Cumulative Distribution and Survival Functions

From a Loss Model to Threshold Probabilities

A loss model describes more than the chance of one particular loss amount: it also lets an actuary answer questions such as “What is the probability that the loss is at most this amount?” or “What is the probability that it exceeds the amount?” The cumulative distribution function and survival function provide two complementary ways to organize those probabilities.

These functions apply to discrete and continuous losses, and they remain useful when a distribution has both point masses and a continuous component. The key practical issue is the treatment of the boundary: “greater than” and “greater than or equal to” need not have the same probability when the loss can equal the threshold. This tutorial is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Definition: For a loss random variable $X$, the cumulative distribution function is $F_X(x)=P(X\leq x)$. The survival function is $S_X(x)=P(X>x)$. Thus, the survival function uses a strict inequality: it measures the probability of exceeding $x$, not the probability of meeting or exceeding $x$.

Both functions are defined for every real threshold $x$, even if only certain loss amounts are possible. The cumulative distribution function records probability accumulated at or below the threshold. The survival function records probability remaining strictly above it.

$$ S_X(x)=1-F_X(x) $$

This complement relationship follows because the events $X\leq x$ and $X>x$ cannot occur together, and together they include every possible outcome. Their probabilities therefore sum to one. A cumulative distribution function is nondecreasing, takes values between zero and one, and is right-continuous. Its value can jump at a loss amount that has positive probability.

Constructing the Functions

For a discrete loss, the probability mass at each possible value is accumulated as the threshold passes that value. If the possible losses are $x_1,x_2,\ldots$, with probabilities $P(X=x_j)$, then

F_X(x)=\sum_{x_j\leq x}P(X=x_j).

The survival function can be obtained by summing the probabilities of all possible losses strictly greater than $x$. For a continuous loss with density $f_X$, the cumulative distribution function is the integral of the density up to the threshold, and the survival function is the integral above it:

F_X(x)=\int_{-\infty}^{x} f_X(t)\,dt, \qquad S_X(x)=\int_x^{\infty} f_X(t)\,dt.

These density integrals apply when the loss distribution is continuous and has a density over the relevant range. They do not capture a point mass by themselves. For a model with both a point mass and a continuous component, the cumulative probability must include both the mass at or below $x$ and the integrated continuous probability at or below $x$.

Useful identities: For any $a<b$, $P(a<X\leq b)=F_X(b)-F_X(a)$. If $F_X(a-)$ denotes the left-hand limit of the cumulative distribution function at $a$, then $P(X=a)=F_X(a)-F_X(a-)$ and $P(X\geq a)=1-F_X(a-)$. In a continuous distribution, $P(X=a)=0$, so strict and inclusive threshold probabilities are equal.

The interval identity follows by subtracting the probability accumulated through $a$ from that accumulated through $b$: the outcomes left over are exactly those with $a<X\leq b$. At a point mass, however, $F_X(a)$ includes the probability at $a$, while $F_X(a-)$ does not. This is why $1-F_X(a)$ gives $P(X>a)$, not $P(X\geq a)$.

Worked Examples

Worked Example: A Discrete Loss With an Inclusive Threshold

A hypothetical equipment insurer models one annual claim payment $X$, in dollars, as follows: $P(X=0)=0.50$, $P(X=1{,}000)=0.25$, $P(X=4{,}000)=0.15$, and $P(X=10{,}000)=0.10$. Construct $F_X(x)$, then find (a) the probability the payment exceeds $2,500 and (b) the probability it is at least $4,000.

Method. First check that the probability masses sum to one: $0.50+0.25+0.15+0.10=1.00$. Then add the masses at or below each successive possible payment.

