Deriving The Survival And Distribution Relationship

Why the Threshold Convention Matters

An empirical cumulative distribution and an empirical survival function describe the same observed losses from opposite sides of a threshold. The conversion is simple only when the event being counted is stated precisely. “Loss exceeds 250” excludes a loss of exactly 250; “loss is at least 250” includes it. When records contain ties at the threshold, those two counts differ.

This distinction matters whenever an actuarial calculation refers to a threshold, such as a loss amount used to classify observations. The central question is whether the event is $X>x$ or $X\ge x$. The empirical distribution function uses the inclusive event $X\le x$, so its complement is the strict upper tail $X>x$, not the inclusive upper tail.

Definition: For $n$ complete observed losses $x_1,\ldots,x_n$, the empirical cumulative distribution function at $x$ is the proportion of records at or below $x$: $$ F_n(x)=\frac{1}{n}\sum_{i=1}^{n} I(x_i\le x). $$ The empirical strict survival function is the proportion strictly above $x$: $$ \widehat S_n(x)=\frac{1}{n}\sum_{i=1}^{n} I(x_i>x). $$ Here $I(A)$ equals 1 when statement $A$ is true and 0 otherwise. These definitions describe the observed data exactly. Interpreting them as estimates of a population distribution requires a suitable sampling model, commonly independent observations from the same distribution.

The empirical cumulative function includes any observations equal to $x$. Therefore, when each observation is either at or below $x$, or strictly above $x$, the two proportions add to 1:

F_n(x)+\widehat S_n(x)=1 \qquad\Longrightarrow\qquad \widehat S_n(x)=1-F_n(x).

For an inclusive upper tail, define $\widehat S_n^{\,\ge}(x)$ as the proportion at or above $x$. Its complement consists of observations strictly below $x$. Writing $F_n(x-)$ for the left limit of the empirical cumulative distribution—the proportion strictly below $x$—gives a different identity:

F_n(x-)=\frac{1}{n}\sum_{i=1}^{n} I(x_i<x), \qquad \widehat S_n^{\,\ge}(x)=1-F_n(x-).

Key relationship: $1-F_n(x)$ counts observations strictly greater than $x$. In contrast, $1-F_n(x-)$ counts observations greater than or equal to $x$. The difference between the inclusive and strict upper tails is exactly the empirical mass at $x$.

Why the Complement Formulas Hold

The identities follow by partitioning the records into mutually exclusive groups. For the strict upper tail, every recorded loss satisfies exactly one of $x_i\le x$ and $x_i>x$. Counting both groups accounts for all $n$ records, so their proportions sum to 1. For the inclusive upper tail, the corresponding partition is $x_i<x$ or $x_i\ge x$.

\begin{aligned} \{x_i:x_i\le x\}\,\cup\,\{x_i:x_i>x\}&=\{x_1,\ldots,x_n\},\\ \{x_i:x_i<x\}\,\cup\,\{x_i:x_i\ge x\}&=\{x_1,\ldots,x_n\}. \end{aligned}

Each pair is disjoint, and each covers the full data set. Dividing its group counts by $n$ proves the two complement formulas. The difference between the two cumulative probabilities is the proportion of observations equal to $x$:

F_n(x)-F_n(x-) =\frac{1}{n}\sum_{i=1}^{n}I(x_i=x) =\widehat S_n^{\,\ge}(x)-\widehat S_n(x).

Thus a jump in the empirical cumulative distribution at $x$ is the empirical probability mass at that value. The strict and inclusive tails differ by precisely that jump. If no observed loss equals the threshold, the jump is zero, and the two tail proportions agree.

Worked Examples

Worked Example: Converting at a Tied Loss Amount

A hypothetical portfolio has 12 complete loss records, in dollars: 120, 150, 150, 200, 240, 240, 240, 310, 400, 400, 520, and 700. At a threshold of 240 dollars, calculate $F_n(240)$, the strict survival estimate, the inclusive upper-tail proportion, and the empirical mass at 240 dollars.

Method. Count observations in the four relevant groups: below 240, equal to 240, at or below 240, and strictly above 240. The sample size is $n=12$.

There are four losses below 240: 120, 150, 150, and 200. There are three equal to 240. Hence seven losses are at or below 240:

F_n(240)=\frac{7}{12}=0.5833.

The five losses strictly above 240 are 310, 400, 400, 520, and 700. Thus:

\widehat S_n(240)=\frac{5}{12}=0.4167 =1-\frac{7}{12}.

For the inclusive upper tail, include the three records equal to 240. There are $3+5=8$ losses at or above the threshold:

\widehat S_n^{\,\ge}(240)=\frac{8}{12}=\frac{2}{3}=0.6667.

The empirical mass at 240 is $3/12=0.2500$. The left-limit cumulative proportion is $4/12$, so the second complement check gives $1-4/12=8/12$. Also, $8/12-5/12=3/12$, as required. The common mistake is to use $1-F_n(240)$ for the inclusive tail; that gives $5/12$ and omits all three losses equal to 240.

