Why Boundaries Need Care
In a continuous model, including or excluding a single endpoint does not change an interval probability. In a discrete model, an endpoint can carry positive probability, so changing an inequality can change the answer. A mixed model combines both possibilities. When its density is defined in separate pieces, a shared boundary may also be a point where the density changes value. These features make endpoint notation matter.
The key distinction is between a density value and a probability at a point. A density describes probability accumulated over an interval; its value at one point is not itself the probability of that point. An atom, by contrast, is an outcome with positive probability. This distinction is essential when calculating interval probabilities in piecewise models.
Endpoint Rules for Mixed Models
For a continuous random variable with density \(f_X\), the probability of a single point is zero: \(P(X=c)=0\). Thus, for such a variable, \(P(a<X<b)\), \(P(a\le X\le b)\), and the versions with one endpoint included all equal \(\int_a^b f_X(x)\,dx\). This remains true if the density has a jump at \(c\): a jump in density is not an atom.
For a mixed model, calculate interval probability by integrating the continuous density across the interval and adding the probabilities of atoms that satisfy its endpoint conditions. The standard CDF expressions encode those conditions:
These identities apply to any distribution. The left limit \(F_X(c-)\) excludes a possible atom at \(c\); \(F_X(c)\) includes it. For a mixed model, the same distinctions can be made directly by integrating over the relevant open intervals and then adding atoms at included endpoints.
Why CDF Differences Give Interval Probabilities
The CDF definition makes the endpoint rules precise. The event \(\{a<X\le b\}\) is the part of \(\{X\le b\}\) left after removing \(\{X\le a\}\). These are nested events, so subtracting their probabilities leaves exactly the probability of the interval.
To include \(a\), remove only outcomes strictly below \(a\), whose probability is \(F_X(a-)\). To exclude \(b\), use \(F_X(b-)\) in place of \(F_X(b)\). This reasoning also explains why the probability at a boundary is the CDF jump:
If the distribution is continuous at \(c\), this jump is zero. A density can still have different one-sided values at \(c\); continuity of the density is not required for continuity of the CDF.
Worked Examples
Worked Example: An Atom at a Shared Density Boundary
A hypothetical payment amount \(X\), measured in thousands of dollars, has an atom of probability \(1/4\) at \(2\). Its continuous density is \(1/8\) for \(0<x<2\), \(1/4\) for \(2<x<4\), and zero elsewhere. Find \(P(1\le X\le3)\), \(P(1<X<2)\), and \(P(1<X\le2)\). First verify the model.
Method and model check. The continuous probability on \((0,2)\) is \(2(1/8)=1/4\). On \((2,4)\), it is \(2(1/4)=1/2\). Including the atom gives \(1/4+1/2+1/4=1\), so this is a valid mixed distribution. The density values at exactly \(0,2,4\) do not change any integral; the separate atom at \(2\) does matter.
For \(1\le X\le3\), the continuous contribution from \(1\) to \(2\) is \((2-1)(1/8)=1/8\). The atom at \(2\) contributes \(1/4\). The continuous contribution from \(2\) to \(3\) is \((3-2)(1/4)=1/4\). Therefore,
The event \(1<X<2\) excludes the atom at \(2\), so only the first density piece contributes: \(P(1<X<2)=(2-1)(1/8)=1/8=0.1250\). The event \(1<X\le2\) includes that atom: \(P(1<X\le2)=1/8+1/4=3/8=0.3750\).
Check. The difference between the last two answers is \(3/8-1/8=1/4\), exactly the atom at \(2\). The common mistake avoided here is treating the shared density boundary as though its density value were the point probability.
Worked Example: A Density Jump Without an Atom
A hypothetical continuous variable \(X\) has density \(f_X(x)=1/4\) for \(0<x<1\), \(f_X(x)=3/8\) for \(1<x<3\), and zero elsewhere. Find \(P(1\le X\le2\mid X>1/2)\), and determine whether \(P(X=1)\) is positive.
Method and model check. The two density pieces have total probability \((1)(1/4)+(2)(3/8)=1/4+3/4=1\). The model is continuous: no separate point mass is specified at \(1\). The density jumps from \(1/4\) to \(3/8\), but that jump does not create an atom.
The numerator is the probability of \(1\le X\le2\). Since a single point has probability zero in this continuous model, the boundary at \(1\) contributes no extra probability. The density over the interval is \(3/8\), so \(P(1\le X\le2)=(2-1)(3/8)=3/8\).
