How to Use the Method of Moments for a Gamma Distribution

The method of moments estimates a distribution’s parameters by matching theoretical moments to moments calculated from the data. For a two-parameter gamma distribution, matching the first two moments gives estimates of the shape and scale. The main convention to watch is whether the gamma model uses a scale or a rate, and whether the sample variance uses denominator $n$ or $n-1$.

Define the gamma parameterization

Use the shape–scale form $X\sim\operatorname{Gamma}(\alpha,\theta)$, where $\alpha>0$ is the shape and $\theta>0$ is the scale. Its density is

$$f(x)=\frac{x^{\alpha-1}e^{-x/\theta}}{\Gamma(\alpha)\theta^\alpha},\qquad x>0.$$

The population mean and variance are

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

Some references instead use the shape–rate form, with rate $\lambda=1/\theta$. In that notation, the density contains $e^{-\lambda x}$, and the mean and variance are $\alpha/\lambda$ and $\alpha/\lambda^2$.

Derive the method-of-moments estimates

Suppose $x_1,\ldots,x_n$ are observations from the gamma distribution. Let the sample mean be

$$\bar{x}=\frac{1}{n}\sum_{i=1}^n x_i.$$

For the second moment, the method of moments uses the empirical raw moment $n^{-1}\sum_i x_i^2$. Equivalently, after subtracting the squared sample mean, it uses the variance with denominator $n$:

$$v_n=\frac{1}{n}\sum_{i=1}^n(x_i-\bar{x})^2 =\frac{1}{n}\sum_{i=1}^n x_i^2-\bar{x}^2.$$

Match these quantities to the gamma mean and variance:

$$\bar{x}=\alpha\theta, \qquad v_n=\alpha\theta^2.$$

Dividing the variance equation by the mean equation gives $\theta=v_n/\bar{x}$. Substituting this into the mean equation gives the estimates

$$\hat{\alpha}_{\mathrm{MOM}}=\frac{\bar{x}^{\,2}}{v_n}, \qquad \hat{\theta}_{\mathrm{MOM}}=\frac{v_n}{\bar{x}}.$$

For the shape–rate parameterization, take the reciprocal of the scale estimate:

$$\hat{\lambda}_{\mathrm{MOM}}=\frac{\bar{x}}{v_n}.$$

These formulas require a positive sample mean and positive sample variance. If all observations are identical, then $v_n=0$ and the formulas do not give a finite, positive gamma shape and scale.

Worked example

Suppose the observed data are $2,4,6$. The sample mean is

$$\bar{x}=\frac{2+4+6}{3}=4.$$

The variance using denominator $n=3$ is

$$v_n=\frac{(2-4)^2+(4-4)^2+(6-4)^2}{3} =\frac{8}{3}.$$

Therefore,

$$\hat{\alpha}_{\mathrm{MOM}}=\frac{4^2}{8/3}=6, \qquad \hat{\theta}_{\mathrm{MOM}}=\frac{8/3}{4}=\frac{2}{3}.$$

The fitted model is $\operatorname{Gamma}(6,2/3)$ in the shape–scale convention. In the shape–rate convention, it is $\operatorname{Gamma}(6,3/2)$, since the rate is the reciprocal of the scale. Checking the fitted moments confirms the result: the mean is $6(2/3)=4$, and the variance is $6(2/3)^2=8/3$.

Which sample variance should you use?

The strict method-of-moments calculation matches the first two empirical raw moments, so it uses $v_n$, with denominator $n$. Do not automatically substitute the usual unbiased sample variance

$$s^2=\frac{1}{n-1}\sum_{i=1}^n(x_i-\bar{x})^2.$$

Some applied discussions or software may use $s^2$ when matching a theoretical mean and variance. That convention gives $\bar{x}^{\,2}/s^2$ for shape and $s^2/\bar{x}$ for scale; it is a different finite-sample calculation. Since $s^2=\frac{n}{n-1}v_n$, its estimates differ from the raw-moment method. Check which convention a problem or implementation specifies.

Common mistakes

• Mixing up scale and rate: The rate is the reciprocal of the scale. A scale estimate of $2/3$ corresponds to a rate estimate of $3/2$, not $2/3$.
• Using the wrong variance denominator: For the usual raw-moment method, use denominator $n$. Using $n-1$ changes the estimates.
• Treating the gamma shape as the mean: The mean is the product $\alpha\theta$; the shape alone is not generally the mean.
• Ignoring the data assumptions: A gamma random variable is positive. If observations include zeros because of censoring, rounding, or a separate probability mass at zero, the ordinary two-parameter gamma model may not describe the data appropriately.