Worked Example: A Threshold Between Observed Losses

A hypothetical set of 10 complete losses, in dollars, is 90, 110, 110, 140, 180, 210, 260, 300, 300, and 450. Calculate the strict and inclusive upper-tail proportions at 200 dollars and at 210 dollars. Explain why the answers behave differently at these two thresholds.

At 200 dollars. Five losses are at or below 200: 90, 110, 110, 140, and 180. Therefore $F_n(200)=5/10=0.5000$, and:

\widehat S_n(200)=1-F_n(200)=\frac{5}{10}=0.5000.

No observation equals 200. The same five losses are strictly below 200, so the number at or above 200 is also five. Consequently $\widehat S_n^{\,\ge}(200)=5/10=0.5000$. The tails agree because the empirical mass at 200 is zero.

At 210 dollars. Six losses are at or below 210, so $F_n(210)=6/10=0.6000$. Four losses—260, 300, 300, and 450—are strictly above 210:

\widehat S_n(210)=1-\frac{6}{10}=\frac{4}{10}=0.4000.

One loss equals 210, so five losses are at or above 210. Thus $\widehat S_n^{\,\ge}(210)=5/10=0.5000$. The difference, $0.5000-0.4000=0.1000$, is exactly the empirical mass $1/10$ at 210. This comparison checks both the no-tie case and the tied-threshold case.

Worked Example: Recovering Both Tails and the Tie Mass

A hypothetical sample contains eight complete losses, in dollars: 75, 100, 100, 125, 200, 200, 275, and 400. At a threshold of 200 dollars, find the empirical cumulative proportion, the left-limit cumulative proportion, both upper tails, and the mass at the threshold.

Count at or below 200. The losses 75, 100, 100, 125, 200, and 200 meet this condition. Therefore $F_n(200)=6/8=0.7500$.

Count strictly below 200. The four losses 75, 100, 100, and 125 are strictly below the threshold. Hence $F_n(200-)=4/8=0.5000$.

The strict upper tail contains 275 and 400, giving $2/8=0.2500$. This also follows from $1-F_n(200)=1-6/8=2/8$. The inclusive upper tail contains both 200-dollar losses as well as 275 and 400, giving $4/8=0.5000$; equivalently, $1-F_n(200-)=1-4/8=4/8$.

Finally, exactly two losses equal 200, so the empirical mass is $2/8=0.2500$. As a check, the cumulative jump is $6/8-4/8=2/8$, and the tail difference is $4/8-2/8=2/8$. Both give the same tie mass.

Reading the Empirical Step at a Threshold

The empirical cumulative distribution is a step function. At a value observed in the data, it includes all records equal to that value; its left limit excludes them. The strict survival proportion $1-F_n(x)$ therefore takes the value after the cumulative step at $x$, while the inclusive upper tail $1-F_n(x-)$ includes the mass at that step.

This relationship does not require losses to be distinct. Ties may arise naturally in discrete-valued data or because amounts have been recorded to a fixed precision. Nor does the identity require a particular parametric distribution: it follows directly from how the empirical proportions are defined. For a population distribution that has no probability mass at $x$, the population strict and inclusive tails at that exact point coincide. That fact should not be used to ignore observed ties in an empirical calculation.

Exam trap: If a question asks for the proportion with loss “above $x$,” calculate $1-F_n(x)$. If it asks for “at least $x$,” calculate $1-F_n(x-)$. Do not decide which formula to use from the word “tail” alone; identify whether equality belongs to the event.

A reliable check is to compute the threshold mass separately. The inclusive tail must equal the strict tail plus that mass. If there are no observations equal to the threshold, both tails are equal. These checks catch complement errors before a proportion is used in a later calculation.

Check Your Understanding

For each problem, pay attention to whether the threshold is included in the event.

A complete sample of six losses is 40, 60, 60, 90, 120, and 150 dollars. At 60 dollars, find $F_n(60)$, the strict upper-tail proportion, and the inclusive upper-tail proportion. (Answer: $F_n(60)=3/6$, strict tail $=3/6$, inclusive tail $=5/6$.)
Eight losses are 20, 50, 80, 120, 120, 160, 200, and 240 dollars. At 100 dollars, find the strict and inclusive upper-tail proportions and explain their equality. (Answer: both are $5/8$; no observation equals 100.)
Ten losses are 100, 150, 200, 250, 300, 300, 350, 400, 450, and 500 dollars. At 300 dollars, calculate $F_n(300)$, the strict upper tail, the inclusive upper tail, and the mass at 300. (Answer: $F_n(300)=6/10$, strict tail $=4/10$, inclusive tail $=6/10$, mass $=2/10$.)
At a threshold $a$, suppose $F_n(a)=0.72$ and the empirical mass at $a$ is $0.08$. Find the strict and inclusive upper-tail proportions. (Answer: strict tail $=0.28$; inclusive tail $=0.36$.)
Twelve losses are 30, 40, 50, 50, 60, 70, 70, 70, 80, 90, 100, and 110 dollars. At 70 dollars, find the proportions strictly below, at or below, strictly above, and at or above the threshold. (Answer: strictly below $=4/12$, at or below $=7/12$, strictly above $=5/12$, at or above $=8/12$.)

What You'll Learn