For the conditioning event, the support begins at zero and \(X\) has no atom at \(1/2\). Thus, \(P(X>1/2)=1-P(X\le1/2)\). The probability from \(0\) to \(1/2\) is \((1/2)(1/4)=1/8\), giving \(P(X>1/2)=7/8\). Hence,
Also, \(P(X=1)=0\). Check. The numerator is no greater than the denominator, and the conditional probability is between zero and one. The mistake avoided is adding probability at \(1\) merely because the density has a boundary there.
Worked Example: Conditional Probability With Two Boundary Atoms
A hypothetical claim amount \(X\), in thousands of dollars, has atoms \(P(X=1)=1/5\) and \(P(X=3)=1/10\). Its density is \(1/5\) on \(0<x<1\), \(1/10\) on \(1<x<3\), \(3/20\) on \(3<x<5\), and zero elsewhere. Find \(P(1<X\le3\mid X\le3)\). Verify the model and identify which endpoint atom belongs in each probability.
Method and model check. The continuous masses are \((1)(1/5)=1/5\), \((2)(1/10)=1/5\), and \((2)(3/20)=3/10\). Adding both atoms gives \(1/5+1/5+3/10+1/5+1/10=1\).
For the numerator \(P(1<X\le3)\), the atom at \(1\) is excluded, the density on \((1,3)\) contributes \(2(1/10)=1/5\), and the atom at \(3\) is included with probability \(1/10\). Therefore, the numerator is \(1/5+1/10=3/10\).
For the denominator \(P(X\le3)\), include the density on \((0,1)\), the atom at \(1\), the density on \((1,3)\), and the atom at \(3\). Their probabilities sum to \(1/5+1/5+1/5+1/10=7/10\). Thus,
Check. The numerator event is contained in the denominator event, and \(3/10\le7/10\). The common mistake avoided is either including the atom at \(1\) in the numerator despite the strict inequality, or excluding the atom at \(3\) despite the inclusive inequality.
A Reliable Boundary-Checking Process
For a piecewise or mixed model, write the event first, then check its boundaries before integrating. The following sequence keeps atoms and continuous pieces separate.
Record each point \(c\) with \(P(X=c)>0\). A density breakpoint alone is not evidence of an atom.
For each atom at an endpoint, decide directly whether it satisfies the event. A strict inequality excludes equality; a non-strict inequality includes it.
Split the interval wherever the density formula changes, and calculate each piece’s area separately.
Add the included atoms to the continuous areas. For conditional probabilities, divide by the probability of the conditioning event and verify that the denominator is positive.
Common Mistakes and Why They Matter
- Reading density height as point probability. A density value such as \(f_X(2)=1/4\) is not \(P(X=2)=1/4\). Point probability must come from an atom, or equivalently a CDF jump.
- Assuming every piecewise boundary has an atom. A density that changes formula at \(c\) can still describe a continuous distribution with \(P(X=c)=0\).
- Ignoring a stated atom because the density is also specified nearby. An atom and a continuous component can both occur at the same boundary in a mixed model; the point mass must be added separately.
- Using the wrong CDF limit. \(F_X(c)\) includes a possible atom at \(c\), while \(F_X(c-)\) excludes it. Select the expression that matches the event’s endpoint.
- Forgetting to check total probability. Integrate every density piece over its full support and add all atoms. If the result is not one, the model or the calculation needs review.
Check Your Understanding
For each problem, check the endpoint conditions before calculating. Unless stated otherwise, report probabilities to four decimal places.
- A variable has an atom of probability \(1/4\) at \(2\), density \(1/8\) on \(0<x<2\), and density \(1/4\) on \(2<x<4\). Find \(P(2\le X<4)\). (Answer: \(0.7500\))
- A continuous variable has density \(1/4\) on \(0<x<1\) and \(3/8\) on \(1<x<3\). Find \(P(1\le X\le2\mid X>1/2)\). (Answer: \(0.4286\))
- A variable has atoms \(P(X=1)=1/5\), \(P(X=3)=1/10\), and density \(1/5\) on \(0<x<1\), \(1/10\) on \(1<x<3\), and \(3/20\) on \(3<x<5\). Find \(P(X<3\mid X\le3)\). (Answer: \(0.8571\))
- A variable has an atom of probability \(0.15\) at \(4\), density \(0.10\) on \(0<x<2\), \(0.125\) on \(2<x<4\), and \(0.20\) on \(4<x<6\). Find \(P(2\le X\le4)\). (Answer: \(0.4000\))
- Use the model in Question 3 to find \(P(0<X\le1\mid X\le3)\). (Answer: \(0.5714\))