Point Masses and Jumps in Severity Distributions

Why a CDF Can Jump

A severity CDF does more than summarize probabilities below a threshold. At a particular loss amount, it can also show the probability that the loss is exactly that amount. This matters whenever claim amounts are recorded or modeled in discrete increments, such as fixed benefit amounts or losses rounded to a specified unit.

For a loss random variable $X$, the cumulative distribution function is $F_X(x)=P(X\le x)$. If some probability is concentrated at one exact amount, the CDF increases abruptly there. The size of that increase is the probability of the exact amount. To identify it correctly, compare the CDF value at the amount with its left-hand limit.

Definition: For a real number $x$, the left-hand limit of the severity CDF is $F_X(x-)=\lim_{t\uparrow x}F_X(t)=P(X<x)$. The probability mass at exactly $x$ is $$ P(X=x)=F_X(x)-F_X(x-). $$

The distinction between $P(X\le x)$ and $P(X<x)$ is essential. The first includes any probability at $x$; the second excludes it. A positive difference between these probabilities appears as a jump in the CDF. If there is no probability at $x$, the CDF has no jump there.

How to Read a Jump

A discrete severity distribution assigns probabilities to specified loss amounts. Its CDF is a nondecreasing step function: it stays constant between possible amounts and rises at each amount with positive probability. At a possible loss amount $x$, $F_X(x-)$ is the cumulative probability just before including that amount, while $F_X(x)$ includes it.

The jump formula follows directly from the events involved. The event $\{X<x\}$ is contained in $\{X\le x\}$. The part added when moving from the first event to the second is precisely $\{X=x\}$. Therefore,

P(X\le x)=P(X<x)+P(X=x),

and, because $F_X(x)=P(X\le x)$ and $F_X(x-)=P(X<x)$, subtracting gives $P(X=x)=F_X(x)-F_X(x-)$. This argument also explains why the CDF value at $x$ alone is not the probability of an exact loss: it includes all probability accumulated below $x$.

For a purely discrete severity, the CDF can be written as a cumulative sum over the possible amounts. If $X$ takes amounts $x_1<x_2<\cdots<x_n$, with probabilities $p_1,\ldots,p_n$, then at $x_j$, the CDF includes $p_j$ and all probabilities at smaller amounts. Its left limit at $x_j$ includes only amounts smaller than $x_j$, so the difference is $p_j$. The probabilities must be nonnegative and sum to 1.

This relationship is consistent with the earlier material on density and severity probabilities: a density describes probability over ranges for a continuous model, whereas a CDF jump directly records probability at one exact value. A density value at a single point is not the probability of an exact severity.

Worked Example: Finding Exact-Amount Probabilities From a Discrete Severity

A hypothetical equipment policy has a claim severity $X$, measured in dollars, with the following finite discrete distribution: $P(X=200)=0.20$, $P(X=500)=0.50$, and $P(X=1{,}000)=0.30$. Find $F_X(500-)$, $F_X(500)$, and the probability of a claim of exactly $500. Then check that the CDF jump gives the stated probability.

Model and method. The severity takes only the three listed positive amounts, with the listed probabilities. They sum to $0.20+0.50+0.30=1.00$. Use the cumulative probabilities below and at $500.

Before $500 is included, only the $200 amount has accumulated. Thus,

$$ F_X(500-)=P(X<500)=P(X=200)=0.20. $$

At $500, both the $200 and $500 amounts are included:

F_X(500)=P(X\le 500)=P(X=200)+P(X=500)=0.20+0.50=0.70.

The jump is therefore

$$ F_X(500)-F_X(500-)=0.70-0.20=0.50. $$

So $P(X=500)=0.50$, matching the model. The value $F_X(500)=0.70$ is the probability of a claim no greater than $500, not the probability of exactly $500.

Check. The CDF reaches 1.00 at $1,000, since $0.20+0.50+0.30=1.00$. The jump at $500 is 0.50, as required.

Common mistake avoided. Treating $F_X(500)$ as the exact-amount probability would give 0.70. The correction is to subtract the left limit: $0.70-0.20=0.50$.

Endpoints and CDF Differences

Jump calculations also clarify interval probabilities. For real values $a<b$, the probability that the severity lies in an interval depends on whether each endpoint is included. In particular, $F_X(b)$ includes an atom at $b$, while $F_X(a-)$ excludes an atom at $a$. These conventions give:

P(a\le X\le b)=F_X(b)-F_X(a-).

The subtraction removes everything strictly below $a$ from everything at or below $b$, leaving the probability from $a$ through $b$, inclusive. If the lower endpoint is excluded instead, use $F_X(a)$ as the amount to subtract:

P(a<X\le b)=F_X(b)-F_X(a).

Thus the choice between $F_X(a-)$ and $F_X(a)$ is not a notation detail. It determines whether probability at the lower endpoint remains in the interval.

Worked Example: Using a Jump to Handle Interval Endpoints

A hypothetical repair-cost severity $X$ takes the values $100, $400, $700, and $1,000, with probabilities 0.10, 0.25, 0.40, and 0.25, respectively. Find (a) the probability of exactly $400, (b) $P(400\le X\le700)$, and (c) $P(400<X\le700)$. Assume the finite discrete distribution is complete; its probabilities sum to 1.00.

Model and method. The CDF increases by the probability assigned to each amount. First calculate the CDF values and left limit needed for the jump and interval calculations.

