Why The Empirical Distribution Is A Step Function

Why the Empirical Distribution Has Steps

An empirical distribution function summarizes a finite set of observed losses by reporting the fraction of observations at or below each threshold. Because each record contributes either zero or one to that fraction, the function changes only when the threshold reaches an observed value. Between observed values, the fraction is constant. This is why the empirical distribution is a step function rather than a smooth curve.

The definition also gives more than a convenient calculation rule. It lets us establish directly that the function is nondecreasing, right-continuous, and has jumps whose sizes reflect the observed data. These properties are useful when reading or checking an empirical distribution, especially at thresholds that equal an observed loss.

Definition: Let $X_1,\ldots,X_n$ be $n\geq 1$ complete observed losses, each a finite real number. The empirical distribution function is $$ F_n(t)=\frac{1}{n}\sum_{i=1}^{n}\mathbf{1}_{\{X_i\leq t\}}, $$ where $\mathbf{1}_{\{X_i\leq t\}}$ equals 1 when $X_i\leq t$, and equals 0 otherwise. Thus $F_n(t)$ is the proportion of the $n$ observations that are less than or equal to $t$.

The inequality in the indicator is inclusive: an observation exactly equal to $t$ is counted in $F_n(t)$. The definition presumes that all $n$ records are observed as specified. It is not, by itself, a method for adjusting incomplete or censored records.

What the Indicator Definition Implies

For a fixed observation $X_i$, its indicator as a function of $t$ is 0 for $t<X_i$, then becomes 1 at $t=X_i$ and remains 1 thereafter. Adding the $n$ indicators and dividing by $n$ gives the empirical distribution. The properties of the whole function follow from this simple contribution made by each record.

Theorem: For a nonempty finite sample of complete real-valued observations, $F_n(t)$ is nondecreasing and right-continuous. It is constant between distinct observed values. Its jump at any value $a$ is the number of observations equal to $a$, divided by $n$. Also, $$ \lim_{t\to-\infty}F_n(t)=0 \qquad\text{and}\qquad \lim_{t\to+\infty}F_n(t)=1. $$

Monotonicity

Take any two thresholds $s$ and $t$ with $s\leq t$. If an observation satisfies $X_i\leq s$, it also satisfies $X_i\leq t$. Therefore each indicator at threshold $s$ is less than or equal to the corresponding indicator at threshold $t$:

\mathbf{1}_{\{X_i\leq s\}}\leq \mathbf{1}_{\{X_i\leq t\}} \quad\text{for every }i.

Summing these $n$ inequalities and dividing by the positive number $n$ gives $F_n(s)\leq F_n(t)$. Thus the empirical distribution cannot decrease as the threshold increases. It may stay flat over an interval or increase when the threshold passes observed values.

Why It Is Right-Continuous

Right-continuity means that as a threshold approaches $t$ from values greater than $t$, the function values approach $F_n(t)$. For a finite sample, choose a sufficiently small positive interval immediately to the right of $t$ that contains no observed values strictly greater than $t$. Such an interval exists because there are only finitely many observations. Throughout that interval, no new indicator changes from 0 to 1. Any observation equal to $t$ is already counted at $t$, since the definition uses $X_i\leq t$. Therefore $F_n$ stays equal to $F_n(t)$ just to the right of $t$, which proves right-continuity.

At an observed value, the function may have a different value immediately to the left. The difference is its jump. If $a$ is any real value, let $m(a)$ be the number of observations equal to $a$. Immediately below $a$, observations equal to $a$ have not yet been counted; at $a$, all of them are counted. Hence:

F_n(a)-\lim_{t\uparrow a}F_n(t)=\frac{m(a)}{n}.

This formula includes the case $m(a)=0$: at a value not observed in the sample, there is no jump. It also shows why repeated observations produce a larger jump than a single observation at the same sample size. The total of all jump sizes is 1, because every one of the $n$ observations contributes $1/n$ once the threshold has passed its value.

Constant Intervals and Limiting Values

If no observed value lies between two thresholds, the indicators at those thresholds are identical for every record. Thus $F_n$ is constant between successive distinct observed values. Below the smallest observation, no record has been counted, so the function is zero. At or above the largest observation, all records have been counted, so the function is one. These facts give the limits at negative and positive infinity in the theorem.

Worked Example: Building the Steps From Distinct Losses

A hypothetical portfolio records four complete losses, in dollars: $80, 160, 300,$ and $520$. Construct $F_n(t)$, evaluate it at $t=159$ and $t=300$, and identify its jumps and limiting values.

Method. Use the indicator definition with $n=4$. For each range of thresholds, count how many of the four losses are less than or equal to the threshold.

For $t<80$, none of the losses is at or below $t$, so $F_4(t)=0/4=0$. For $80\leq t<160$, exactly one loss is counted, giving $1/4=0.2500$. For $160\leq t<300$, two are counted, giving $2/4=0.5000$. For $300\leq t<520$, three are counted, giving $3/4=0.7500$. At $t\geq520$, all four are counted, giving $4/4=1.0000$. Therefore,

F_4(t)= \begin{cases} 0, & t<80,\\ \frac14, & 80\leq t<160,\\ \frac12, & 160\leq t<300,\\ \frac34, & 300\leq t<520,\\ 1, & t\geq520. \end{cases}

At $t=159$, the losses at or below the threshold are $80$ only. Thus $F_4(159)=1/4=0.2500$. At $t=300$, the losses counted are $80,160,300$, including the loss exactly at the threshold. Thus $F_4(300)=3/4=0.7500$.

