# Can anyone help explain this basic example of posterior

I am having trouble understanding the authors reasoning here. It is from "The Bayesian Choice"

I am confused about why the posterior is initially written without depending on the data, and why we integrate the numerator.

It is,

Consider one observation $$x$$, from a normal $$N(\frac{\theta_{1}+\theta_{2}}{2},1)$$

Then (From the book, page 24).

Sorry for being confusing! The joint posterior distribution on $$(\xi_1,\xi_2)$$ is $$\pi(\xi_1,\xi_2|x)\propto \exp\{-(x-\xi_1)^2/2\}\pi_1(2\xi_1)\pi_2(2\xi_2)$$ Therefore the marginal posterior on $$\xi_2$$ is given by the marginal of the above, up to a constant, that is, $$\pi(\xi_2|x)\propto \int\exp\{-(x-\xi_1)^2/2\}\pi_1(2\xi_1)\pi_2(2\xi_2)\,\text{d}\xi_1$$ which does not depend on $$x$$. This is a case, albeit an artificial case, when the posterior and the prior are equal.

• Oh that is no problem! Thanks so much. I am very new to the topic so I am likely at a slightly lower level then your average reader. Oct 31, 2018 at 19:23

I am not providing another answer to the question, but instead another example of the same sort as presented in the question. The example I present came up in the course of my research (as part of a Gibbs sampler within a Dirichlet process mixture model); in that sense it is not "contrived."

There is one observation $$x$$ drawn from an order-statistic distribution. The probability density for the $$k$$-th order statistic from a sample of $$n$$ iid draws from the uniform distribution on the unit interval is given by $$$$p(x|k,n) = \textsf{Beta}(x|k,n-k+1) ,$$$$ where \begin{align} n &\in \{1, 2, \ldots \} \\ k &\in \{1, \ldots, n\} . \end{align} Let the prior for $$(k,n)$$ be given by $$p(k|n)\,p(n)$$, where $$p(k|n) = 1/n$$ for all $$k$$ and $$p(n)$$ is an arbitrary distribution.

The posterior distribution for $$(k,n)$$ is characterized by $$$$p(k,n|x) \propto p(x|k,n)\,p(k,n) = \frac{\textsf{Beta}(x|k,n-k+1)}{k}\,p(n) .$$$$ The marginal posterior for $$n$$ can be computed by integrating out (i.e., summing out) $$k$$: $$$$p(n|x) = \sum_{k=1}^n \frac{\textsf{Beta}(x|k,n-k+1)}{k}\,p(n) = p(n) .$$$$ We see that the posterior distribution for $$n$$ is unchanged from its prior distribution.

The result is delivered by the adding-up property of order statistics. For example, suppose $$n$$ draws are made from the uniform distribution and sorted in order. Then one of the sorted draws is chosen at random (i.e., with probability $$1/n$$). This random choice "undoes" the effect of sorting so the distribution of the draw chosen via this mechanism is simply the underlying distribution, which in this case is the uniform distribution.

For completeness, note that the posterior distribution for $$k$$ conditional on $$n$$ can be expressed in terms of the following density: $$$$p(k|x,n) = \textsf{Binomial}(k-1|n-1,x) .$$$$