Gamma Shape-Rate and Shape-Scale Parameterizations

Why Two Gamma Parameterizations?

A gamma model describes a positive, continuous quantity using a shape parameter and one additional parameter. That second parameter may be stated as a rate or as a scale. The formulas look different, but the models are the same when the rate and scale are reciprocals. Confusing the two can produce a mean that is off by a reciprocal, so identify the convention and its units before calculating.

For example, a hypothetical waiting time might be described by a gamma model with shape 4 and rate 0.5 per hour. Its equivalent scale is 2 hours, not 0.5 hours. This tutorial sets out both conventions and shows how to move between them without changing the underlying distribution.

Definition: A random variable $X$ has a gamma distribution with shape $\alpha>0$ and rate $\lambda>0$ if its density is $$ f_X(x)=\frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x}, \qquad x>0. $$ The equivalent shape-scale form uses a scale $\theta>0$: $$ f_X(x)=\frac{1}{\Gamma(\alpha)\theta^\alpha}x^{\alpha-1}e^{-x/\theta}, \qquad x>0. $$ Here $\Gamma(\alpha)=\int_0^\infty t^{\alpha-1}e^{-t}\,dt$ is the gamma function. The two forms describe the same distribution when $\theta=1/\lambda$.

Rate and Scale: Read the Units

The shape $\alpha$ is dimensionless. If $X$ is measured in hours, the rate $\lambda$ is measured per hour, while the scale $\theta$ is measured in hours. Their reciprocal relationship is also a useful units check: a quantity measured in hours cannot be equal to a rate measured per hour, but its reciprocal can.

Parameter notation varies between books and software. A notation such as “Gamma($\alpha,\beta$)” is ambiguous unless the source states whether $\beta$ is a rate or a scale. In calculations, write the convention in words or label the parameters explicitly. The central conversion is

\theta=\frac{1}{\lambda} \qquad\text{and}\qquad \lambda=\frac{1}{\theta}.

A larger rate means a smaller scale and, for a fixed shape, a smaller mean. A larger scale means a larger mean. The conversion does not change the shape or the model; it only expresses the same positive distribution using the other parameter convention.

Quantity	Shape-rate form	Shape-scale form	Units when $X$ is in hours
Shape	$\alpha>0$	$\alpha>0$	Dimensionless
Second parameter	Rate $\lambda>0$	Scale $\theta>0$	Per hour for $\lambda$; hours for $\theta$
Conversion	$\lambda=1/\theta$	$\theta=1/\lambda$	Reciprocal units
Mean	$\alpha/\lambda$	$\alpha\theta$	Hours
Variance	$\alpha/\lambda^2$	$\alpha\theta^2$	Hours squared

Worked Example: Convert Parameters and Calculate Moments

A hypothetical processing time $X$, measured in hours, has a gamma distribution with shape $\alpha=4$ and rate $\lambda=0.5$ per hour. Convert to the scale convention, find the mean and variance, and calculate the density at $x=6$ hours. Use the shape-rate formula for the density, and check it using the scale form.

Method and parameter conversion. The stated convention is shape-rate. The scale is the reciprocal rate: $$ \theta=\frac{1}{0.5\text{ per hour}}=2\text{ hours}. $$

Mean and variance. Under the rate form, $E[X]=\alpha/\lambda$ and $\operatorname{Var}(X)=\alpha/\lambda^2$. Therefore, $$ E[X]=\frac{4}{0.5}=8\text{ hours}, \qquad \operatorname{Var}(X)=\frac{4}{(0.5)^2}=\frac{4}{0.25}=16\text{ hours}^2. $$

Density at 6 hours. Since $\Gamma(4)=3!=6$, $$ f_X(6)=\frac{(0.5)^4}{6}(6)^{4-1}e^{-(0.5)(6)} =\frac{0.0625(216)e^{-3}}{6} =2.25e^{-3} \approx 0.1120\text{ per hour}. $$

Check using the scale form. Substituting $\theta=2$ hours gives $$ f_X(6)=\frac{6^3e^{-6/2}}{6(2^4)} =\frac{216e^{-3}}{96} =2.25e^{-3} \approx 0.1120\text{ per hour}. $$ The two density calculations agree. The common mistake would be to use $0.5$ as a scale; that would change the model and would not reproduce these results.

