The following question is motivated by material from pages 358-364 of Blitzstein and Hwang's Introduction to Probability. (NB: I have modified the book's notation in various places to make my question clearer.)

First, suppose that $\{N_t\}_{t \in \mathbb{R}^+}$ is a family of r.v.s indexed by the set $\mathbb{R}^+$ of positive reals, and that for each $t \in \mathbb{R}^+$, we have $N_t \sim \mathrm{Poisson}(\lambda t)$, where $\lambda$ is an unknown parameter, independent of $t$.

Or, to put it in Bayesian terms, we have

$$ N_t\mid\lambda \sim \mathrm{Poisson}(\lambda t) $$

Now, take1 $\mathrm{Gamma}(n_0, t_0)$ as the (conjugate) prior distribution for $\lambda$. I.e.

$$ \lambda \sim \mathrm{Gamma}(n_0, t_0) $$

Starting from these assumptions, the authors derive (p. 362-364) the posterior distribution for $\lambda$ as

$$ \lambda \mid (N_t = n) \sim \mathrm{Gamma}(n_0 + n, t_0 + t) $$

Furthermore, authors derive (p. 364) the posterior expectation for $\lambda$ as

$$ \operatorname{E}(\lambda \mid N_t = n) = \frac{n_0 + n}{t_0 + t} $$

These results suggest that the first parameter of a $\mathrm{Gamma}$ distribution is akin to a count (of occurrences), and the second one is akin to an length of time.

Under this interpretation, the prior distribution $\mathrm{Gamma}(n_0, t_0)$ would be obtained from the total number of occurrences ($n_0$) observed over one or more intervals of time adding up to a total of $t_0$ (time units). To get the posterior distribution, we update the parameters of the prior's $\mathrm{Gamma}$ by adding a number of occurrences ($n$) observed during a new interval of time of length $t$, so that now we have a total of $n_0 + n$ occurrences observed in a cumulative interval of length $t_0 + t$.

On the other hand, on p. 358 the authors note that a $\mathrm{Gamma}(1, \nu)$ distribution is equivalent to an $\mathrm{Exponential}(\nu)$ distribution, defined as the distribution whose PDF is $f(t) = \nu e^{-\nu t}$. In this case, the second parameter of the $\mathrm{Gamma}$ distribution seems to be behaving like a rate (occurrences per unit time), rather than a length of time.

I'm puzzled by these two radically different interpretations of the $\mathrm{Gamma}$ distribution's second parameter (first as a length of time, and later as a rate).

Is my reasoning above wrong? If not, is there some way to unify or rationalize such divergent interpretations?

1 The book uses the convention that $\mathrm{Gamma}(a, b)$ is the distrubtion whose PDF is $f(x) = \frac{(b x)^a e^{-b x}}{x \Gamma(a)},\; x > 0$.

  • $\begingroup$ If we have a total of $n_0+n$ occurrences observed in a cumulative interval of length $t_0+t$, then $\frac{n_0 + n}{t_0 + t}$ looks like a "rate" to me. Am I missing something? $\endgroup$ – Just_to_Answer Aug 9 '17 at 1:30
  • $\begingroup$ @Just_to_Answer: Yes, but that expression refers to the posterior expectation of $\lambda$, a Poisson parameter. In this post I'm discussing the interpretation of the second parameter for the Gamma. $\endgroup$ – kjo Aug 9 '17 at 9:01

You are, more or less, looking at different coordinate systems.

The conjugate prior distribution is for a rate parameter, while the other is the distribution for a time random variable.

$$\bbox[2ex]{\underbrace{\quad \lambda}_{\text{a rate}} \sim \mathcal{Gamma}(n_0,\underbrace{t_0\qquad}_{\text{time interval}})} \\ \text{versus}\\\bbox[2ex]{\underbrace{\qquad\quad X}_{\text{a time interval}} \sim \mathcal{Gamma}(1 ,\underbrace{\nu\quad}_{\text{a rate}})}$$

That is all.


Two conventional ways of parametrizing the family of Gamma distributions are these: \begin{align} & \frac 1 {\Gamma(\alpha)} \left( \frac t \mu \right)^{\alpha-1} e^{-t/\mu} \, \frac{dt} \mu \qquad \text{for } t\ge0 \\[12pt] & \frac 1 {\Gamma(\alpha)} (\lambda t)^{\alpha-1} e^{-\lambda t} \, (\lambda\, dt) \qquad \text{for } t\ge 0 \end{align}

If $\lambda = \dfrac 1 \mu$ then these are both the same distribution, with expected value $\alpha\mu = \dfrac\alpha \lambda.$ If $t$ is measured in units of time, then so is $\mu,$ and $\lambda$ is then in units of frequency (i.e. $\lambda$ is a rate).

  • $\begingroup$ That's not the issue here. They're using the same distribution (shape-rate parameters); just for different coordinate systems. $\endgroup$ – Graham Kemp Aug 9 '17 at 3:54

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