Solving Gamma Rate MLEs With Known Shape

Why the Gamma Rate MLE Matters

Gamma models describe positive quantities that can accumulate over time or arise from several stages of a process. In a hypothetical service study, for example, a recorded completion time might be modeled as gamma distributed. If the shape is fixed by the model and the rate is unknown, maximum likelihood estimation reduces the observations to a simple statistic: their total, or equivalently their sample mean.

The key is to be precise about the parameterization. This tutorial uses the rate parameter, not the scale. The rate has reciprocal units to the observations, and the resulting estimate must be reported in those units. The derivation is aligned with the SOA VEE Mathematical Statistics syllabus.

Model: Let $X_1,\ldots,X_n$ be independent observations from a gamma distribution with known shape $\alpha>0$ and unknown rate $\lambda>0$. The density of each observation is $$ f_X(x;\lambda)=\frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x},\qquad x>0, $$ and is zero for $x\leq 0$. Here $\Gamma(\alpha)$ is the gamma function. Under this rate parameterization, $E[X_i]=\alpha/\lambda$ and $\operatorname{Var}(X_i)=\alpha/\lambda^2$.

The assumptions matter. The observations must be independent, share the same rate, and follow the stated gamma model; the shape $\alpha$ is fixed and known rather than estimated. The support $x>0$ does not depend on $\lambda$, so for observed positive values the support contributes no extra parameter constraint. A recorded value of zero is not a valid observation under this continuous model.

Some references use a scale parameter $\theta=1/\lambda$, giving mean $\alpha\theta$. That is a different parameterization. The formula derived here estimates $\lambda$, whose units are the inverse of the measurement units.

Constructing the Likelihood

For observed values $x_1,\ldots,x_n>0$, independence means that their joint density is the product of the individual densities. Treating the data as fixed and the rate as variable gives the likelihood.

L(\lambda)=\prod_{i=1}^{n} f_X(x_i;\lambda) =\frac{\lambda^{n\alpha}}{\Gamma(\alpha)^n} \left(\prod_{i=1}^{n}x_i^{\alpha-1}\right) \exp\!\left(-\lambda\sum_{i=1}^{n}x_i\right),\qquad \lambda>0.

The observations appear through $\prod x_i^{\alpha-1}$, which does not involve $\lambda$, and through their sum. When maximizing with respect to $\lambda$, the first factor is constant. Taking logarithms turns the powers and product into a sum and does not change which positive value of $\lambda$ maximizes the likelihood.

\ell(\lambda)=\log L(\lambda) =n\alpha\log\lambda-n\log\Gamma(\alpha) +(\alpha-1)\sum_{i=1}^{n}\log x_i -\lambda\sum_{i=1}^{n}x_i.

Deriving and Checking the Rate MLE

Differentiate the log-likelihood with respect to the unknown rate, set the score equal to zero, and solve. The known shape and the observed sum are constants in this differentiation.

Differentiate.
The score is $U(\lambda)=\frac{d\ell}{d\lambda}=\frac{n\alpha}{\lambda}-\sum_{i=1}^{n}x_i$.

Set the score to zero.
Solving $\frac{n\alpha}{\lambda}=\sum_{i=1}^{n}x_i$ gives $\widehat{\lambda}=\frac{n\alpha}{\sum_{i=1}^{n}x_i}$.

Check the maximum.
The second derivative is $\ell''(\lambda)=-\frac{n\alpha}{\lambda^2}<0$ for every $\lambda>0$. Thus the log-likelihood is strictly concave and the stationary point is its unique maximum.

Theorem: For independent gamma observations with common known shape $\alpha>0$ and unknown rate $\lambda>0$, the maximum likelihood estimator is $$ \widehat{\lambda}_{\mathrm{MLE}} =\frac{n\alpha}{\sum_{i=1}^{n}X_i} =\frac{\alpha}{\overline{X}}, $$ where $\overline{X}=\frac{1}{n}\sum_{i=1}^{n}X_i$. For observed data, substitute $x_i$ for $X_i$.

