I am working on Question 2 of this exam from Duke: First Year Exam - 2017

Let $X\mid\theta\sim \mathcal{U}(0,\theta)$ be a uniform likelihood and $\theta\mid\lambda\sim\text{Exp}(\lambda)$ be a prior on the upper limit of the uniform distribution. My question is, What is the posterior distribution of $\theta\mid X$?

I know Bayes' Theorem says, \begin{align} p(\theta\mid X)&\propto p(X\mid\theta)p(\theta) \\ &= \left(\frac{1}{\theta}\right)\left(\lambda\cdot\exp\left(-\lambda \theta\right)\right). \end{align} So far, so good. But I am having trouble because now I might try to find the normalizing constant of this posterior distribution: \begin{align} C &= \int_0^\infty p\left(\theta\mid X\right) \, d\theta \\ &=\int_0^\infty \left(\frac{1}{\theta}\right)\left(\lambda\cdot\exp\left(-\lambda \theta\right)\right) \, d\theta. \end{align} But according to WolframAlpha, this integral does not converge! How can one proceed to characterize this posterior distribution? Judging from the examination, you'd need to find the posterior mean to state what the Bayes estimator is under squared loss, but this expectation is proportional to the normalizing constant that doesn't seem to exist.

  • $\begingroup$ You have a bit more information you aren't including yet, if you know that $X \sim \mathcal{U}(0, \theta)$ and you observe $X$ what can you say about $P(\theta < X)$? $\endgroup$ – Nadiels May 7 '18 at 18:49
  • $\begingroup$ Ah! Should the bounds of integration in the normalizing constant calculation be from $X$ to $\infty$? $\endgroup$ – 5d41402abc4 May 7 '18 at 18:58
  • $\begingroup$ Why isn't the answer accepted? $\endgroup$ – Ramanujan Feb 17 at 17:07

Why is your integral going from $0$ to $\infty$?

$\theta$ cannot be less than $X.$ Work with the interval $\theta\in [X,\infty).$

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