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Let $\mathcal{N} = \{1, 2, \ldots\}$ be the set of positive integers and let $\mathcal{F}$ be the $\sigma$-field of all subsets of $\mathcal{N}$. Let $X$ be a random variable taking values in $(\mathcal{N}, \mathcal{F})$ and let its unknown probability measure be $P$.

Every probability measure $P$ on $\mathcal{N}$ is given by a infinite-dimensional vector $(p_1 = P(\{1\}), p_2 = P(\{2\}, \ldots))$, satisfying the following condition $$ \sum_{n=1}^\infty p_n = 1. $$ For example, we can set $p_n = (\frac{1}{2})^n$, for $n = 1, 2, \cdots$.

To define a random probability measure on $\mathcal{N}$, we need to make sure that $p_1, p_2, \ldots$ are themselves random, which means that we need to define a probability measure on $\{(p_1, p_2, \ldots): 0 \leq p_1 \leq 1, 0 \leq p_2 \leq 1, \sum {p_n} = 1\}$. One possible way is through a stick-breaking process, which depends on a set of Beta-distributed random variables. Is there a general way to construct the random measure on the positive integers?

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  • $\begingroup$ Which sort of description do you expect? $\endgroup$
    – Did
    Commented Oct 15, 2013 at 14:09
  • $\begingroup$ Something like a definition: what is such a random measure look like, or a motivation: why is this so. $\endgroup$
    – fishiwhj
    Commented Oct 15, 2013 at 14:25

1 Answer 1

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Let $\Delta$ denote the set of $(p_n)_{n\in\mathbb N}$ such that $p_n\geqslant0$ for every $n$ and $\sum\limits_np_n=1$. Every way of drawing at random an element of $\Delta$ corresponds to the following construction.

Let $\Delta_0=\{\varnothing\}$. For every $k\geqslant1$, let $\Delta_k$ denote the set of $(p_n)_{1\leqslant n\leqslant k}$ such that $p_n\geqslant0$ for every $1\leqslant n\leqslant k$ and $\sum\limits_{1\leqslant n\leqslant k}p_n\leqslant1$. Each measure on $\Delta$ is described uniquely by a collection of measures $\mu_p$ on $[0,1]$, indexed by $p$ in $\bigcup\limits_{k\geqslant0}\Delta_k$, as follows:

  • First, $p_1=q_1$ where $q_1$ is drawn according to $\mu_\varnothing$.
  • Then, for every $k\geqslant1$, conditionally on $(p_1,\ldots,p_k)$, $p_{k+1}=q_{k+1}(1-p_1-\cdots-p_k)$, where $q_{k+1}$ is picked in $[0,1]$ according to $\mu_{(p_1,\ldots,p_k)}$.

Dirichlet processes correspond to the case when every $\mu_p$ is the same beta distribution $(1,\alpha)$, for some positive $\alpha$. In the general case, $\mu_p$ may depend on $p$ as long as the infinite product $\prod\limits_k(1-q_k)$ diverges (that is, is zero) almost surely.

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  • $\begingroup$ Could you give a example to show how $u_p$ depends on $p$? $\endgroup$
    – fishiwhj
    Commented Oct 17, 2013 at 9:12
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    $\begingroup$ An extreme case is when every $\mu_p$ is a Dirac mass at $p_1$, then the lengths of the parts are $p_1(1-p_1)^k$ for every $k\geqslant0$, where $p_1$ is random distributed according to $\mu_\varnothing$. Or, for every $(p_1,\ldots,p_k)$, $\mu_{(p_1,\ldots,p_k)}$ could depend only on $q_k$ (which is measurable with respect to $(p_1,\ldots,p_k)$), making $(q_k)$ a Markov chain on $[0,1]$. Or... $\endgroup$
    – Did
    Commented Oct 17, 2013 at 9:54

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