#StackBounty: #bayesian #bootstrap Understanding Bayesian Bootstrap theory

Bounty: 50

I’m trying to understand the theory in section 4 of Rubin (1981) paper on Bayesian Bootstrap (BB):

$textbf{Theory:}$ Let $d=left(d_{1}, ldots, d_{K}right)$ be the vector of all possible distinct values of $X$, and let $pi=left(pi_{1}, cdots, pi_{K}right)$ be the associated vector of probabilities
Pleft(X=d_{k} mid piright)=pi_{k}, quad sum pi_{k}=1

Let $x_{1}, ldots, x_{n}$ be an i.i.d. sample from the equation above and let $n_{k}$ be the number of $x_{i}$ equal to $d_{k}$. If the prior distribution of $pi$ is proportional to
prod_{k=1}^{K}pi_{k}^{l_k}quad left(0right. text { if } left.sumpi_{k} neq 1right)

then the posterior distribution of $pi$ is the $K-1$ variate Dirichlet distribution $Dleft(n_{1}+l_{1}+1,right.$ $left.ldots, n_{K}+l_{K}+1right)$ which is proportional to
quad prod_{k=1}^{K} pi_{k}^{left(n_{k}+l_{k}right)} quadleft(0right. text{ if } x_{imath} neq d_{k} text{for some } i, k text{ or if} left.sum pi_{k} neq 1right)

  • What does $K-1$ variate mean?

This posterior distribution can be simulated using $m-1$ independent uniform random numbers, where $m=n+K+sum_{1}^{K} l_{k}$.

  • Where does this come from?

Let $u_{1}, cdots, u_{m-1}$ be i.i.d. $U(0,1),$ and let $g_{1}, cdots, g_{m}$ be the $m$ gaps generated by the ordered $u_{imath}$. Partition the $g_{1}, cdots, g_{m}$ into $K$ collections, the $k$-th having $n_{k}+l_{k}+1$ elements,

  • Is element referring to $u$‘s or gaps? I think gaps because $sum_1^K(n_{k}+l_{k}+1)=m$. If so, is partitioning mean to group adjacent gaps together? Something like the bottom line below for $m=7$ and $K=3$?

enter image description here

and let $P_{k}$ be the sum of the $g_{i}$ in the $k$-th collection, $k=1, cdots, K$.

  • Does this mean $P_{k}$ is the size of collection $k$? Does "sum of the $g_{i}$" mean sum of the length of $g_{i}$‘s?

Then $left(P_{1}, ldots, P_{K}right)$ follows the $K-1$ variate $Dleft(n_{1}+l_{1}+1, ldots, n_{K}+l_{K}+1right)$ distribution. Consequently, the BB which assigns one gap to each $x_{i}$

  • But we have $m$ gaps vs. $n$ $x_i$‘s. How does this work?

is simulating

  • What does simulating mean in this context?

the posterior distribution of $pi$ and thus of a parameter $phi=Phi(pi, d)$ under the improper prior distribution proportional to $prod_{k=1}^{K} pi_{k}^{-1}$.

  • Where did the $l_k=-1$ come from?

Simulations corresponding to other prior distributions with integer $l_{k}$ can also be performed; for example, with a uniform prior distribution on $pi$, (i.e., all $l_{k}=0$ ) generate $n+K-1$ uniform random variables, form $n+K$ gaps, add the first $left(n_{1}+1right)$ gaps together to yield the simulated value of $pi_{1}$, add the second $left(n_{2}+1right)$ gaps together to yield the simulated value of $pi_{2}$, and so on. However, when using a proper prior distribution, all a priori possible values of $X$ must be specified because they have positive posterior probability.

  • What does "all a priori possible values of $X$ must be specified" mean and how is this different from the previous case of improper prior with $l_k=-1$?

Get this bounty!!!

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