Erlang Severity As A Sum Of Exponentials

Why a Sum of Exponential Losses Matters

A claim can arise from several successive loss-generating stages. For example, a hypothetical equipment repair might involve separate component failures whose costs are modeled as independent exponential amounts. If each stage has the same exponential distribution, the total cost is not exponential: it is an Erlang severity. Its integer shape counts the stages, and its rate describes the size of each stage.

The Erlang distribution is also a useful special case of the gamma family. Here the emphasis is on its construction as a sum, which gives a direct interpretation of the shape parameter and a particularly convenient formula for survival probabilities.

Definition: Let $X_1,\ldots,X_k$ be independent, identically distributed exponential random variables with rate $\lambda>0$, where $k$ is a positive integer. Their sum $X=\sum_{j=1}^{k}X_j$ has an Erlang distribution with shape $k$ and rate $\lambda$. Each $X_j$ has density $f_{X_j}(x)=\lambda e^{-\lambda x}$ for $x\geq 0$, mean $1/\lambda$, and variance $1/\lambda^2$.

We use the rate parameterization: the mean of each exponential component is $1/\lambda$, not $\lambda$. If the loss is measured in dollars, $\lambda$ has units of inverse dollars; if loss is measured in thousands of dollars, $\lambda$ has units of inverse thousands of dollars. The shape $k$ is a count and has no units.

Deriving the Density from the Sum

The Erlang density follows by repeated convolution. For $k=1$, the sum is just one exponential variable, with density $\lambda e^{-\lambda x}$. Suppose the sum of $n$ components has density $f_n(t)=\lambda^n t^{n-1}e^{-\lambda t}/(n-1)!$ for $t\geq0$. Adding one independent exponential component gives the following density for the sum of $n+1$ components:

\begin{aligned} f_{n+1}(x) &=\int_0^x f_n(t)\lambda e^{-\lambda(x-t)}\,dt\\ &=\frac{\lambda^{n+1}e^{-\lambda x}}{(n-1)!} \int_0^x t^{n-1}\,dt\\ &=\frac{\lambda^{n+1}x^n e^{-\lambda x}}{n!},\qquad x\geq0. \end{aligned}

The induction starts at $n=1$, so it proves the formula for every positive integer $k$. In particular, the density and distribution function are

f_X(x)=\frac{\lambda^k x^{k-1}e^{-\lambda x}}{(k-1)!},\quad x\geq0, \qquad F_X(x)=\int_0^x f_X(t)\,dt.

The density is zero for $x<0$. The factor $x^{k-1}$ distinguishes the sum from a single exponential: for $k>1$, the density starts at zero and initially rises before eventually declining. The exponential case is recovered when $k=1$.

Moments of an Erlang Severity

The sum representation makes the first two moments immediate, provided the component losses are independent and have the same rate. Expected values add, and independence means the variances add:

E[X]=\sum_{j=1}^k E[X_j]=\frac{k}{\lambda}, \qquad \operatorname{Var}(X)=\sum_{j=1}^k\operatorname{Var}(X_j) =\frac{k}{\lambda^2}.

A direct density calculation gives every nonnegative integer raw moment. For an integer $r\geq0$, substitute $u=\lambda x$ in the integral for $E[X^r]$. The resulting integral is the factorial integral $\int_0^\infty u^{k+r-1}e^{-u}\,du=(k+r-1)!$:

\begin{aligned} E[X^r] &=\frac{\lambda^k}{(k-1)!}\int_0^\infty x^{k+r-1}e^{-\lambda x}\,dx\\ &=\frac{(k+r-1)!}{(k-1)!\lambda^r}. \end{aligned}

For example, $E[X^2]=k(k+1)/\lambda^2$, and subtracting $E[X]^2=k^2/\lambda^2$ gives $\operatorname{Var}(X)=k/\lambda^2$. This agrees with the variance obtained by adding the independent components.

Useful parameter check: If the mean $\mu$ and variance $\sigma^2$ are known to come from an Erlang model, then $k=\mu^2/\sigma^2$ and $\lambda=k/\mu$. This identifies an Erlang model only when $\mu^2/\sigma^2$ is a positive integer and the assumed common-rate, independence model is appropriate.