Means and Variances Under Either Convention

For a gamma distribution with shape $\alpha$, rate $\lambda$, and scale $\theta=1/\lambda$, the first two moments are

E[X]=\frac{\alpha}{\lambda}=\alpha\theta, \qquad \operatorname{Var}(X)=\frac{\alpha}{\lambda^2}=\alpha\theta^2.

The rate and scale versions are not separate moment rules. They are the same formulas after substituting $\theta=1/\lambda$. For instance, $\alpha/\lambda=\alpha\theta$, and $\alpha/\lambda^2=\alpha\theta^2$. The units provide a quick check: a mean has the units of $X$, and a variance has the squared units of $X$.

Justification of the moment formulas: Starting with the rate-form density, for $r\geq 0$ the $r$th raw moment is obtained by integrating $x^r f_X(x)$. Use the substitution $t=\lambda x$, so $x=t/\lambda$ and $dx=dt/\lambda$: $$ E[X^r] =\frac{\lambda^\alpha}{\Gamma(\alpha)} \int_0^\infty x^{\alpha+r-1}e^{-\lambda x}\,dx =\frac{\Gamma(\alpha+r)}{\Gamma(\alpha)\lambda^r}. $$ Integration by parts in the defining gamma-function integral gives $\Gamma(a+1)=a\Gamma(a)$ for $a>0$. Thus $E[X]=\alpha/\lambda$ and $E[X^2]=\alpha(\alpha+1)/\lambda^2$. Subtracting the squared mean gives $$ \operatorname{Var}(X)=E[X^2]-(E[X])^2 =\frac{\alpha(\alpha+1)-\alpha^2}{\lambda^2} =\frac{\alpha}{\lambda^2}. $$ Replacing $1/\lambda$ by $\theta$ gives the scale-form moments. This calculation uses the gamma density as given; establishing its normalization is a separate question.

Worked Example: Start With a Scale and Find a Density

A hypothetical instrument records a positive delay $X$ in minutes. Suppose $X$ has shape $\alpha=3$ and scale $\theta=5$ minutes. Find the equivalent rate, mean, variance, and density at $x=10$ minutes.

Method and conversion. The given second parameter is a scale, so the rate is its reciprocal: $$ \lambda=\frac{1}{5\text{ minutes}}=0.2\text{ per minute}. $$

Moments. Using the scale formulas, $$ E[X]=\alpha\theta=3(5)=15\text{ minutes}, \qquad \operatorname{Var}(X)=\alpha\theta^2=3(5^2)=75\text{ minutes}^2. $$

Density. Since $\Gamma(3)=2!=2$, the scale density at 10 minutes is $$ f_X(10)=\frac{10^{3-1}e^{-10/5}}{2(5^3)} =\frac{100e^{-2}}{250} =0.4e^{-2} \approx 0.0541\text{ per minute}. $$

Check in rate form. The rate density gives $$ f_X(10)=\frac{(0.2)^3(10)^2e^{-(0.2)(10)}}{2} =\frac{0.008(100)e^{-2}}{2} =0.4e^{-2} \approx 0.0541\text{ per minute}. $$ The results agree. A common mistake is to multiply by 5 to obtain the rate; the rate is $1/5$ per minute, not 5 per minute.

Changing the Units of the Random Variable

A rate and a scale must change when the unit used to measure $X$ changes. Suppose $X$ is measured in days and $H=24X$ is the same quantity measured in hours. If $X$ has shape $\alpha$, rate $\lambda_{\text{day}}$ per day, and scale $\theta_{\text{day}}$ days, then

\lambda_{\text{hour}}=\frac{\lambda_{\text{day}}}{24}, \qquad \theta_{\text{hour}}=24\theta_{\text{day}}.

The rate is divided by 24 because one day contains 24 hours: an event rate of 0.25 per day is $0.25/24$ per hour. The scale is multiplied by 24 because a duration of one day is 24 hours. The shape does not change. The converted parameters also preserve the reciprocal relationship.

Worked Example: Convert a Waiting-Time Model From Days to Hours

In a hypothetical monitoring system, a waiting time $X$ has shape $\alpha=2$ and rate $\lambda_{\text{day}}=0.25$ per day. Convert the model to hours, calculate its mean and variance in both units, and find the density at 48 hours.

Method and parameter conversion. Convert the rate by dividing by 24: $$ \lambda_{\text{hour}}=\frac{0.25}{24} =\frac{1}{96}\text{ per hour}. $$ The scale is 4 days, since $1/0.25=4$. In hours it is $$ \theta_{\text{hour}}=24(4)=96\text{ hours}. $$ These converted parameters are reciprocal: $1/(1/96)=96$.

