A problem with the expectation of a Pareto

Question

My course notes (3rd-year module in Bayesian Statistics, unpublished) contain the following section.

Assume we have data on the number of people queuing at an ATM at a specific hour for several randomly chosen days, and that we are interested in the mean time between two consecutive arrivals to the queue at some arbitrary point within this time.

First, let $\textbf{x} = \{x_1,\dots,x_n\}$ represent the observed data and assume these are a random sample from $\text{Po}(x_i\;|\;\lambda)$. We will use the conjugate prior $\text{Ga}(\lambda\;|\;a,b)$ to represent our prior knowledge, thus $$\pi(\lambda\;|\;\textbf{x})\propto\left[\lambda^se^{-n\lambda}\right]\times\left[\lambda^{a-1}e^{-b\lambda}\right]=\text{Ga}(\lambda\;|\;a^\ast,b^\ast),$$ where $a^\ast=a+s$ and $b^\ast=b+n$, with $s=\sum_{i=1}^nx_i$ a sufficient statistic for $x_i$.

Using standard queuing theory, the time within two consecutive arrivals, $y$, is distributed $\text{Ex}(y\;|\;\lambda)$. Thus, the predictive distribution is $$f(y\;|\;\textbf{x})=\int_0^\infty\lambda e^{-y\lambda}\frac{(b^\ast)^{a^\ast}}{\Gamma(a^\ast)}\lambda^{a^\ast-1}e^{-\lambda b^\ast}\;\text{d}\lambda=\frac{a^\ast}{b^\ast}\left(1+\frac{y}{b^\ast}\right)^{-(a^\ast+1)}.$$ Finally, the expected predictive time between consecutive arrivals is $$\mathbb{E}[y\;|\;\textbf{x}]=\int_0^\infty y\frac{a^\ast}{b^\ast}\left(1+\frac{y}{b^\ast}\right)^{-(a^\ast+1)}\;\text{d}y=\frac{b^\ast}{a^\ast-1}.$$

But, by my reasoning, \begin{align} & \begin{aligned} f(y\;|\;\textbf{x}) & = \frac{a^\ast}{b^\ast}\left(1+\frac{y}{b^\ast}\right)^{-(a^\ast+1)} \\ & = \frac{a^\ast b^{\ast a^\ast}}{(y+b^\ast)^{a^\ast+1}} \\ & = \text{Pa}(y+b^\ast\;|\;a^\ast,b^\ast) \end{aligned} \\[1em] \therefore \hspace{1em} & \begin{aligned}[t] \mathbb{E}[y\;|\;\textbf{x}] & = \mathbb{E}[y+b^\ast\;|\;y\sim\text{Pa}(a^\ast,b^\ast)] \\ & = \mathbb{E}[y\;|\;y\sim\text{Pa}(a^\ast,b^\ast)] + b^\ast \\ & = \frac{a^\ast b^\ast}{a^\ast-1} + b^\ast \\ & = \frac{(2a^\ast-1)b^\ast}{a^\ast-1} \\ & \neq \frac{b^\ast}{a^\ast-1} \text{ (unless $a^\ast=1$)}. \end{aligned} \end{align}

What's gone wrong?

Answer 1

Your basic insight of recognising the Pareto distribution is a good way to look at this, but you are just making a small error in your reasoning about the density, which is messing things up. Since $f(y|\mathbf{x}) = \text{Pa}(y+b^* | a^*,b^*)$ you therefore have:$^\dagger$

$$y|\mathbf{x} \sim \text{Pa}(a^*,b^*) - b^*.$$

(Note that we subtract $b^*$ from the Pareto variable; we don't add it.) Consequently, you should be using a minus sign for the expectation computation, which gives:

$$\begin{align} \mathbb{E}[y|\textbf{x}] &= \mathbb{E}[r - b^* | r \sim \text{Pa}(a^*,b^*)] \\[12pt] &= \mathbb{E}[r | r \sim \text{Pa}(a^*,b^*)] - b^* \\[6pt] &= \frac{a^* b^*}{a^*-1} - b^* \\[6pt] &= \frac{a^* b^*}{a^*-1} - \frac{a^* b^* - b^*}{a^*-1} \\[6pt] &= \frac{b^*}{a^*-1}. \\[6pt] \end{align}$$

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$^\dagger$ This is a general rule of probability that applies when we shift the location of a random variable by a constant. In general, if we start with a random varable $A$ with density $f_A$ then the random variable $R \equiv A+k$ has distribution function:

$$F_R(r) = \mathbb{P}(R \leqslant r) = \mathbb{P}(A+k \leqslant r) = \mathbb{P}(A \leqslant r-k) = F_A(r-k),$$

which then gives it the density function:

$$f_R(r) = \frac{d}{dr} F_R(r) = \frac{d}{dr} F_A(r-k) = f_A(r-k).$$

As you can see, using the location shift $+k$ on the random variable translates to imposing the shift $-k$ inside the density argument. This is because the point $R=r$ corresponds to $A=r-k$.