Survival Functions And Tail Probabilities

Why Tail Probabilities Matter

For an insurer, a large loss can matter more than the center of a severity distribution. Questions such as “What is the probability that one claim exceeds $2,000?” or “Among claims already above $1,000, what fraction exceed $3,000?” are tail-probability questions. The survival function answers the first directly and, with a conditional-probability calculation, answers the second.

The survival function is defined from the loss random variable itself, so it applies to discrete, continuous, and other severity models. Its relationship with the cumulative distribution function (CDF) is simple, but the strict inequality in its definition is important. At a loss amount that can occur with positive probability, the probability of exceeding that amount differs from the probability of reaching or exceeding it.

Definition: For a loss random variable $X$, its survival function is $S_X(x)=P(X>x)$. Its CDF is $F_X(x)=P(X\le x)$. Since the events $X>x$ and $X\le x$ are complements, $S_X(x)=1-F_X(x)$ for every real threshold $x$.

From the CDF to the Survival Function

The CDF accumulates probability at and below a threshold. The survival function counts probability strictly above it. Subtracting the CDF from one therefore gives the strict upper-tail probability:

S_X(x)=P(X>x)=1-P(X\le x)=1-F_X(x).

This identity works for any distribution; no density or probability mass function is required. If a CDF is given as a formula or table, evaluate it at the threshold and subtract from one. Conversely, if the survival function is given, then $F_X(x)=1-S_X(x)$.

A valid survival function cannot increase as the threshold increases: raising the threshold makes the event $X>x$ more restrictive. It takes values between zero and one. For a finite, nonnegative loss, $S_X(x)=1$ for $x<0$, because every possible loss exceeds a negative amount; and $S_X(x)$ approaches zero as $x$ grows without bound. A proposed survival formula should be checked against these basic properties before it is used.

Key distinction: $S_X(x)=P(X>x)$ is a strict exceedance probability. The inclusive probability is $P(X\ge x)$. If the loss can equal $x$ with positive probability, these quantities are not interchangeable.

To express the inclusive probability using the CDF, write $F_X(x-)=\lim_{y\uparrow x}F_X(y)$, the probability accumulated strictly below $x$. Then

P(X\ge x)=1-F_X(x-).

The difference between the inclusive and strict probabilities is the probability of equality at the threshold:

P(X\ge x)-P(X>x)=P(X=x).

For a continuous severity model with no probability concentrated at $x$, $P(X=x)=0$, so the strict and inclusive probabilities agree at that threshold. For a discrete model, or any model with an atom, check the event wording rather than assuming they agree.

Worked Example: Strict and Inclusive Exceedance for a Discrete Loss

A hypothetical repair-cost severity $X$, in dollars, has the following distribution: $P(X=0)=0.15$, $P(X=400)=0.25$, $P(X=900)=0.35$, and $P(X=1{,}800)=0.25$. Find (a) the probability a loss exceeds $900, (b) the probability it is at least $900, and (c) the probability it exceeds $600 given that it is at least $900.

Method. Use the stated probabilities to identify which outcomes satisfy each event; for the conditional probability, divide the probability of the intersection by the probability of the conditioning event. No interest or mortality assumption is involved because this is a single-loss severity model.

(a) “Exceeds $900” excludes a loss exactly equal to $900. The only qualifying outcome is $1,800, so

$$ P(X>900)=P(X=1{,}800)=0.25. $$

This also follows from the CDF: $F_X(900)=0.15+0.25+0.35=0.75$, hence $S_X(900)=1-0.75=0.25$.

(b) “At least $900” includes both $900 and $1,800. Therefore

P(X\ge 900)=0.35+0.25=0.60.

P(X>600\mid X\ge 900) =\frac{P(X>600,\ X\ge 900)}{P(X\ge 900)} =\frac{0.60}{0.60}=1.

Check. The conditional answer is one because the condition $X\ge900$ guarantees $X>600$. The common mistake is to report $0.25$ for part (b) by using $S_X(900)$; that is the strict exceedance probability and omits the probability $0.35$ at exactly $900.

Conditional Tail Probabilities

A tail probability can also be used as the denominator in a conditional probability. For thresholds $y>x$, the event $X>y$ is contained in $X>x$. Consequently, provided $S_X(x)>0$,

P(X>y\mid X>x) =\frac{P(X>y,\ X>x)}{P(X>x)} =\frac{S_X(y)}{S_X(x)}.

The condition $y>x$ matters: it makes the numerator event a subset of the conditioning event. If the events are not nested, the intersection must be found separately. Also, the denominator must be positive; conditional probability given an event of probability zero is not defined by this ratio.

Worked Example: Conditional Exceedance From a Specified CDF

A hypothetical equipment-loss severity $X$, in dollars, has CDF $F_X(x)=0$ for $x<0$, $F_X(x)=x/5{,}000$ for $0\le x\le5{,}000$, and $F_X(x)=1$ for $x>5{,}000$. Find the probability a loss exceeds $1,250 and the probability it exceeds $3,750 given that it exceeds $1,250.

Method. First convert each CDF value into a strict tail probability using $S_X(x)=1-F_X(x)$. Then apply the nested-event conditional-tail ratio. The given CDF has no jumps, so there is no positive probability at either threshold. This is a severity model only; no interest rate or mortality assumption is needed.

At $1,250,

F_X(1{,}250)=\frac{1{,}250}{5{,}000}=0.25, \qquad S_X(1{,}250)=1-0.25=0.75.

