both events, leaves the probability of equality.
To see why \(F(a-)=P(X<a)\), consider values \(x\) approaching \(a\) from below. The events \(\{X\le x\}\) grow toward \(\{X<a\}\); their limiting probability is \(P(X<a)\). Thus the identity holds for any random variable and any real \(a\), not only for a discrete random variable.
The size of this difference is called the jump at \(a\). If \(F(a)=F(a-)\), there is no jump and \(P(X=a)=0\). A CDF can still increase across an interval even when it has no jump at a particular point. In a mixed distribution, for example, some probability may occur as jumps and some may contribute to increases between jumps.
Reading Jumps in a Finite-Support CDF
For a finite-support distribution, the CDF is constant on intervals containing no possible values of \(X\), and it jumps at the values that have positive probability. This familiar stepwise pattern is one useful case, not a description of every discrete distribution. For instance, a distribution can assign positive probability to a dense set of values, in which case there need not be any interval on which its CDF is constant.
Worked Example: Recovering Probabilities From a Stepwise CDF
Distribution and parameters. A random variable \(X\) has support \(\{0,2,5\}\), with \(P(X=0)=0.20\), \(P(X=2)=0.45\), and \(P(X=5)=0.35\). Its CDF is \(F(x)=0\) for \(x<0\), \(F(x)=0.20\) for \(0\le x<2\), \(F(x)=0.65\) for \(2\le x<5\), and \(F(x)=1\) for \(x\ge5\). Find the point mass at each support value using the jumps.
Method. At each support value \(a\), subtract the left limit \(F(a-)\) from the CDF value \(F(a)\).
At \(a=0\), \(F(0)=0.20\) and \(F(0-)=0\), so \(P(X=0)=0.20-0=0.20\).
At \(a=2\), \(F(2)=0.65\) and \(F(2-)=0.20\), so \(P(X=2)=0.65-0.20=0.45\).
At \(a=5\), \(F(5)=1\) and \(F(5-)=0.65\), so \(P(X=5)=1-0.65=0.35\).
Check. The recovered masses sum to \(0.20+0.45+0.35=1.00\), as required for this distribution, whose support consists of these three points. The common mistake avoided here is treating \(F(a)\) itself as the point mass; it includes probability at every value up to and including \(a\).
Jumps in a Mixed CDF
In a mixed distribution, the CDF may rise between points as well as jump at them. A jump still gives the point mass, but the total probability need not be the sum of the jumps alone. Probability represented by increases between jumps must also be accounted for.
Worked Example: Separating Point Masses From Increases Between Them
Distribution and parameters. A random variable \(X\) has CDF \(F(x)=0\) for \(x<0\); \(F(x)=0.20+0.10x\) for \(0\le x<4\); \(F(x)=0.75+0.05(x-4)\) for \(4\le x<7\); and \(F(x)=1\) for \(x\ge7\). Find \(P(X=0)\), \(P(X=4)\), and \(P(X=7)\), and explain why their sum is less than one.
Method. Find the left limit and the CDF value at each point, then subtract. On the first interval, the CDF approaches \(0.20+0.10(4)=0.60\) as \(x\) approaches \(4\) from below. On the second interval, it approaches \(0.75+0.05(7-4)=0.90\) as \(x\) approaches \(7\) from below.
At \(0\), \(F(0)=0.20\) while \(F(0-)=0\), giving \(P(X=0)=0.20\).
At \(4\), \(F(4)=0.75\) and \(F(4-)=0.60\), giving \(P(X=4)=0.75-0.60=0.15\).
At \(7\), \(F(7)=1\) and \(F(7-)=0.90\), giving \(P(X=7)=1-0.90=0.10\).
The three point masses sum to \(0.20+0.15+0.10=0.45\). This is less than one because the CDF also increases between the jumps: it rises by \(0.60-0.20=0.40\) from just after \(0\) to just before \(4\), and by \(0.90-0.75=0.15\) from \(4\) to just before \(7\). Those increases total \(0.55\), and \(0.45+0.55=1.00\).
Check. The CDF is nondecreasing, its endpoint values progress from zero to one, and the point masses plus the between-jump increases account for the full probability. The common mistake avoided is taking \(F(4)=0.75\) as the mass at \(4\); the correct jump is \(0.75-0.60=0.15\).
Endpoint Conventions and Interval Probabilities
The point-mass identity also clarifies why interval endpoints matter. For \(a<b\), \(P(a<X\le b)=F(b)-F(a)\): subtracting \(F(a)\) removes all outcomes at or below \(a\), while \(F(b)\) includes outcomes through \(b\). If the left endpoint is included instead, then \(P(a\le X\le b)=F(b)-F(a-)\). The two expressions differ by \(P(X=a)\).
