Geometric Severity Models

Introduction

A severity model describes the size of an individual loss. Some losses are recorded in whole-number units: for example, a hypothetical repair cost might be measured in fixed increments, or a claim count might represent the number of damaged components. The geometric distribution is a simple model for a positive integer-valued severity when each successive unit represents another trial that may end the process.

The parameter convention matters. In this tutorial, a severity $X$ has support $1,2,3,\ldots$, and $p$ is the probability that a trial ends the process. Thus the probability of a severity of one unit is $p$, not zero. This is the positive-integer geometric model.

Definition: Let $X$ be the number of independent Bernoulli trials up to and including the first success, with success probability $p$, where $0<p\leq 1$. Then $X$ has a geometric distribution on the positive integers: $$ P(X=k)=p(1-p)^{k-1},\qquad k=1,2,3,\ldots $$ Write $q=1-p$ for the probability of failure on one trial. The probability mass function is then $P(X=k)=pq^{k-1}$.

The trial interpretation is one way to motivate the model; in a severity application, the units need not literally be repeated trials. The mathematical requirements are that the support is the positive integers and that the probabilities follow this geometric pattern for a specified $p$. If the possible severities include zero, or if the possible values are not equally spaced integer units, this model does not directly describe them without a change of variable or model.

Probabilities and the Tail

To calculate a point probability, substitute the specified positive integer $k$ into the probability mass function. Summing the probabilities from 1 through $k$ gives the cumulative distribution function. Using the geometric-series formula, the cumulative and tail probabilities are:

F_X(k)=P(X\leq k)=1-q^k,\qquad P(X>k)=q^k,\qquad k=0,1,2,\ldots

At $k=0$, $F_X(0)=0$, as required because $X$ is always positive. For an integer interval, it is often quicker to subtract tail probabilities than to add several individual masses. For integers $1\leq r\leq s$,

P(r\leq X\leq s)=P(X>r-1)-P(X>s) =q^{r-1}-q^s.

For a noninteger threshold $a$, the event must be translated to the integer support before using these formulas. For example, $X>a$ is equivalent to $X>\lfloor a\rfloor$, where $\lfloor a\rfloor$ is the greatest integer no larger than $a$. This support check prevents an off-by-one error.

Worked Example: Point, Cumulative, and Tail Probabilities

A hypothetical repair severity $X$ is the number of fixed repair units required. Assume $X$ follows the positive-integer geometric model with $p=0.30$. Find $P(X=4)$, $P(X\leq4)$, $P(X>4)$, the mean, and the variance. Give probabilities to four decimals and moments to four decimals.

Method. Use $q=1-p$, then apply the positive-integer probability, cumulative, and tail formulas. The model assumption is that each trial has the same success probability $p=0.30$, independently of the others.

First, $q=1-0.30=0.70$. For a severity of exactly four units, the first three trials fail and the fourth succeeds:

P(X=4)=pq^{4-1} =0.30(0.70)^3 =0.30(0.343) =0.1029.

The cumulative probability through four units and the probability of exceeding four units are:

P(X\leq4)=1-(0.70)^4 =1-0.2401 =0.7599, $$ $$ P(X>4)=(0.70)^4=0.2401.

The two probabilities sum to $0.7599+0.2401=1.0000$, which checks that the cumulative event and its complement have been handled consistently.

The mean and variance are $E[X]=1/p$ and $\operatorname{Var}(X)=q/p^2$, derived below. Thus

E[X]=\frac{1}{0.30}=3.3333,\qquad \operatorname{Var}(X)=\frac{0.70}{(0.30)^2} =\frac{0.70}{0.09} =7.7778.

The mean is measured in repair units and the variance in squared repair units. A common mistake is to use $pq^4$ for $P(X=4)$, which shifts the exponent by one and gives the probability of five units, not four.

Mean and Variance: Why the Formulas Hold

A short derivation is useful both for understanding the moments and for checking the parameter convention. For a nonnegative integer-valued variable, the probability generating function is $G_X(s)=E[s^X]$. For the positive-integer geometric model, sum the probability mass function as a geometric series:

G_X(s)=\sum_{k=1}^{\infty}pq^{k-1}s^k =ps\sum_{j=0}^{\infty}(qs)^j =\frac{ps}{1-qs},\qquad |qs|<1.