For $x<0$, no possible payment is at or below $x$, so $F_X(x)=0$. For $0\leq x<1{,}000$, only the zero payment is included, giving $F_X(x)=0.50$. For $1{,}000\leq x<4{,}000$, add the $1,000 mass: $0.50+0.25=0.75$. For $4{,}000\leq x<10{,}000$, add the $4,000 mass: $0.75+0.15=0.90$. At and above $10,000, the total is $1.00$. Therefore,

F_X(x)= \begin{cases} 0, & x<0,\\ 0.50, & 0\leq x<1{,}000,\\ 0.75, & 1{,}000\leq x<4{,}000,\\ 0.90, & 4{,}000\leq x<10{,}000,\\ 1.00, & x\geq10{,}000. \end{cases}

(a) Since $2,500 lies between $1,000 and $4,000, $F_X(2{,}500)=0.75$. Thus $P(X>2{,}500)=1-0.75=0.25$, or 0.2500 to four decimal places.

(b) The payments at least $4,000 are $4,000 and $10,000. Adding their probabilities gives $P(X\geq4{,}000)=0.15+0.10=0.25$, or 0.2500. The same result follows from the left-hand limit: $1-F_X(4{,}000-)=1-0.75=0.25$.

Check. The event $X\geq4{,}000$ contains exactly the two stated payment outcomes, whose total probability is 0.25. The common mistake is to calculate $1-F_X(4{,}000)=1-0.90=0.10$; that is $P(X>4{,}000)$, which excludes the mass at $4,000.

Worked Example: A Continuous Loss From Its Density

A hypothetical small-business policy has annual covered loss $X$, measured in thousands of dollars, uniformly distributed from 0 to 12. Thus its density is $f_X(x)=1/12$ for $0\leq x\leq12$, and zero otherwise. Find $F_X(x)$, $S_X(x)$, $P(3<X\leq9)$, and $P(X\geq8)$.

Method. Integrate the density across the portion of its support at or below $x$. Since the density is constant, an interval of width $w$ has probability $w/12$.

For $x<0$, the cumulative probability is zero. Within the support, $0\leq x\leq12$, the density accumulated from 0 to $x$ gives $\int_0^x(1/12)\,dt=x/12$. Above 12, all probability has accumulated. Hence,

F_X(x)= \begin{cases} 0, & x<0,\\ x/12, & 0\leq x\leq12,\\ 1, & x>12, \end{cases} \qquad S_X(x)= \begin{cases} 1, & x<0,\\ (12-x)/12, & 0\leq x\leq12,\\ 0, & x>12. \end{cases}

For $3<X\leq9$, subtract the cumulative probabilities at the endpoints: $F_X(9)=9/12=0.75$ and $F_X(3)=3/12=0.25$. Thus $P(3<X\leq9)=0.75-0.25=0.50$, or 0.5000.

For $X\geq8$, the distribution has no point mass at 8, so the inclusive and strict probabilities agree. The survival probability is $S_X(8)=(12-8)/12=4/12=1/3$, or 0.3333 to four decimal places. The loss amounts in the model are in thousands of dollars; for example, the threshold 8 represents $8,000.

Check. The interval from 3 to 9 has width 6, so its probability is $6/12=0.50$. The interval above 8 has width 4, giving $4/12=1/3$. The endpoint distinction does not change either result because a continuous distribution assigns zero probability to any single point.

Worked Example: A Loss With a Point Mass and a Continuous Component

A hypothetical service-contract loss $X$, in thousands of dollars, is zero with probability 0.40. With probability 0.60, it is uniformly distributed over positive amounts from 0 to 10. Find $F_X(x)$, $S_X(x)$, $P(X>0)$, $P(X\geq5)$, and $P(0<X\leq5)$.

Method. Treat the zero-loss probability as a point mass, then add the continuous probability accumulated from 0 to $x$. Conditional on a positive loss, the density is $1/10$; weighted by the 0.60 chance of a positive loss, the unconditional density over the positive range is $0.60(1/10)=0.06$ per thousand dollars.