The calculation establishes both the stationary point and the global maximum. In particular, the score is positive when $\lambda<\widehat{\lambda}_{\mathrm{MLE}}$ and negative when $\lambda>\widehat{\lambda}_{\mathrm{MLE}}$, since the score decreases strictly with $\lambda$. This confirms that the likelihood rises up to the estimate and falls after it.

The form $\alpha/\overline{X}$ also gives an interpretation. The fitted gamma mean is $\alpha/\widehat{\lambda}_{\mathrm{MLE}}=\overline{X}$: the estimated model mean matches the observed sample mean. A larger sample mean therefore produces a smaller estimated rate. This is a useful check, but it does not mean that the rate itself equals the reciprocal of the sample mean unless $\alpha=1$.

Worked Example: A Gamma Model for Completion Times

In a hypothetical process study, four independent completion times, in hours, are $1,2,3,$ and $4$. Suppose each time has a gamma distribution with known shape $\alpha=2$ and unknown rate $\lambda$ per hour. Estimate the rate by maximum likelihood and check that it maximizes the likelihood.

Method. Use the gamma rate likelihood and its score equation. The observed total is $1+2+3+4=10$ hours, with $n=4$ and $n\alpha=4(2)=8$. Therefore $$ \widehat{\lambda}_{\mathrm{MLE}}=\frac{8}{10}=0.8\text{ per hour}. $$

The fitted mean completion time is $\alpha/\widehat{\lambda}=2/0.8=2.5$ hours, which equals the sample mean $10/4=2.5$ hours. For an explicit likelihood check, $\Gamma(2)=1$ and $\prod x_i=24$, so $L(\lambda)=24\lambda^8e^{-10\lambda}$. At the estimate, $\ell(0.8)=\log 24+8\log(0.8)-8\approx-6.6071$. At the comparison value $\lambda=1$, $\ell(1)=\log24-10\approx-6.8219$, which is smaller. The second derivative is negative for every positive rate, so $0.8$ is the unique maximum, not merely a better value than this one comparison.

Common mistake avoided. Using $1/\overline{x}=1/2.5=0.4$ would estimate the scale when the shape is 2, not the rate. The rate MLE includes the known shape: $\alpha/\overline{x}=2/2.5=0.8$ per hour.

Worked Example: A Noninteger Known Shape

A hypothetical laboratory records three independent positive processing times, in minutes: $0.5,1,$ and $2.5$. The model has known shape $\alpha=1.5$ and unknown rate $\lambda$ per minute. Find the rate MLE and evaluate the log-likelihood there.

Method. First calculate the sample total and the numerator in the estimator: $\sum x_i=0.5+1+2.5=4$ minutes, $n=3$, and $n\alpha=3(1.5)=4.5$. Hence $$ \widehat{\lambda}_{\mathrm{MLE}}=\frac{4.5}{4}=1.125\text{ per minute}. $$ The fitted mean is $1.5/1.125=1.3333$ minutes, matching the sample mean $4/3=1.3333$ minutes.

For the log-likelihood, $\sum\log x_i=\log(0.5)+\log(1)+\log(2.5)=\log(1.25)$. Also, $\Gamma(1.5)=\sqrt{\pi}/2$. Substituting into the log-likelihood gives $$ \ell(1.125)=4.5\log(1.125)-3\log(\sqrt{\pi}/2) +0.5\log(1.25)-1.125(4)\approx -3.4982. $$ The score at this value is $4.5/1.125-4=4-4=0$, and $\ell''(1.125)=-4.5/(1.125)^2<0$. Thus the estimate is the unique maximum.

Common mistake avoided. The shape need not be an integer. Replacing $\Gamma(1.5)$ with a factorial would be incorrect; the gamma function is defined for positive noninteger shape values as well.

Worked Example: Converting the Measurement Time Unit

A hypothetical monitoring system records three independent durations as $6,12,$ and $18$ hours. Assume a gamma model with known shape $\alpha=3$ and unknown rate. Estimate the rate first per hour, then express the same model using durations measured in days.