Each observed value occurs once, so each jump is $1/4=0.2500$. The function is zero below 80 and one at and above 520, giving limits 0 and 1 at negative and positive infinity, respectively. As a check, the four jumps sum to $4(1/4)=1$, matching the full rise of the function.

Common mistake avoided: At $t=300$, counting only losses strictly below 300 would give $2/4$. The empirical distribution counts losses less than or equal to the threshold, so the correct value is $3/4$.

Worked Example: Finding a Jump When Values Repeat

A hypothetical group of six insured repair costs, in dollars, is $90, 140, 140, 260, 310,$ and $470$. Find the empirical distribution’s value at $140$, its value just below $140$, its jump at $140$, and its value at $260$.

Method. Apply the indicator definition and count observations at or below each threshold. The sample size is $n=6$.

At $t=140$, the counted costs are $90,140,140$, so the count is $3$. Therefore $F_6(140)=3/6=1/2=0.5000$. Immediately below 140, only the cost of 90 is counted, so the left-hand limit is $1/6=0.1667$, rounded to four decimal places.

The jump at 140 is the difference between the value at the point and the left-hand limit:

F_6(140)-\lim_{t\uparrow140}F_6(t) =\frac{3}{6}-\frac{1}{6} =\frac{2}{6} =\frac{1}{3} =0.3333.

There are exactly two observations equal to 140, and each contributes $1/6$, so the jump is $2/6$. At $t=260$, the costs $90,140,140,260$ are counted. Thus $F_6(260)=4/6=2/3=0.6667$. This is at or above the value $1/2$ at 140, as monotonicity requires.

Check. The total observations counted at 260 are four out of six, not five: 310 and 470 exceed 260. The computed value is therefore consistent with both the data and the nondecreasing property.

Worked Example: Checking Boundary Values and Right-Continuity

A hypothetical sample of five complete losses, in dollars, is $60, 110, 180, 360,$ and $540$. A reviewer claims that $F_5(110)=1/5$, the value immediately below 110 is $2/5$, and $F_5(181)=4/5$. Check each claim, find the jump at 110, and state the two limits.

Method. Count losses using the inclusive condition $X_i\leq t$; separately consider the count immediately below an observed value.

At $t=110$, both 60 and 110 are counted. Thus $F_5(110)=2/5=0.4000$, so the reviewer’s first claim is false. Immediately below 110, only 60 is counted, giving a left-hand limit of $1/5=0.2000$. The claim that the value immediately below is $2/5$ is also false.

The jump at 110 is $2/5-1/5=1/5=0.2000$, corresponding to the single observation equal to 110. At $t=181$, the losses 60, 110, and 180 are counted, so $F_5(181)=3/5=0.6000$; the claim of $4/5$ is false. Below 60 the function is zero, and at or above 540 it is one. Therefore the limits as $t\to-\infty$ and $t\to+\infty$ are 0 and 1.

Check. The sample has five distinct observations, each contributing one jump of $1/5$. The five jumps total 1, consistent with the stated limiting values.

Common Exam Traps

A frequent error is to replace the defining condition $X_i\leq t$ with $X_i<t$. That changes the function at observed values. For a hypothetical sample $100,200,300$, the correct empirical distribution at 200 is

F_3(200)=\frac{\mathbf{1}_{\{100\leq200\}}+ \mathbf{1}_{\{200\leq200\}}+ \mathbf{1}_{\{300\leq200\}}}{3} =\frac{1+1+0}{3} =\frac23 =0.6667.

The incorrect strict count would be $1/3=0.3333$, because it excludes the observation equal to 200. It also gives a different continuity behavior at observed points. For the empirical distribution defined here, the value at a step is the level after the jump, matching the right-hand limit.

Do not confuse the value at an observation with the level immediately before it. The observation at the threshold is included in $F_n(t)$.
Do not assume every jump has size $1/n$. That is true for a value occurring once. If $m$ records have the same value, the jump is $m/n$.
Do not expect the function to rise between observations. With no new observation to count, the indicators do not change and the function stays flat.
Do not reverse the limits. A threshold far below all finite observations counts none; a threshold far above them counts all.

Key takeaway: The empirical distribution is a normalized sum of threshold indicators. That definition directly gives nondecreasing steps, right-continuity, jump size equal to the fraction of observations at the jump value, and limits of 0 and 1.

Check Your Understanding

For each problem, use the inclusive empirical distribution definition and check that your values agree with the observed records.

A hypothetical complete sample of four losses is $35, 90, 150, 280$. Find $F_4(90)$ and the jump at 150. (Answer: $F_4(90)=2/4=0.5000$; jump at 150 $=1/4=0.2500$.)
A sample of five complete costs is $40, 80, 80, 190, 300$. Find $F_5(80)$ and the left-hand limit at 80. (Answer: $F_5(80)=3/5=0.6000$; left-hand limit $=1/5=0.2000$.)
For the sample $25, 100, 175, 400$, state the value of $F_4(t)$ for $100\leq t<175$, and give the limits at negative and positive infinity. (Answer: $F_4(t)=2/4=0.5000$; the limits are 0 and 1, respectively.)
A sample has eight observations, three of which equal 225. If five observations are strictly below 225 and none of the remaining observations are below it, find the jump at 225 and $F_8(225)$. (Answer: jump $=3/8=0.3750$; $F_8(225)=8/8=1.0000$.)
For the complete sample $50, 120, 240, 500$, a proposed calculation gives $F_4(240)=2/4$. Identify the error and give the correct value and jump at 240. (Answer: it excluded the observation equal to 240; $F_4(240)=3/4=0.7500$, and the jump is $1/4=0.2500$.)

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