Gamma Parameterizations And Erlang Special Cases

Why Gamma Parameterizations Need Care

A gamma severity model is commonly written with either a scale parameter or a rate parameter. These are reciprocal quantities, not interchangeable labels: using one in a formula for the other changes the density, the units, and the resulting probabilities. A second useful fact is that when the gamma shape is a positive integer, the model has the special name Erlang. This special case gives finite-sum probability formulas and a direct interpretation as several exponential stages.

The focus here is the conversion between the two parameterizations and the integer-shape special case. Throughout, $X$ is a continuous, nonnegative loss, and the gamma parameters satisfy $\alpha>0$ and $\theta>0$, or equivalently $\alpha>0$ and $\lambda>0$.

Definition: A gamma random variable with shape $\alpha$ and scale $\theta$ has density $$ f_X(x)=\frac{x^{\alpha-1}e^{-x/\theta}}{\Gamma(\alpha)\theta^\alpha},\qquad x>0. $$ The equivalent rate parameterization uses $\lambda=1/\theta$, so the same density is $$ f_X(x)=\frac{\lambda^\alpha x^{\alpha-1}e^{-\lambda x}}{\Gamma(\alpha)},\qquad x>0. $$ Here $\Gamma(\alpha)$ is the gamma function; for a positive integer $n$, $\Gamma(n)=(n-1)!$. The density is zero for $x\leq 0$.

Converting Between Scale and Rate

The conversion follows by matching the exponential terms in the two density forms. The scale form contains $e^{-x/\theta}$, while the rate form contains $e^{-\lambda x}$. Since these must agree for every $x>0$, their coefficients of $x$ in the exponent must be equal: $\lambda=1/\theta$. The factors outside the exponential then agree as well, because $\theta^{-\alpha}=\lambda^\alpha$.

The scale has the same units as the loss, while the rate has reciprocal-loss units. For example, if loss is measured in dollars, a scale is measured in dollars and a rate in inverse dollars. This units check is a useful way to detect a mistaken conversion.

Equivalent parameterizations: A gamma model with shape $\alpha$ and scale $\theta$ is the same distribution as a gamma model with shape $\alpha$ and rate $\lambda=1/\theta$. Conversely, $\theta=1/\lambda$. The shape does not change during the conversion.

The standard gamma moment formulas provide another check on the conversion. In scale form, $E[X]=\alpha\theta$ and $\operatorname{Var}(X)=\alpha\theta^2$. In rate form, the same quantities are $E[X]=\alpha/\lambda$ and $\operatorname{Var}(X)=\alpha/\lambda^2$. The formulas agree after substituting $\theta=1/\lambda$. The variance has squared loss units, as it should.

Worked Example: Converting a Gamma Loss Model

A hypothetical claim amount $X$ is gamma with shape $\alpha=3$ and scale $\theta=2{,}000$ dollars. Express the model using a rate parameter, check that both density forms agree, and calculate its mean and variance. Give the variance in dollars squared.

Method. Keep the shape fixed and take the reciprocal of the scale. Because the scale is in dollars, the resulting rate is in inverse dollars: $$ \lambda=\frac{1}{\theta}=\frac{1}{2{,}000}=0.0005\text{ per dollar}. $$

For the scale form, $\Gamma(3)=2!=2$, so the density is $$ f_X(x)=\frac{x^2e^{-x/2000}}{2(2000)^3} =\frac{x^2e^{-x/2000}}{16{,}000{,}000{,}000},\qquad x>0. $$

For the rate form, $\lambda^3=(0.0005)^3=0.000000000125$, and therefore $$ f_X(x)=\frac{(0.0005)^3x^2e^{-0.0005x}}{2} =\frac{x^2e^{-0.0005x}}{16{,}000{,}000{,}000},\qquad x>0. $$ The exponents agree because $x/2000=0.0005x$, and the coefficients also agree.

The mean and variance are $$ E[X]=\alpha\theta=3(2{,}000)=6{,}000\text{ dollars}, \qquad \operatorname{Var}(X)=\alpha\theta^2=3(2{,}000)^2=12{,}000{,}000\text{ dollars}^2. $$

Check. Using the rate gives $E[X]=\alpha/\lambda=3/0.0005=6{,}000$ dollars and $\operatorname{Var}(X)=\alpha/\lambda^2=3/(0.0005)^2=12{,}000{,}000$ dollars squared. Both parameterizations give the same results. The common mistake avoided here is treating $0.0005$ as a scale; its units and reciprocal relationship show it is a rate.