The first derivative at $s=1$ gives $E[X]$. The second derivative gives the second factorial moment $E[X(X-1)]$. Since $G_X'(s)=p/(1-qs)^2$ and $G_X''(s)=2pq/(1-qs)^3$, substituting $s=1$ yields:

E[X]=G_X'(1)=\frac{p}{(1-q)^2}=\frac{1}{p}, \qquad E[X(X-1)]=G_X''(1)=\frac{2q}{p^2}.

Because $X^2=X(X-1)+X$, these results give the second raw moment and variance:

E[X^2]=\frac{2q}{p^2}+\frac{1}{p} =\frac{2q+p}{p^2} =\frac{2-p}{p^2}, \qquad \operatorname{Var}(X)=E[X^2]-(E[X])^2 =\frac{1-p}{p^2} =\frac{q}{p^2}.

The formulas have sensible limiting behavior. If $p=1$, every trial succeeds immediately, so $X=1$ with probability 1; the formulas give mean 1 and variance 0. As $p$ decreases, long sequences of failures become more likely, increasing both the mean and the spread.

Theorem: For a positive-integer geometric severity with $0<p\leq1$ and $q=1-p$, $E[X]=1/p$ and $\operatorname{Var}(X)=q/p^2$. These formulas apply to the support $1,2,\ldots$; they are not the formulas for a geometric variable counting failures before the first success.

Quantiles and Conditional Tail Probabilities

For $0<\alpha<1$, define the $\alpha$-quantile as the smallest value $x$ for which $F_X(x)\geq\alpha$. Since this severity takes integer values, its quantile is an integer. For $0<p<1$, the condition is $1-q^k\geq\alpha$, or $q^k\leq1-\alpha$. Taking logarithms and accounting for $\ln(q)<0$ gives:

k\geq\frac{\ln(1-\alpha)}{\ln(q)}, \qquad x_\alpha=\left\lceil\frac{\ln(1-\alpha)}{\ln(q)}\right\rceil.

The ceiling function rounds up to the next integer, or leaves an integer unchanged. Always verify the result by checking that $F_X(x_\alpha)\geq\alpha$ and $F_X(x_\alpha-1)<\alpha$. If the ratio inside the ceiling is exactly an integer, that integer is the quantile.

The geometric model also has a useful conditional-tail property. For nonnegative integers $m,n$, provided $P(X>m)>0$,

P(X>m+n\mid X>m) =\frac{P(X>m+n)}{P(X>m)} =\frac{q^{m+n}}{q^m} =q^n.

Thus, conditional on the severity already exceeding $m$ units, the probability of exceeding it by at least another $n$ units is the same as the original probability of exceeding $n$ units. This is a direct consequence of the geometric tail formula, rather than an assumption that the remaining severity is zero.

Worked Example: Recovering the Parameter and Finding a Quantile

A hypothetical positive-integer severity has a geometric distribution. An analyst is told that the probability it exceeds three units is $0.512$. Find $p$, calculate the probability that it is between four and seven units inclusive, and find its 90th percentile. Give probabilities to four decimals.

Method. Use the tail probability to identify $q$, then use the difference of tails for the interval and the inverse CDF for the quantile.

Since $P(X>3)=q^3$,

q^3=0.512,\qquad q=\sqrt[3]{0.512}=0.8,\qquad p=1-q=0.2.

For the inclusive interval from four through seven, subtract the probability of exceeding seven from the probability of exceeding three:

P(4\leq X\leq7)=P(X>3)-P(X>7) =(0.8)^3-(0.8)^7 =0.512-0.2097152 =0.3022848 \approx0.3023.

For the 90th percentile, set $\alpha=0.90$:

\frac{\ln(1-0.90)}{\ln(0.8)} =\frac{\ln(0.10)}{\ln(0.8)} \approx10.3189, \qquad x_{0.90}=\lceil10.3189\rceil=11.

Check the cutoff on either side:

F_X(10)=1-(0.8)^{10}=0.8926258<0.90, \qquad F_X(11)=1-(0.8)^{11}=0.9141007\geq0.90.

Therefore the 90th percentile is 11 units. The check confirms the smallest-integer definition of a quantile. A common mistake is to round $10.3189$ down to 10; that gives a cumulative probability below 0.90.

Applying the Model to Monetary Severity

A positive-integer model can represent monetary severity when each integer corresponds to a fixed monetary unit. If $X$ is the number of units and each unit is worth $c$ dollars, define the dollar severity $Z=cX$. Then $E[Z]=cE[X]$ and $\operatorname{Var}(Z)=c^2\operatorname{Var}(X)$. The square on $c$ is essential for variance.