Mean and variance in days. $$ E[X]=\frac{2}{0.25}=8\text{ days}, \qquad \operatorname{Var}(X)=\frac{2}{(0.25)^2} =\frac{2}{0.0625}=32\text{ days}^2. $$

Mean and variance in hours. Using the converted rate, $$ E[H]=\frac{2}{1/96}=192\text{ hours}, \qquad \operatorname{Var}(H)=\frac{2}{(1/96)^2} =2(96^2)=18{,}432\text{ hours}^2. $$ The mean is $8(24)=192$ hours, and the variance is $32(24^2)=18{,}432$ hours squared, confirming the unit conversion.

Density at 48 hours. Since $\Gamma(2)=1$, $$ f_H(48)=\left(\frac{1}{96}\right)^2(48)e^{-48/96} =\frac{1}{192}e^{-1/2} \approx 0.0032\text{ per hour}. $$ The density has units per hour. The common mistake is to multiply the rate by 24; that would give 6 per hour, inconsistent with the original rate of 0.25 per day and with the converted scale of 96 hours.

Common Parameterization Traps

The gamma distribution is continuous, so its density at a point is not the probability of taking exactly that value. For example, $f_X(10)=0.0541$ per minute is a density, not a 5.41% probability that $X=10$. Probabilities require an interval or a cumulative probability calculation.

When a problem gives a density formula, inspect the exponential term to identify the convention. The term $e^{-\lambda x}$ signals a rate $\lambda$, while $e^{-x/\theta}$ signals a scale $\theta$. Then check the leading factor: it is $\lambda^\alpha/\Gamma(\alpha)$ in rate form and $1/(\Gamma(\alpha)\theta^\alpha)$ in scale form.

Identify the convention.
Look for an explicit label such as “rate” or “scale”; if the notation is unstated, use the density’s exponential term and coefficient to determine it.

Record units.
For $X$ measured in a time unit, the rate is per time unit and the scale is in that time unit.

Convert only if useful.
Use $\theta=1/\lambda$, preserve $\alpha$, and confirm that the converted units are reciprocal.

Check the answer’s dimensions.
The mean has the units of $X$; the variance has squared units; the density has inverse units.

These checks are especially useful in exam calculations because rate-versus-scale errors can leave the algebra internally consistent while changing the model. The gamma parameterization conventions and calculations covered here are aligned with the SOA VEE Mathematical Statistics syllabus.

Check Your Understanding

For each problem, label the convention before calculating and check that the units agree.

A positive duration $X$ in hours has shape 2 and rate 0.4 per hour. Find its equivalent scale, mean, and variance. (Answer: scale $2.5$ hours; mean $5$ hours; variance $12.5$ hours$^2$.)
A hypothetical equipment delay has shape 5 and scale 3 days. Convert to the rate form and find its mean and variance. (Answer: rate $1/3$ per day; mean $15$ days; variance $45$ days$^2$.)
A continuous time $X$ in minutes has shape 2 and rate 1.5 per minute. Calculate the density at 2 minutes, using $\Gamma(2)=1$. (Answer: $4.5e^{-3}\approx 0.2240$ per minute.)
A hypothetical lifetime in days has shape 4 and rate 0.08 per day. Convert the rate to per hour and calculate the mean and variance in hours. (Answer: rate $0.003333$ per hour; mean $1{,}200$ hours; variance $360{,}000$ hours$^2$.)
A positive measurement $X$ in seconds has shape $1.5$ and scale 4 seconds. Find the equivalent rate, mean, variance, and density at 4 seconds. Use $\Gamma(1.5)=\sqrt{\pi}/2$. (Answer: rate $0.25$ per second; mean $6$ seconds; variance $24$ seconds$^2$; density $e^{-1}/(2\sqrt{\pi})\approx 0.1038$ per second.)

Quantity	Shape-rate form	Shape-scale form	Units when \(X\) is in hours
Shape	\(\alpha>0\)	\(\alpha>0\)	Dimensionless
Second parameter	Rate \(\lambda>0\)	Scale \(\theta>0\)	Per hour for \(\lambda\); hours for \(\theta\)
Conversion	\(\lambda=1/\theta\)	\(\theta=1/\lambda\)	Reciprocal units
Mean	\(\alpha/\lambda\)	\(\alpha\theta\)	Hours
Variance	\(\alpha/\lambda^2\)	\(\alpha\theta^2\)	Hours squared

What You'll Learn