Worked Example: Using Jumps to Handle Inclusive Endpoints
Distribution and parameters. Let \(X\) have the stepwise CDF \(F(x)=0\) for \(x<0\), \(F(x)=0.10\) for \(0\le x<2\), \(F(x)=0.35\) for \(2\le x<4\), \(F(x)=0.60\) for \(4\le x<6\), and \(F(x)=1\) for \(x\ge6\). Find \(P(X=4)\) and \(P(2<X\le6)\).
Method. First use the jump formula for equality. Then use the CDF difference for an interval open on the left and closed on the right.
At \(4\), \(F(4)=0.60\) and \(F(4-)=0.35\). Therefore, \(P(X=4)=0.60-0.35=0.25\).
For the interval, \(F(6)=1\) and \(F(2)=0.35\). Thus \(P(2<X\le6)=F(6)-F(2)=1-0.35=0.65\).
Check. The interval contains the values \(4\) and \(6\), whose masses are \(0.60-0.35=0.25\) and \(1-0.60=0.40\). Their sum is \(0.65\), matching the CDF calculation. The common mistake avoided is subtracting \(F(2-)\), which would include the mass at \(2\), even though the event requires \(X>2\).
Common Mistakes and Practical Checks
- Using the CDF value alone. \(F(a)\) includes all probability at values below \(a\), not just the probability at \(a\). Subtract \(F(a-)\).
- Using the wrong left limit. For a piecewise CDF, evaluate the expression on the interval immediately to the left of \(a\), or calculate its limit as \(x\) approaches \(a\) from below.
- Assuming every CDF consists only of jumps. Some CDFs increase between jumps. The sum of selected point masses may therefore be less than one.
- Ignoring endpoint inclusion in an interval. \(F(b)-F(a)\) gives \(P(a<X\le b)\), not \(P(a\le X\le b)\). To include \(a\), use \(F(b)-F(a-)\).
A useful check is to confirm that every calculated jump is nonnegative, because a CDF is nondecreasing. For a distribution known to put all its probability on a finite list of points, the jumps at those points must sum to one. For a mixed distribution, also account for the increases between jumps before checking the total.
Check Your Understanding
For each problem, identify the relevant left limit before calculating. Answers are included so you can check your work.
- A random variable has CDF \(F(x)=0\) for \(x<-1\), \(F(x)=0.30\) for \(-1\le x<2\), \(F(x)=0.55\) for \(2\le x<5\), and \(F(x)=1\) for \(x\ge5\). Find the point masses at \(-1\), \(2\), and \(5\), then check their sum. (Answer: \(0.30,\ 0.25,\ 0.45\); sum \(=1.00\))
- A mixed CDF is \(F(x)=0\) for \(x<0\), \(F(x)=0.20+0.10x\) for \(0\le x<4\), \(F(x)=0.70+0.05(x-4)\) for \(4\le x<8\), and \(F(x)=1\) for \(x\ge8\). Find \(P(X=0)\), \(P(X=4)\), and \(P(X=8)\), and calculate \(P(X\le4)\). (Answer: \(0.20,\ 0.10,\ 0.10\); \(P(X\le4)=0.70\))
- A CDF has \(F(1-)=0.18\), \(F(1)=0.42\), \(F(3-)=0.65\), and \(F(3)=0.80\). Find the masses at \(1\) and \(3\), then find \(P(1\le X\le3)\). (Answer: \(P(X=1)=0.24,\ P(X=3)=0.15,\ P(1\le X\le3)=0.62\))
- A random variable has CDF \(F(x)=0\) for \(x<-2\), \(F(x)=0.25\) for \(-2\le x<0\), \(F(x)=0.40\) for \(0\le x<2\), \(F(x)=0.40+0.10(x-2)\) for \(2\le x<5\), and \(F(x)=1\) for \(x\ge5\). Find the masses at \(-2\), \(0\), \(2\), and \(5\), and the total mass at these points. (Answer: \(0.25,\ 0.15,\ 0,\ 0.30\); total \(=0.70\))
- A CDF satisfies \(F(2-)=0.12\), \(F(2)=0.37\), \(F(6-)=0.71\), and \(F(6)=0.88\). Find both point masses and \(P(2<X\le6)\). (Answer: \(P(X=2)=0.25,\ P(X=6)=0.17,\ P(2<X\le6)=0.88-0.37=0.51\))