For $x<0$, $F_X(x)=0$. At $x=0$, the point mass is included, so $F_X(0)=0.40$. For $0\leq x\leq10$, the cumulative probability is the zero-loss mass plus the integrated positive-loss density: $0.40+\int_0^x0.06\,dt=0.40+0.06x$. Above 10, all probability has accumulated. Therefore,

F_X(x)= \begin{cases} 0, & x<0,\\ 0.40+0.06x, & 0\leq x\leq10,\\ 1, & x>10, \end{cases} \qquad S_X(x)= \begin{cases} 1, & x<0,\\ 0.60-0.06x, & 0\leq x\leq10,\\ 0, & x>10. \end{cases}

At zero, $F_X(0)=0.40$, so $P(X>0)=S_X(0)=1-0.40=0.60$. At 5, $F_X(5)=0.40+0.06(5)=0.70$. There is no point mass at 5, so $P(X\geq5)=P(X>5)=1-0.70=0.30$. Finally, $P(0<X\leq5)$ excludes the zero-loss mass; it is the continuous probability from 0 to 5, $0.06(5)=0.30$. Each reported probability is to four decimal places: 0.6000, 0.3000, and 0.3000.

Check. The positive-loss probability is 0.60. Its uniform distribution splits evenly at 5, so each half has probability $0.60/2=0.30$. The jump in $F_X$ at zero is 0.40, matching the probability of no loss.

Boundary Conventions and a Common Exam Trap

The choice between $>$ and $\geq$ is not a matter of notation alone. At a threshold with positive probability, the two events differ by the probability of equality:

P(X\geq a)=P(X>a)+P(X=a).

Consequently, $S_X(a)=1-F_X(a)$ always gives $P(X>a)$. For an inclusive threshold, use $P(X\geq a)=1-F_X(a-)$, or add the mass $P(X=a)$ to the strict exceedance probability. For a continuous loss, there is no point mass at $a$, so $P(X\geq a)=P(X>a)$.

A frequent error is to read “at least $a$” as “above $a$” and use $1-F_X(a)$. The discrete worked example shows the consequence: at $4,000, this gives 0.10, but that excludes the 0.15 probability of a $4,000 payment. The correct inclusive probability is 0.25. Always translate the wording into an event first, then select the corresponding function or boundary convention.

Key takeaway: $F_X(x)$ includes probability at $x$; $S_X(x)$ excludes it. For intervals, $F_X(b)-F_X(a)$ gives $P(a<X\leq b)$. Check whether an endpoint can have positive probability before replacing an inclusive inequality with a strict one.

Check Your Understanding

For each problem, identify the relevant distribution-function boundary and calculate the requested probability.

A hypothetical deductible reimbursement $X$, in dollars, takes values 0, 500, and 2,000 with probabilities 0.30, 0.45, and 0.25. Find $F_X(500)$ and $P(X\geq500)$. (Answer: $F_X(500)=0.7500$; $P(X\geq500)=0.7000$.)
A continuous hypothetical repair loss is uniform from 0 to 20 thousand dollars. Find $S_X(7)$ and $P(4<X\leq14)$. (Answer: $S_X(7)=0.6500$; $P(4<X\leq14)=0.5000$.)
A hypothetical benefit payment is zero with probability 0.20; otherwise it is uniform from 0 to 8 thousand dollars. Find $F_X(0)$, $P(X>0)$, and $P(X>6)$. (Answer: $F_X(0)=0.2000$; $P(X>0)=0.8000$; $P(X>6)=0.2000$.)
A discrete hypothetical loss has probability 0.10 at 0, 0.50 at 3,000, and 0.40 at 9,000 dollars. Find $P(0<X\leq3{,}000)$ using the cumulative distribution function, and explain why $F_X(3{,}000)$ alone is not the answer. (Answer: $0.6000-0.1000=0.5000$; $F_X(3{,}000)$ also includes the probability of a zero loss.)
A hypothetical mixed loss is zero with probability 0.25 and, otherwise, uniform on positive losses up to 12 thousand dollars. Find $P(X\geq9)$, and state whether the inclusive boundary changes the result. (Answer: $0.1875$; it does not change the result because there is no point mass at 9.)

What You'll Learn