Thus the probability of exceeding $1,250 is $0.75$. At $3,750,

F_X(3{,}750)=\frac{3{,}750}{5{,}000}=0.75, \qquad S_X(3{,}750)=1-0.75=0.25.

Since $3{,}750>1{,}250$, the event $X>3{,}750$ is contained in $X>1{,}250$. Therefore

P(X>3{,}750\mid X>1{,}250) =\frac{0.25}{0.75} =\frac{1}{3} \approx 0.3333.

Check. The conditional probability is smaller than one, as the higher threshold is more restrictive. A common error is to divide the lower-threshold tail by the higher-threshold tail, which reverses the conditioning and gives an impossible value greater than one.

Working Directly With a Survival Model

Sometimes a model is stated in terms of $S_X$ rather than $F_X$. The same steps apply: substitute the threshold into the survival function, then interpret the result as $P(X>x)$. A useful check is to verify that the function starts at a plausible value for the loss support, stays between zero and one, and does not increase with $x$.

Worked Example: Comparing Two High-Loss Thresholds

A hypothetical nonnegative claim severity $X$, measured in dollars, is specified by $S_X(x)=(1+x/1{,}000)^{-2}$ for $x\ge0$, with $S_X(x)=1$ for $x<0$. Calculate the probability a claim exceeds $1,000, the probability it exceeds $3,000, and the probability it exceeds $3,000 given that it exceeds $1,000.

Method. Evaluate the specified survival function at each positive threshold and use the conditional-tail ratio. The model is fully specified by this survival function; no separate frequency, interest, or mortality model is part of the question.

At $1,000, the expression inside the power is $1+1{,}000/1{,}000=2$, so

S_X(1{,}000)=2^{-2}=\frac{1}{4}=0.25.

At $3,000, the expression is $1+3{,}000/1{,}000=4$, giving

S_X(3{,}000)=4^{-2}=\frac{1}{16}=0.0625.

Because $3,000 is a higher threshold than $1,000, the conditional probability is

P(X>3{,}000\mid X>1{,}000) =\frac{S_X(3{,}000)}{S_X(1{,}000)} =\frac{1/16}{1/4} =\frac{1}{4}=0.2500.

Check. The unconditional probability of exceeding $3,000, $0.0625$, is smaller than the probability of exceeding $1,000, $0.25$, as it must be. A common mistake is to use $1-S_X(x)$ as the exceedance probability; that quantity is the CDF $F_X(x)$, the probability of being at or below the threshold.

Common Interpretation Traps

The main practical risk is not the subtraction itself but matching the formula to the event. The word “exceeds” means $>$, while “at least” means $\ge$. If a threshold has positive probability, using the wrong version can materially change the answer. In a model with no probability at the threshold, the two answers coincide, but that should follow from the model rather than be assumed.

Using the CDF as an exceedance probability. $F_X(x)$ is $P(X\le x)$; subtract it from one to obtain $P(X>x)$.
Ignoring equality at an atom. $S_X(x)$ excludes outcomes exactly equal to $x$. For $P(X\ge x)$, account for the probability below $x$, or add $P(X=x)$ to $S_X(x)$.
Reversing a conditional ratio. For $y>x$, $P(X>y\mid X>x)=S_X(y)/S_X(x)$, not the reverse.
Using the ratio without nested events. The tail ratio applies directly when $y>x$. Otherwise, calculate the intersection required by the conditional-probability definition.

Key takeaway: Start with the event wording. For a strict exceedance, use $S_X(x)=1-F_X(x)$. For an inclusive threshold, check whether probability is concentrated at the threshold. For nested strict tails with $y>x$, divide $S_X(y)$ by $S_X(x)$ to obtain the conditional exceedance probability.

Check Your Understanding

For each hypothetical severity model, identify whether the event is strict or inclusive before calculating. Probabilities in the answers are given to four decimal places where appropriate.

A repair loss is $0 with probability $0.10$, $600 with probability $0.30$, $1,400 with probability $0.40$, and $5,000 with probability $0.20$. Find $P(X>1{,}400)$ and $P(X\ge1{,}400)$. (Answer: $0.2000$ and $0.6000$)
A hypothetical severity has $F_X(x)=x/8{,}000$ for $0\le x\le8{,}000$. Find $P(X>2{,}000)$ and $P(X>6{,}000\mid X>2{,}000)$. (Answer: $0.7500$ and $0.3333$)
A loss is $0, $1,000, $3,000, or $6,000 with probabilities $0.15, 0.30, 0.40,$ and $0.15$, respectively. Find $P(X>1{,}000)$ and $P(X\ge1{,}000)$. (Answer: $0.5500$ and $0.8500$)
A hypothetical nonnegative severity has $S_X(x)=(1+x/2{,}000)^{-1}$ for $x\ge0$. Find $P(X>2{,}000)$, $P(X>6{,}000)$, and $P(X>6{,}000\mid X>2{,}000)$. (Answer: $0.5000$, $0.2500$, and $0.5000$)
A hypothetical loss has $F_X(500)=0.20$ and $F_X(2{,}000)=0.70$, and the model assigns no probability between these two CDF values at either endpoint. Find the probability of exceeding $500 and the conditional probability of exceeding $2,000 given that the loss exceeds $500. (Answer: $0.8000$ and $0.3750$)

What You'll Learn