Integer Shape: The Erlang Distribution

A gamma distribution becomes an Erlang distribution when its shape parameter is a positive integer. Writing that integer as $n$, the density has a factorial in place of the gamma function. This is a special case of the gamma model, not a different parameter conversion.

Definition: An Erlang random variable with positive integer shape $n$ and rate $\lambda>0$ has density $$ f_X(x)=\frac{\lambda^n x^{n-1}e^{-\lambda x}}{(n-1)!},\qquad x>0. $$ Its equivalent scale is $\theta=1/\lambda$. The Erlang model is the gamma model with $\alpha=n$.

An Erlang variable can also represent the total time or amount accumulated across $n$ independent exponential stages. To see why, suppose $T_n$ is the sum of $n$ independent exponential random variables, each with rate $\lambda$. For $n=1$, the density is $\lambda e^{-\lambda t}$. If the sum of $n-1$ stages has density $\lambda^{n-1}s^{n-2}e^{-\lambda s}/(n-2)!$, convolution with the final exponential gives, for $t>0$, $$ f_{T_n}(t)=\int_0^t \frac{\lambda^{n-1}s^{n-2}e^{-\lambda s}}{(n-2)!} \lambda e^{-\lambda(t-s)}\,ds =\frac{\lambda^n e^{-\lambda t}}{(n-2)!}\int_0^t s^{n-2}\,ds =\frac{\lambda^n t^{n-1}e^{-\lambda t}}{(n-1)!}. $$ This is the Erlang density. Independence and a common rate are essential to this particular sum interpretation. If $n=1$, Erlang reduces to the exponential distribution; for $n>1$, it is not itself exponential.

Erlang Probabilities as Finite Sums

For a nonnegative threshold $x$, the Erlang cumulative probability and survival probability can be evaluated without leaving an integral. Set $z=\lambda x$, a dimensionless quantity. The resulting finite sum is especially convenient when the shape is an integer.

Theorem: If $X$ is Erlang with positive integer shape $n$ and rate $\lambda$, then for $x\geq 0$, $$ P(X>x)=e^{-\lambda x}\sum_{j=0}^{n-1}\frac{(\lambda x)^j}{j!}, \qquad F_X(x)=P(X\leq x)=1-e^{-\lambda x}\sum_{j=0}^{n-1}\frac{(\lambda x)^j}{j!}. $$

To justify the survival formula, change variables in the density integral from $t$ to $u=\lambda t$. This gives $$ P(X>x)=\int_{\lambda x}^{\infty}\frac{u^{n-1}e^{-u}}{(n-1)!}\,du. $$ For $n=1$, this integral is $e^{-\lambda x}$. For a larger integer $n$, integration by parts separates a boundary term from the same integral with shape one lower: $$ P(X>x)=e^{-\lambda x}\frac{(\lambda x)^{n-1}}{(n-1)!}+P(X_{n-1}>x), $$ where the final term denotes the corresponding unit-rate integral with power $n-2$, evaluated from $\lambda x$ to infinity. Repeating this step down to shape one yields the stated sum. The CDF formula follows because $P(X\leq x)=1-P(X>x)$ for a continuous distribution.

Worked Example: A Two-Stage Erlang Severity

A hypothetical repair loss $X$, measured in thousands of dollars, follows a gamma distribution with shape $2$ and scale $2$ thousand dollars. Find the probability that the loss exceeds $4$ thousand dollars, and calculate the mean and variance.

Method. Since the shape is the positive integer $2$, this is an Erlang distribution. Convert the scale to a rate: $$ \lambda=\frac{1}{2}=0.5\text{ per thousand dollars}. $$ At $x=4$ thousand dollars, the dimensionless product is $\lambda x=0.5(4)=2$. Apply the Erlang survival formula with $n=2$: $$ P(X>4)=e^{-2}\left(1+\frac{2^1}{1!}\right) =e^{-2}(1+2) =3e^{-2} =3(0.1353353) =0.4060059. $$ Thus the probability, to four decimals, is $0.4060$. The probability of a loss at most $4$ thousand dollars is the complement, $1-0.4060059=0.5939941$, or $0.5940$ to four decimals.

