Question 10. Beta downtime
The proportion of downtime has density
f(x\mid\theta)=\theta(1-x)^{\theta-1}, \quad 0<x<1,\ \theta>0,
and \theta\sim\operatorname{Gamma}(\alpha,\beta) with density \beta^{\alpha}\theta^{\alpha-1}e^{-\beta\theta}/\Gamma(\alpha). Let x_1,\ldots,x_n be an iid sample.
(a) State the prior mean. [1 mark]
(b) Obtain the posterior distribution of \theta and identify it. [8 marks]
(c) State the posterior mean. [2 marks]
(d) Explain the posterior mean as n\to\infty; as \alpha,\beta\to 0; and as \alpha,\beta\to\infty with \alpha/\beta fixed. [9 marks]
(e) Complete the sentence: “The effect of the prior hyperparameters \alpha and \beta on the posterior mean of \theta is like…” [5 marks]
Definitions. The observations are downtime proportions, while \theta>0 is the unknown shape parameter. The prior \operatorname{Gamma}(\alpha,\beta) uses rate \beta, so its mean is \alpha/\beta and variance \alpha/\beta^2. Define the transformed exposure S=-\sum_i\log(1-x_i)>0; it is not the sum of the original downtime proportions.
(a)
\mathbb{E}(\theta)=\alpha/\beta.
(b) The likelihood is \theta^{n}\exp\bigl(\theta\sum_{i=1}^{n}\log(1-x_i)\bigr) times a factor free of \theta. Multiplying by the gamma prior gives
\theta^{\alpha+n-1} \exp\Bigl(-\theta\Bigl(\beta-\sum_{i=1}^{n}\log(1-x_i)\Bigr)\Bigr).
Since \log(1-x_i)<0, the rate is positive.
\boxed{\theta\mid{\boldsymbol{x}}\sim\operatorname{Gamma}\Bigl(\alpha+n,\;\beta-\sum_{i=1}^{n}\log(1-x_i)\Bigr).}
(c)
\mathbb{E}(\theta\mid{\boldsymbol{x}}) = \frac{\alpha+n}{\beta-\sum_{i=1}^{n}\log(1-x_i)}.
(d) Write S_n=-\sum_{i=1}^n\log(1-x_i)>0. The log-likelihood is n\log\theta-S_n\theta+\text{constant}, whose derivative gives \hat\theta_{\mathrm{MLE}}=n/S_n. The posterior mean is
\frac{\alpha+n}{\beta+S_n} =\frac{1+\alpha/n}{S_n/n+\beta/n}.
If observations come from the stated model at \theta_0>0, the transformed values -\log(1-X_i) are iid \operatorname{Exp}(\theta_0), so S_n/n\to1/\theta_0 and the posterior mean tends to \theta_0, as does the MLE. With the observed data fixed, \alpha,\beta\to0 gives n/S_n; this is a limit of proper priors, whose limiting kernel 1/\theta is improper. If \alpha,\beta\to\infty while \alpha/\beta\to m>0, the posterior mean tends to m, the limiting prior mean. Without a limit for \alpha/\beta, there is no unique limiting posterior mean.
(e) The effect is like adding \alpha to the observation count and \beta to the transformed total exposure S=-\sum\log(1-x_i), not to the raw downtime sum. Indeed,
\boxed{ \mathbb{E}(\theta\mid{\boldsymbol{x}}) =\frac{\beta}{\beta+S}\frac{\alpha}{\beta} +\frac{S}{\beta+S}\frac nS.}
The prior mean \alpha/\beta and the MLE n/S therefore receive nonnegative weights that sum to one.
The transformation explains the form of the update. For Z=-\log(1-X), we have x=1-e^{-z} and dx/dz=e^{-z}, so
f_Z(z\mid\theta) =\theta(e^{-z})^{\theta-1}e^{-z} =\theta e^{-\theta z},\qquad z>0.
Thus the problem is exponential-rate inference on the transformed scale: n counts observations and S is their total exposure. The prior adds information in exactly those two positions. In the strong-prior limit with \alpha/\beta=m fixed, the prior variance is m/\beta\to0, explaining why the posterior mean is forced towards m.