Gamma Response Parameterizations

Why Two Gamma Parameterizations?

Gamma responses are useful when a quantity is continuous, strictly positive, and often right-skewed, as can happen with waiting times, costs, or durations. The same gamma distribution is commonly written using either a scale parameter or a rate parameter. The formulas look different, so distinguishing these conventions is essential when interpreting parameters or translating a formula from one source to another.

We use \(\alpha\) for shape, \(\theta\) for scale, and \(\beta\) for rate. Shape controls the form of the distribution; scale and rate describe the same dimensional aspect in reciprocal units. The parameterization changes, but the probability distribution does not.

Definitions and Conversion

For \(\alpha>0\), the gamma function is defined by the integral below. It supplies the normalizing constant in the gamma density.

The integral is finite for \(\alpha>0\). A random variable \(X\) has a gamma distribution with shape \(\alpha\) and scale \(\theta\) if its density has the following form.

In the shape-rate parameterization, the positive rate \(\beta\) appears in the exponential as \(-\beta x\).

The two notations use the same symbol \(\operatorname{Gamma}\); therefore, a formula should state whether its second parameter is scale or rate. Their relationship is reciprocal.

To verify the conversion, substitute \(\theta=1/\beta\) into the shape-scale density. Its normalizing factor becomes \(1/(\Gamma(\alpha)(1/\beta)^\alpha)=\beta^\alpha/\Gamma(\alpha)\), and its exponential term becomes \(e^{-x/(1/\beta)}=e^{-\beta x}\). The resulting density is precisely the shape-rate density. Since the densities agree at every \(x>0\), they define the same distribution. Both are zero for \(x\leq 0\), so their supports agree as well.

Moments and Skewness

The key calculations follow from the gamma integral. First, integration by parts gives the recurrence \(\Gamma(u+1)=u\Gamma(u)\) for \(u>0\). Indeed, take \(t^u\) as the part to differentiate and \(e^{-t}\,dt\) as the part to integrate. The boundary term \(-t^ue^{-t}\) is zero at both endpoints: \(t^u\) tends to zero as \(t\) decreases to zero, and exponential decay dominates the power as \(t\) tends to infinity. Therefore,

This recurrence lets us derive the raw moments rather than memorize separate formulas. For a nonnegative integer \(m\), integrate \(x^m\) against the shape-scale density and substitute \(t=x/\theta\). The integral is finite, and the substitution gives

Applying the recurrence repeatedly yields the first three raw moments. These moments determine the mean, variance, and third central moment.

For completeness, the derivation is as follows. The recurrence gives \(E[X]=\alpha\theta\), \(E[X^2]=\alpha(\alpha+1)\theta^2\), and \(E[X^3]=\alpha(\alpha+1)(\alpha+2)\theta^3\). Subtracting the squared mean from the second raw moment gives

Writing \(\mu=E[X]\), expand \((X-\mu)^3\) and take expectations. The third central moment is \(E[X^3]-3\mu E[X^2]+2\mu^3\), so

Skewness is the third central moment divided by the variance to the power \(3/2\). Here that gives \((2\alpha\theta^3)/(\alpha\theta^2)^{3/2}=2/\sqrt{\alpha}\), since \(\alpha,\theta>0\). This completes the derivation. \(\square\)

For the rate form, use \(\theta=1/\beta\). The mean and variance become \(\alpha/\beta\) and \(\alpha/\beta^2\), while skewness remains \(2/\sqrt{\alpha}\). The skewness does not depend on scale or rate: it depends only on shape.

Interpreting the Parameters Carefully

The shape parameter affects skewness: as \(\alpha\) increases, \(2/\sqrt{\alpha}\) decreases. Scale and rate affect the units and magnitude of \(X\), but not this standardized measure of asymmetry. For example, changing scale while holding shape fixed changes the mean and variance but leaves skewness unchanged.

A common mistake is to treat a larger rate as if it meant a larger scale. The formulas show the reverse: increasing \(\beta\) decreases the mean \(\alpha/\beta\), whereas increasing \(\theta\) increases the mean \(\alpha\theta\). A reliable check is to ask what happens to \(X\) when the parameter changes: a higher rate makes the exponential factor \(e^{-\beta x}\) decay faster, while a higher scale stretches the distribution toward larger values.

Another possible confusion concerns the density near zero. When \(0<\alpha<1\), the factor \(x^{\alpha-1}\) grows without bound as \(x\) approaches zero from above. This does not put probability at zero: the support remains \(x>0\), and a density value at a point is not a point probability. The distribution still assigns probability zero to the single point \(X=0\).