The mean and variance are $$ E[X]=n\theta=2(2)=4\text{ thousand dollars}, \qquad \operatorname{Var}(X)=n\theta^2=2(2^2)=8\text{ thousand-dollars}^2. $$ Check. The two probabilities sum to $0.4060059+0.5939941=1$, and the mean is $4$ thousand dollars, the same threshold used in the question. The method avoids confusing the exceedance probability with the CDF.

Worked Example: A Three-Stage Exponential Total

In a hypothetical process, completion requires three independent stages. The duration of each stage is exponential with rate $0.5$ per day. Let $T$ be the total duration. Find $P(T\leq 2)$, then check the result using the survival probability.

Method. The stages are independent and have a common rate, so their sum is Erlang with shape $n=3$ and rate $\lambda=0.5$ per day. At $t=2$ days, $\lambda t=0.5(2)=1$. The CDF is $$ P(T\leq 2) =1-e^{-1}\left(1+\frac{1^1}{1!}+\frac{1^2}{2!}\right) =1-e^{-1}(1+1+0.5) =1-2.5(0.3678794) =0.0803015. $$ To four decimals, $P(T\leq 2)=0.0803$.

The survival probability is $P(T>2)=e^{-1}(2.5)=0.9196985$. Checking the complement gives $0.0803015+0.9196985=1.0000000$, as required. As a further parameter check, the mean duration is $n/\lambda=3/0.5=6$ days and the variance is $n/\lambda^2=3/(0.5)^2=12$ days squared. A common error is to use the one-stage exponential CDF $1-e^{-\lambda t}$; that formula would ignore the two additional stages.

Common Parameterization and Shape Traps

When reading a problem, first identify whether the stated second gamma parameter is a scale or a rate. A numerical value alone does not identify its role. Use the accompanying terminology, units, and any supplied density or moment information. Then write down the reciprocal conversion before substituting into a formula.

For example, if a loss is measured in dollars and a problem gives rate $\lambda=0.002$ per dollar, the corresponding scale is $\theta=1/0.002=500$ dollars. The incorrect calculation $\theta=0.002$ dollars both fails the reciprocal relation and produces the wrong units. Substituting that mistaken scale into the mean formula would give $\alpha(0.002)$ dollars instead of the correct $\alpha(500)$ dollars.

A separate trap is to call every gamma distribution Erlang. The shape must be a positive integer. A gamma model with shape $2.5$, for example, still has a valid gamma density, but it is not Erlang and the finite sum with upper index $n-1$ does not apply. Conversely, a positive integer shape makes the Erlang formulas available whether or not the model is described as a sum of stages.

Key takeaway: Convert rate and scale by taking a reciprocal, while keeping the shape unchanged. Use Erlang formulas only when the gamma shape is a positive integer; in that case, the survival probability is an exponential factor times a finite sum.

These distinctions matter in severity calculations because a reciprocal error can distort both the scale of the modeled losses and the tail probability. Recognizing the Erlang case can simplify a probability calculation substantially, but the formula is reliable only when the shape, rate, units, and requested tail or cumulative event have all been identified correctly. This treatment is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, identify the parameterization before calculating. Probabilities are rounded to four decimals.

A hypothetical loss has gamma shape $4$ and scale $250$ dollars. State the equivalent rate, mean, and variance. (Answer: rate $0.004$ per dollar; mean $\$1{,}000$; variance $250{,}000$ dollars squared.)
A gamma severity has shape $3$ and rate $0.25$ per thousand dollars. Give its scale and mean, in thousands of dollars. (Answer: scale $4$ thousand dollars; mean $12$ thousand dollars.)
A duration is Erlang with shape $2$ and rate $0.4$ per hour. Find the probability it exceeds $5$ hours. (Answer: $e^{-2}(1+2)=0.4060$.)
A hypothetical claim amount, measured in thousands of dollars, is Erlang with shape $3$ and rate $1$ per thousand dollars. Find the probability it is at most $1$ thousand dollars. (Answer: $1-e^{-1}(1+1+1/2)=0.0803$.)
Three independent stages each have exponential duration with rate $0.5$ per day. Find the mean and variance of their total duration, and the probability the total exceeds $4$ days. (Answer: mean $6$ days; variance $12$ days squared; $P(T>4)=e^{-2}(1+2+2)=0.6767$.)

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