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It is well-known, that any Borel probability distribution on $[0,1]$ can be obtained starting from the probability space $([0,1],\mathscr B([0,1]),\lambda)$ where $\lambda$ is the Lebesgue measure. I wonder, whether a similar result exists for the case of discrete distributions on $\Bbb N_0$.

Namely, let the sample space be $\Omega = \Bbb N_0$.

  1. Does there exist a probability measure $\mu$ on $(\Omega,2^\Omega)$ such that for any other probability measure $\nu$ on the latter space there exists a measurable function $f_{\mu\nu}$ such that $\nu = (f_{\mu\nu})_*\mu$?

  2. If there are several measures $\mu$ with the latter property, are there any sufficient conditions which assure the existence of $f_{\mu\nu}$? In particular, it is clearly necessary that $\mu(\omega)>0$ for infinitely many $\omega$. Is it also a sufficient condition?

Inspired by this problem.

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No. A measure defined on $2^\Omega$ is neccesarily purely atomic. Let $n$ be such that $\mu(\{n\})>0$. Let $\nu$ be a probability measure with the property that all atoms have measure smaller than $\mu(\{n\})$. But $f(n)$ will have distribution measure $\mu(\{n\})$.

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  • $\begingroup$ Hm, that was easy - you mean that we wouldn't be able to 'split' $\mu(\{n\})$ into smaller peaces that require representing $\nu$? For example, if we let $k$ be any integer such that $\frac{1}{k}\leq \frac12\mu(\{n\})$ and further let $\nu$ be the uniform distribution on $[1;k]$? $\endgroup$ – Ilya Jun 24 '13 at 9:51
  • $\begingroup$ @Ilya Exactly!${}$ $\endgroup$ – Michael Greinecker Jun 24 '13 at 9:58

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