Survival Probabilities for Integer Shape

For severity calculations, the survival function is $S_X(x)=P(X>x)=1-F_X(x)$. Erlang survival probabilities have a finite-sum form. Write $z=\lambda x$. Substitution into the density integral gives $S_X(x)=\int_z^\infty t^{k-1}e^{-t}\,dt/(k-1)!$. Integration by parts reduces this integral to a lower power of $t$, and repeating the reduction yields

S_X(x)=e^{-\lambda x}\sum_{j=0}^{k-1}\frac{(\lambda x)^j}{j!}, \qquad x\geq0.

To see why the finite sum results, define $I_n(z)=\int_z^\infty t^n e^{-t}\,dt$. Integration by parts gives $I_n(z)=z^n e^{-z}+nI_{n-1}(z)$, with $I_0(z)=e^{-z}$. Applying this recurrence through $n=k-1$ gives $I_{k-1}(z)=e^{-z}(k-1)!\sum_{j=0}^{k-1}z^j/j!$. Dividing by $(k-1)!$ proves the survival formula. The sum ends at $k-1$, not $k$.

Check the model.
Confirm that the loss is a sum of a positive integer number of independent exponential components with one common rate.

Set the units and parameters.
Use the component rate $\lambda$, and express the threshold $x$ in the same loss units used for that rate.

Calculate the requested quantity.
Use $k/\lambda$ for the mean, $k/\lambda^2$ for the variance, or the finite survival sum with $z=\lambda x$.

Worked Example: Total Cost from Four Repair Stages

A hypothetical equipment policy models total repair cost $X$, in thousands of dollars, as the sum of four independent exponential stage costs. Each stage has rate $\lambda=2$ per thousand dollars. Find the mean, variance, and probability that total cost exceeds $2,000. There is no interest or mortality assumption: this is an undiscounted severity model.

Method. Use the sum-of-components moments and the Erlang survival formula. The model is Erlang with $k=4$ and $\lambda=2$ per thousand dollars. The threshold is $x=2$ thousand dollars, so $z=\lambda x=2(2)=4$.

The mean is $k/\lambda=4/2=2$ thousand dollars, or $2,000. The variance is $k/\lambda^2=4/2^2=1$ thousand-dollars-squared. Thus the standard deviation is $1$ thousand dollars, or $1,000.

For the tail probability, substitute $k=4$ and $z=4$:

\begin{aligned} P(X>2) &=e^{-4}\left(1+4+\frac{4^2}{2!}+\frac{4^3}{3!}\right)\\ &=e^{-4}\left(1+4+8+\frac{64}{6}\right)\\ &=e^{-4}\left(\frac{71}{3}\right)\\ &\approx 0.4335. \end{aligned}

The result is a probability between zero and one. The common mistake avoided is treating the four-stage sum as exponential and using $e^{-\lambda x}=e^{-4}$ alone. That expression is the survival function for one exponential component, not the four-component total.

Worked Example: Recovering Shape and Rate from Moments

For a hypothetical portfolio, an undiscounted claim severity $X$, measured in thousands of dollars, is assumed to be Erlang with unknown integer shape and rate. Its mean is 6 thousand dollars and its variance is 18 thousand-dollars-squared. Determine the parameters and calculate $P(X>6)$. No interest or mortality assumption is involved.

Method. Use the ratio $\mu^2/\sigma^2$ to identify the shape. With $\mu=6$ and $\sigma^2=18$, both measured in consistent thousand-dollar units,

k=\frac{\mu^2}{\sigma^2}=\frac{6^2}{18}=2.

This is a positive integer, as required for an Erlang model. Then $\lambda=k/\mu=2/6=1/3$ per thousand dollars. Check the fitted moments: $k/\lambda=2/(1/3)=6$ thousand dollars, and $k/\lambda^2=2/(1/9)=18$ thousand-dollars-squared.

At $x=6$ thousand dollars, $\lambda x=(1/3)(6)=2$. With $k=2$, the finite survival sum has terms $j=0,1$:

P(X>6)=e^{-2}\left(1+2\right)=3e^{-2}\approx 0.4060.