Worked Example: Fixed-Dollar Units and a Conditional Tail

A hypothetical equipment policy records covered severity in units of $c=\$250$. Assume the number of units $X$ is geometric on $1,2,\ldots$ with $p=0.40$. Find the probability that the dollar severity $Z=250X$ exceeds $\$1{,}500$, the probability it exceeds $\$1{,}750$ given that it already exceeds $\$1{,}000$, and the mean and variance of $Z$. Give probabilities to four decimals and money moments in dollars and dollars squared.

Method. Translate dollar cutoffs into integer-unit events first. Here $q=1-0.40=0.60$.

The event $Z>\$1{,}500$ means $250X>1{,}500$, or $X>6$. Therefore

P(Z>1{,}500)=P(X>6)=(0.60)^6=0.046656\approx0.0467.

For the conditional probability, $Z>\$1{,}750$ is $X>7$, and $Z>\$1{,}000$ is $X>4$. Hence

P(Z>1{,}750\mid Z>1{,}000) =\frac{P(X>7)}{P(X>4)} =\frac{(0.60)^7}{(0.60)^4} =(0.60)^3 =0.2160.

The mean and variance follow by scaling $X$:

E[X]=\frac{1}{0.40}=2.5,\qquad \operatorname{Var}(X)=\frac{0.60}{(0.40)^2} =\frac{0.60}{0.16}=3.75, $$ $$ E[Z]=250(2.5)=\$625.00,\qquad \operatorname{Var}(Z)=250^2(3.75) =62{,}500(3.75)=\$234{,}375.00^2.

The event translations are checked at the boundaries: exactly six units is $\$1{,}500$, so exceeding $\$1{,}500$ requires at least seven units; exceeding $\$1{,}750$ requires at least eight. The variance uses $250^2$, not 250, because variance is measured in squared dollars.

Common Convention Trap

Some references define a geometric variable as the number of failures before the first success. That variable has support $0,1,2,\ldots$, with probability $P(X=k)=pq^k$, mean $q/p$, and variance $q/p^2$. It is not the convention used here. If $N$ counts failures before the first success, then the positive-integer count in this tutorial is $X=N+1$. Consequently $E[X]=E[N]+1=1/p$, while the variance is unchanged by adding one.

The most reliable exam procedure is to write down the support and the meaning of $p$ before calculating. For the convention here, check that $P(X=1)=p$, that the probabilities sum to one, and that $P(X>k)=q^k$. These checks expose an accidental shift in the exponent or a zero-based formula.

Key takeaway: For positive-integer geometric severity, $P(X=k)=pq^{k-1}$, $P(X>k)=q^k$, $E[X]=1/p$, and $\operatorname{Var}(X)=q/p^2$. Translate monetary thresholds to integer units before applying the formulas, and round quantiles up only after checking the CDF cutoff.

Check Your Understanding

For each problem, use the positive-integer convention $P(X=k)=pq^{k-1}$. Answers are provided at the end of each question for checking.

A hypothetical claim severity in unit counts has $p=0.40$. Find $P(X=3)$, $P(X>5)$, $E[X]$, and $\operatorname{Var}(X)$. Give probabilities to four decimals. (Answer: $0.1440,\ 0.0778,\ 2.5,\ 3.75$.)
A geometric severity satisfies $P(X>4)=0.4096$. Find $p$ and $P(2\leq X\leq5)$. Give the probability to four decimals. (Answer: $p=0.20$, probability $0.4723$.)
For a positive-integer geometric severity with $p=0.25$, find the 75th percentile and verify that the preceding integer has cumulative probability below 0.75. (Answer: $x_{0.75}=5$, with $F_X(4)=0.6836<0.75$ and $F_X(5)=0.7627\geq0.75$.)
A hypothetical loss is $Z=100X$ dollars, where $X$ is geometric with $p=0.50$. Find $P(Z>\$700)$, $E[Z]$, and $\operatorname{Var}(Z)$. (Answer: $0.0078,\ \$200.00,\ \$20{,}000.00^2$.)
For a geometric severity with $p=0.30$, calculate $P(X>6\mid X>2)$ and explain the number of additional units represented by the conditional event. (Answer: $0.2401$; it is the probability of exceeding the current threshold by four additional units.)

What You'll Learn