Method. In hours, the total is $6+12+18=36$, so $\widehat{\lambda}_{\mathrm{hour}}=3(3)/36=9/36=0.25$ per hour. The fitted mean is $3/0.25=12$ hours, equal to the sample mean.

To use days, divide each observation by 24: the durations are $0.25,0.5,$ and $0.75$ days, with total $1.5$ days. The rate estimate in the day parameterization is $\widehat{\lambda}_{\mathrm{day}}=9/1.5=6$ per day. The two numerical rates differ because their units differ: $6$ per day is $6/24=0.25$ per hour. The fitted mean is $3/6=0.5$ day, or $12$ hours, so both parameterizations agree.

Common mistake avoided. Changing the data from hours to days changes the numerical rate in the opposite direction to the measurement unit: the rate is multiplied by 24 when converting from per hour to per day. Leaving the rate at $0.25$ per day would give a fitted mean of 12 days, contradicting the observed durations.

Interpretation and Common Traps

The estimator depends on the data through the total $\sum X_i$, or equivalently through $\overline{X}$. For a fixed known shape, samples with larger average observations yield smaller rate estimates. This behavior agrees with the gamma mean $\alpha/\lambda$. It also makes clear why stating the units is part of the answer, not an optional detail.

Omitting $n$. The numerator is $n\alpha$, not just $\alpha$, when the denominator is the total $\sum x_i$. The equivalent formula $\alpha/\overline{x}$ already accounts for sample size.
Confusing rate and scale. The scale estimate is the reciprocal of the rate estimate. Do not report $1/\overline{x}$ as the rate unless the known shape is 1.
Using a wrong parameterization. The density here contains $e^{-\lambda x}$ and $\lambda^\alpha$. A density written with $e^{-x/\theta}$ uses scale $\theta$, where $\theta=1/\lambda$.
Ignoring units. If observations are measured in hours, the rate is per hour. Convert the observations and rate together when changing the time unit.
Assuming the score equation alone is enough. Check that the solution lies in the parameter space $\lambda>0$ and verify it is a maximum. Here the negative second derivative proves strict concavity.

Key takeaway: For independent observations from a gamma distribution with known shape $\alpha$ and unknown rate $\lambda$, maximize the likelihood to obtain $\widehat{\lambda}_{\mathrm{MLE}}=n\alpha/\sum X_i=\alpha/\overline{X}$. Keep the rate parameterization and reciprocal units explicit, then check the estimate against the fitted mean.

Check Your Understanding

For each problem, assume independent gamma observations with the stated known shape and an unknown common rate. Verify the units in your calculation.

Five hypothetical durations, measured in days, are $0.5,1,1.5,2,$ and $5$. The known shape is $\alpha=2$. Find the rate MLE and the fitted mean. (Answer: $\widehat{\lambda}=1$ per day; fitted mean $=2$ days.)
Three hypothetical positive measurements, in units of a laboratory scale, are $1,2,$ and $3$, with known shape $\alpha=4$. Find the rate MLE. (Answer: $\widehat{\lambda}=2$ per measurement unit.)
Four hypothetical event durations, in hours, are $0.2,0.4,0.6,$ and $0.8$, with known shape $\alpha=1.5$. Find the rate MLE and fitted mean. (Answer: $\widehat{\lambda}=3$ per hour; fitted mean $=0.5$ hour.)
Three hypothetical service times are $10,20,$ and $30$ minutes, with known shape $\alpha=2$. Find the rate per minute and the equivalent rate per hour. (Answer: $0.1$ per minute, equivalent to $6$ per hour.)
Two hypothetical positive observations are $1$ and $4$, with known shape $\alpha=3$. Find the rate MLE and state whether the score is positive or negative at $\lambda=1$. (Answer: $\widehat{\lambda}=1.2$ per observation unit; the score at $\lambda=1$ is $6-5=1>0$.)

What You'll Learn