The calculation reproduces both given moments and gives a valid probability. The common mistake avoided is setting the rate equal to $1/\mu$ without accounting for the shape. Here $1/\mu=1/6$ per thousand dollars is the rate for one exponential component only if the total had shape one; the fitted Erlang rate is $k/\mu=1/3$.

Worked Example: Probability Between Two Thresholds

A hypothetical service contract has a total loss $X$, in thousands of dollars, modeled as the sum of three independent exponential losses, each with rate $1$ per thousand dollars. Find the probability that $X$ is greater than $1,000 but no greater than $3,000. There is no interest or mortality assumption.

Method. Use the difference of two survival probabilities. The model has $k=3$, $\lambda=1$, and thresholds $x_1=1$, $x_2=3$ in thousands of dollars. Since $P(X>1,\ X\leq3)=P(X>1)-P(X>3)$,

\begin{aligned} P(1<X\leq3) &=e^{-1}\left(1+1+\frac{1^2}{2}\right) -e^{-3}\left(1+3+\frac{3^2}{2}\right)\\ &=2.5e^{-1}-8.5e^{-3}\\ &\approx 0.9197-0.4232\\ &\approx 0.4965. \end{aligned}

The answer is positive and is smaller than either survival probability, as an interval probability must be. The common mistake avoided is subtracting the two cumulative probabilities in reverse order or using the density at the endpoints instead of the probability between them.

Limits of the Erlang Model and Exam Traps

The Erlang structure is useful when repeated stages are a defensible model for the loss, but its assumptions matter. If exponential components are independent but have different rates, their sum is generally not Erlang. If components are dependent, the Erlang density derived by convolution does not apply; in particular, variances cannot simply be added without including covariances. A positive integer shape is essential to the interpretation as a fixed number of identical stages.

Rate versus scale: Under the rate convention, a component mean is $1/\lambda$, and total mean is $k/\lambda$. A scale parameter would be the reciprocal, not the rate itself.
Units: The product $\lambda x$ must be dimensionless. A rate per thousand dollars cannot be multiplied by a threshold expressed in dollars without converting the threshold first.
Tail versus distribution function: The finite sum gives $P(X>x)$. For $P(X\leq x)$, subtract that tail from one.
Shape count: The survival sum stops at $k-1$. Including a $k$th term gives an incorrect tail probability.

For a fitted model, check both the parameters and the result: the inferred shape should be a positive integer, the rate should be positive, and a calculated probability should lie between zero and one. These simple checks often expose a unit or parameterization error before it affects later calculations.

Key takeaway: An Erlang loss is the sum of $k$ independent exponential losses sharing rate $\lambda$. Its mean is $k/\lambda$, its variance is $k/\lambda^2$, and its survival probability is $e^{-\lambda x}\sum_{j=0}^{k-1}(\lambda x)^j/j!$. The formulas depend on the common-rate, independence, and integer-shape assumptions.

Check Your Understanding

For each hypothetical severity model, use the stated loss units and the Erlang formulas. Answers to probabilities are rounded to four decimal places.

A loss is the sum of five independent exponential components, each with rate $1/2$ per thousand dollars. Find the mean and variance of the total loss. (Answer: mean $=10$ thousand dollars; variance $=20$ thousand-dollars-squared.)
A total loss has Erlang shape $2$ and rate $4$ per thousand dollars. Find the probability it exceeds $0.5$ thousand dollars. (Answer: $0.4060$.)
An Erlang loss, measured in thousands of dollars, has mean $8$ and variance $32$. Find its integer shape and rate. (Answer: $k=2$, $\lambda=0.25$ per thousand dollars.)
A total loss is the sum of three independent exponential components, each with rate $1.5$ per thousand dollars. Find the probability the total loss is at most $2$ thousand dollars. (Answer: $0.5768$.)
A total loss is the sum of four independent exponential components, each with rate $0.25$ per thousand dollars. Find the probability the total exceeds $4$ thousand dollars. (Answer: $e^{-1}(1+1+1/2+1/6)\approx0.9810$.)

What You'll Learn