At $400, the cumulative probability strictly below the amount is the probability at $100:

$$ F_X(400-)=0.10. $$

At $400, include the probability at $400:

$$ F_X(400)=0.10+0.25=0.35. $$

(a) The exact-amount probability is the jump:

$$ P(X=400)=F_X(400)-F_X(400-)=0.35-0.10=0.25. $$

At $700, the CDF includes the probabilities at $100, $400, and $700:

$$ F_X(700)=0.10+0.25+0.40=0.75. $$

(b) Both endpoints are included, so subtract the probability strictly below $400$:

P(400\le X\le700)=F_X(700)-F_X(400-)=0.75-0.10=0.65.

P(400<X\le700)=F_X(700)-F_X(400)=0.75-0.35=0.40.

The results check against the amounts in each interval: the inclusive interval contains the $400 and $700 amounts, with probability $0.25+0.40=0.65$; the strict-lower interval contains only $700, with probability 0.40.

Common mistake avoided. Using $F_X(400)$ in part (b) would incorrectly remove the probability at $400, yielding 0.40. For an inclusive lower endpoint, subtract $F_X(400-)$, not $F_X(400)$.

Conditional Probabilities at a Jump

An exact-amount probability may also be used in a conditional probability. The numerator is the probability of the exact amount, obtained from the jump. The denominator is the probability of the conditioning event, calculated with endpoints that match the condition. For example, $X\ge c$ includes a possible atom at $c$, while $X>c$ excludes it.

Worked Example: An Exact Claim Given a Threshold Is Met

A hypothetical appliance-loss severity $X$ has the finite discrete distribution $P(X=200)=0.10$, $P(X=500)=0.20$, $P(X=800)=0.30$, and $P(X=1{,}200)=0.40$. Find $P(X=800\mid X\ge800)$. The listed probabilities sum to 1.00 and specify the entire severity distribution.

Model and method. Use the conditional probability definition. Find the point mass at $800 and the probability of the conditioning event, which includes both $800 and $1,200.

The cumulative probability strictly below $800 is

$$ F_X(800-)=P(X=200)+P(X=500)=0.10+0.20=0.30. $$

At $800, the CDF also includes the probability at $800:

$$ F_X(800)=0.30+0.30=0.60. $$

Therefore, the jump gives

$$ P(X=800)=F_X(800)-F_X(800-)=0.60-0.30=0.30. $$

The event $X\ge800$ contains the $800 and $1,200 outcomes, so

P(X\ge800)=0.30+0.40=0.70.

Using $P(A\mid B)=P(A\cap B)/P(B)$, and noting that $X=800$ implies $X\ge800$, gives

P(X=800\mid X\ge800)=\frac{0.30}{0.70}=\frac{3}{7}=0.4286.

The answer is 0.4286 to four decimal places. As a check, the conditional probabilities for $800 and $1,200 among amounts at least $800 are $0.30/0.70$ and $0.40/0.70$; they sum to 1.0000 before rounding.

Common mistake avoided. The event $X\ge800$ is not $X>800$. Using only the probability at $1,200 in the denominator would omit the $800 outcome, even though the condition includes it.

Common Traps and a Reliable Procedure

The most frequent errors arise from confusing an accumulated probability with an exact-amount probability, or from overlooking an endpoint. A useful procedure is to translate the wording into an event first, then identify whether the CDF value or its left limit matches each boundary.

Identify the event.
Decide whether the question asks for an exact amount, a range, or a conditional probability, and note which endpoints are included.

Choose the correct CDF values.
Use $F_X(x)=P(X\le x)$ for a boundary included through $x$, and $F_X(x-)=P(X<x)$ for amounts strictly below $x$.

Calculate and check.
For an exact value, subtract the left limit from the CDF at the value. Check that the result is between 0 and 1 and agrees with the specified mass if one is given.

In particular, do not assume that a large CDF value means a large probability at that exact amount. The CDF may be large simply because many smaller amounts have already accumulated. Only the increase at the point identifies its exact-amount probability.

Key takeaway: At any loss amount $x$, the probability of that exact severity is the CDF value at $x$ minus its left-hand limit: $P(X=x)=F_X(x)-F_X(x-)$. Use the endpoint convention in the event when applying CDF differences.

Check Your Understanding

For each problem, identify the relevant CDF jump or endpoint convention before calculating.

A hypothetical severity takes values $150, $350, and $900 with probabilities 0.20, 0.50, and 0.30. Find $F_X(350-)$, $F_X(350)$, and $P(X=350)$. (Answer: 0.20, 0.70, and 0.50)
For a discrete severity, $F_X(200-)=0.18$ and $F_X(200)=0.43$. Find the probability of exactly $200. (Answer: 0.25)
A hypothetical severity takes values $100, $250, $500, and $1,000 with probabilities 0.15, 0.35, 0.30, and 0.20. Find $P(250<X\le500)$. (Answer: 0.30)
For a hypothetical severity, $F_X(300-)=0.22$ and $F_X(300)=0.60$. Find both $P(X=300)$ and $P(X>300)$. (Answer: 0.38 and 0.40)
A hypothetical severity takes values $200, $500, $800, and $1,200 with probabilities 0.10, 0.25, 0.40, and 0.25. Find $P(X=800\mid X\ge500)$. (Answer: 0.4444)

What You'll Learn