When to use alternate parametrization of Gamma distribution?

Question

In Loss Models, 4th ed., by Klugman et al., the following parametrization is given for the Gamma distribution:

$$f(x) = \dfrac{(x/\theta)^{\alpha}e^{-x/\theta}}{x\Gamma(\alpha)}\text{.} $$

When $\alpha = 1$, we have the exponential distribution. The example I'm wondering about is this:

Example 5.4 Let $X\mid \Lambda$ have an exponential distribution with parameter $1/\Lambda$. Let $\Lambda$ have a gamma distribution. Determine the unconditional distribution of $X$.

We have (note that the parameter $\theta$ in the Gamma distribution has been replaced by its reciprocal) \begin{align} f_{X}(x) &= \int\limits_{0}^{\infty}f_{X\mid\Lambda}(x\mid\lambda)f_{\Lambda}(\lambda)\text{ d}\lambda = \int\limits_{0}^{\infty}\underbrace{\lambda e^{-\lambda x}}_{f_{X\mid\Lambda}(x\mid\lambda)}\underbrace{\dfrac{\theta^{\alpha}}{\Gamma(\alpha)}\lambda^{\alpha - 1}e^{-\theta \lambda}}_{f_{\Lambda}(\lambda)}\text{ d}\lambda\text{.}\end{align}

Notice the part that I bolded above. Can someone explain to me why I am supposed to use this alternative parametrization?

Answer 1

Because it makes the integral easier! Your integral then evaluates to:
$$\frac{\theta^\alpha}{\Gamma(\alpha)}\int_0^\infty \lambda^{\alpha}e^{-(\theta+x)\lambda} = \frac{\theta^\alpha}{\Gamma(\alpha)}\frac{\Gamma(\alpha+1)}{(\theta+x)^{\alpha+1}} = \alpha\frac{\theta^\alpha}{(\theta+x)^{\alpha+1}} $$

So the density function of the unconditional distribution of $X$ is given by:

$$f_X(x|\alpha,\theta)=\frac{\alpha\theta^\alpha}{(\theta+x)^{\alpha+1}},\qquad x>0$$

Answer 2

With the gamma and exponential distributions (of which the latter is a special case of the former with shape parameter equal to $1$), there are two commonly used parametrizations; either by shape and scale, or by shape and rate.

Parametrization by scale looks like the first formula you wrote: $$f_X(x) = \frac{(x/\theta)^\alpha e^{-x/\theta}}{x \Gamma(\alpha)}, \quad x > 0.$$ Parametrization by rate looks like this: $$f_X(x) = \frac{\lambda^\alpha x^{\alpha-1} e^{-\lambda x}}{\Gamma(\alpha)}, \quad x > 0.$$ That is, we merely have $\lambda \theta = 1$. The scale parametrization has the nice property that ${\rm E}[X] = \alpha\theta$; that is, the scale parameter is proportional to the expected value (and in the case of the exponential distribution, it is equal to the expected value). So it is natural to use this parametrization if we want to calculate a maximum likelihood estimator $\hat \theta$ for a known $\alpha$: it is simply $\hat \theta = \bar X/a$. It is also slightly more convenient in the context of calculating distributional properties of loss variables subject to policy limits or deductibles.

However, in Bayesian statistics, it is frequently easier to use the rate parametrization. For instance, the posterior rate parameter for a Poisson model with gamma conjugate prior will have an additive property: if $X \sim {\rm Poisson}(\lambda)$, and $\lambda \sim {\rm Gamma}(\alpha,\beta)$ with $\beta$ a rate parameter, then the posterior distribution is also gamma but with parameters $(\alpha^* = \alpha + n\bar x, \beta^* = \beta + n)$. This is along the same lines of the exercise whose solution you are asking about, because you are given a hierarchical model and using the law of total probability to compute the unconditional distribution, which is analogous to the Bayesian method of integrating a likelihood function to obtain a marginal distribution (which is the denominator or "normalizing constant" for obtaining the posterior).

A similar situation occurs with the normal distribution; we usually parametrize it by mean $\mu$ and variance $\sigma^2$ (or standard deviation $\sigma$). But in Bayesian statistics, the calculations are easier if we parametrize by mean and precision $\tau$, which is the inverse of variance: $\tau \sigma^2 = 1$. The posterior hyperparameters are more easily expressed, especially when we want to continually update the posterior precision as new observations come in.