Why CDF Properties Matter
A cumulative distribution function is more than a formula for assigning probabilities. Its shape is constrained by probability itself. These constraints let an actuary check whether a proposed model for claim severity is mathematically possible before using it in a calculation. They also clarify what the CDF says about low and high claim amounts.
Let \(X\) denote a real-valued claim severity, measured in a specified unit such as dollars. Its cumulative distribution function is \(F_X(x)=P(X\leq x)\). A severity is ordinarily finite and nonnegative, but the basic CDF properties below hold for any finite real-valued random variable. The nonnegative-severity assumption adds a useful implication, stated separately.
The Four Defining Properties
The properties are necessary conditions: a proposed function that fails even one cannot be a CDF. Together, they are also sufficient. In particular, a function from the real line to \([0,1]\) is a CDF if it is nondecreasing, right-continuous, and has the stated limits. These requirements constrain the whole function, not merely a few values listed in a table.
- Range: \(0\leq F_X(x)\leq 1\) for every \(x\).
- Monotonicity: If \(a\leq b\), then \(F_X(a)\leq F_X(b)\).
- Right-continuity: For every \(x\), \(\lim_{h\downarrow 0}F_X(x+h)=F_X(x)\).
- Limits: \(\lim_{x\to-\infty}F_X(x)=0\) and \(\lim_{x\to+\infty}F_X(x)=1\).
The range condition follows directly because \(F_X(x)\) is a probability. Monotonicity says that the cumulative probability cannot decrease as the threshold increases. Right-continuity specifies how the function behaves when the threshold approaches a value from above. The two limits say that a threshold far below all finite outcomes accumulates probability approaching zero, while a threshold far above them accumulates probability approaching one.
Why the Properties Hold
Each property follows from a basic fact about events and probability. The event-nesting argument proves monotonicity; continuity of probability for nested events proves right-continuity and the limits. The following steps give the full reasoning.
If \(a\leq b\), every outcome satisfying \(X\leq a\) also satisfies \(X\leq b\). Thus \(\{X\leq a\}\subseteq\{X\leq b\}\). Probabilities preserve event inclusion, so \(P(X\leq a)\leq P(X\leq b)\), or \(F_X(a)\leq F_X(b)\).
Fix \(x\), and take any decreasing sequence \(x_n\downarrow x\) with \(x_n>x\). The events \(\{X\leq x_n\}\) decrease as \(n\) increases, and their intersection is exactly \(\{X\leq x\}\). For decreasing events, probability of the intersection equals the limit of their probabilities. Therefore \(\lim_{n\to\infty}F_X(x_n)=F_X(x)\). This sequence criterion gives right-continuity.
For integer \(n\) increasing without bound, the events \(\{X\leq -n\}\) decrease to the empty event, since a finite real value of \(X\) cannot be at most every negative integer. Continuity of probability for decreasing events gives \(\lim_{n\to\infty}F_X(-n)=0\), which establishes the limit as \(x\to-\infty\).
The events \(\{X\leq n\}\) increase to the entire sample space because \(X\) is finite and real-valued. Continuity of probability for increasing events gives \(\lim_{n\to\infty}F_X(n)=1\), establishing the limit as \(x\to+\infty\).
The proof also explains why the assumptions matter. If a model allowed an outcome at infinity, the events \(\{X\leq n\}\) might not eventually cover that outcome, and the positive-infinity limit need not be one. Ordinary claim severity is modeled as a finite real amount, so the stated limit applies.
Worked Example: Checking a Proposed Severity CDF
A hypothetical insurer proposes the following function for a nonnegative claim severity \(X\), in thousands of dollars:
Determine whether \(G\) satisfies the CDF properties. Identify the boundary values that need particular attention.
Method: check range, monotonicity, right-continuity, and the two limits. For \(0\leq x<5\), \(x/10\) ranges from 0 up to, but not including, 0.5. The next branch is 0.8, and the last branch is 1. All values lie in \([0,1]\).
Within each branch the function is nondecreasing. At 0, it takes the value 0, matching the values immediately to the right. At 5, the value is \(G(5)=0.8\); values immediately to the right are also 0.8. The value increases from the left-hand branch, whose values approach 0.5, so the boundary does not create a decrease. At 9, \(G(9)=1\), and the values to the right are 1. Thus the function is nondecreasing and right-continuous at its branch boundaries.
For \(x<0\), \(G(x)=0\), so \(\lim_{x\to-\infty}G(x)=0\). For \(x\geq9\), \(G(x)=1\), so \(\lim_{x\to+\infty}G(x)=1\). The function satisfies all the conditions and is a valid CDF. Its form is consistent with nonnegative severity because it is zero for negative thresholds.
Check: The value at 5 is 0.8, not 0.5; the latter is only the value approached from below. Right-continuity concerns values approached from the right. Common mistake avoided: requiring the two sides of a boundary to agree would incorrectly rule out valid CDFs.
Reading the Properties in a Severity Model
For a claim severity, monotonicity means that increasing the dollar threshold cannot reduce the cumulative probability \(P(X\leq x)\). It does not say that every increase in threshold must increase the CDF strictly: the CDF may remain constant over a range. Right-continuity concerns the value at a threshold and values immediately above it; it does not require continuity from the left.
If \(X\geq0\) with probability 1, then for every \(x<0\), the event \(\{X\leq x\}\) is impossible and \(F_X(x)=0\). This is a consequence of the severity support assumption, not a property that every real-valued CDF must have. In either case, the limits at infinity describe the overall probability accumulation: the CDF approaches 0 far to the left and 1 far to the right.
A useful exam check is to assess each property independently. A function can stay within \([0,1]\) and have the correct limits but still decrease somewhere. It can be nondecreasing and have the correct limits but fail right-continuity. Checking only a formula’s interior branches, while overlooking the values assigned at their endpoints, can miss either defect.
Worked Example: Finding a Monotonicity Failure
A hypothetical analyst proposes \(H\) for a claim severity measured in hundreds of dollars:
Check whether this is a CDF, even though all its values are probabilities and its limits appear appropriate.
Method: test the defining properties separately. The range condition holds: every branch is between 0 and 1. The limit at negative infinity is 0, and the limit at positive infinity is 1. At \(x=3\), \(H(3)=0.68\), and values immediately to the right are 0.68, so the boundary is right-continuous. At \(x=7\), \(H(7)=1\), also matching values immediately to the right.
Monotonicity fails. For example, take \(a=2\) and \(b=3\), so \(a\leq b\). Then \(H(2)=0.72\) while \(H(3)=0.68\), giving \(H(2)>H(3)\). A CDF must instead satisfy \(H(2)\leq H(3)\). Therefore \(H\) is not a valid CDF.
Correction: the issue is not that 0.68 or 0.72 is individually an impossible probability; it is their order at increasing thresholds. Common mistake avoided: checking the range and endpoint limits alone is not enough.
Worked Example: Testing Right-Continuity at a Boundary
For a hypothetical portfolio’s severity in thousands of dollars, a proposed function \(K\) is 0 for \(x<0\), 0.35 for \(0\leq x\leq4\), 0.60 for \(4<x<10\), and 1 for \(x\geq10\). Determine whether \(K\) is a CDF.
Method: examine the boundary value and the limit from the right. The function stays within \([0,1]\). It is nondecreasing: the successive levels are 0, 0.35, 0.60, and 1. Its limits are 0 at negative infinity and 1 at positive infinity. These checks pass.
At \(x=4\), the definition assigns \(K(4)=0.35\), because the 0.35 branch includes 4. For every sufficiently small \(h>0\), \(4+h\) lies in the next branch, so \(K(4+h)=0.60\). Consequently,
The function is not right-continuous at 4, so it is not a valid CDF. At 10, by contrast, \(K(10)=1\) and \(K(10+h)=1\) for \(h>0\), so that boundary passes the right-continuity check.
Common mistake avoided: looking only at the function’s monotone sequence of levels does not establish right-continuity. The assigned boundary value must agree with the limit from the right.
Common Exam Traps
- Confusing nondecreasing with strictly increasing. A CDF is allowed to be flat. The required comparison is \(F_X(a)\leq F_X(b)\) when \(a\leq b\), not necessarily \(F_X(a)<F_X(b)\).
- Checking continuity on the wrong side. Right-continuity compares \(F_X(x)\) with values just greater than \(x\). Agreement with values just less than \(x\) is not the defining condition.
- Assuming every severity CDF has \(F_X(x)=0\) for \(x<0\) without stating the support. That conclusion uses \(X\geq0\) with probability 1. A general real-valued random variable can take negative values.
- Forgetting a limit check. A function may satisfy local shape conditions yet fail to approach 0 on the left or 1 on the right. Both limits are part of the CDF conditions.
Check Your Understanding
For each question, apply the relevant CDF properties and state the conclusion.
- For a CDF \(F_X\), suppose \(F_X(2)=0.41\). What inequality must hold between \(F_X(2)\) and \(F_X(6)\)? (Answer: \(F_X(6)\geq0.41\).)
- A proposed function is 0 for \(x<-1\), 0.20 for \(-1\leq x<2\), 0.55 for \(2\leq x<4\), and 1 for \(x\geq4\). Does it satisfy the CDF conditions? (Answer: Yes; it is nondecreasing, right-continuous, stays in \([0,1]\), and has limits 0 and 1.)
- A proposed function has \(J(1)=0.63\) and \(J(3)=0.58\). Can it be a CDF, given that it is nondecreasing elsewhere and has the correct limits? (Answer: No; \(1<3\) but \(J(1)>J(3)\), violating monotonicity.)
- A proposed function has \(M(5)=0.27\), while \(\lim_{h\downarrow0}M(5+h)=0.44\). Can it be a CDF? (Answer: No; it fails right-continuity at 5.)
- Suppose \(X\) is a nonnegative severity and \(F_X\) is its CDF. What is \(F_X(-2)\), and what is \(\lim_{x\to+\infty}F_X(x)\)? (Answer: \(F_X(-2)=0\) and \(\lim_{x\to+\infty}F_X(x)=1\).)