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Let $\Omega$ be an uncountable set and $(\Omega, \mathcal{F},P)$ be a probability space built on $\Omega$.

Let $S \subset \{A \in \mathcal{F}: P(A)=0,\;|A|=1\}:|S|<\infty$ be a finite subset of the class of singleton P-null sets in $\mathcal{F}$.

I am trying to use $S$ to construct a new probability space from $(\Omega, \mathcal{F},P)$: $(S,\sigma(S),P')$

Since $\sigma(S)$ is discrete, I can define a potential probability measure for $P'$ on the above space by specifying its values for each sample point in $S$: $P'(s_i)=\lim\limits_{n\rightarrow \infty} \frac{P(Q_{in})}{ P \left(\bigcup\limits_{j \in \{1...|S|\}}Q_{jn}\right) }$, where $Q_{in} \subset \mathcal{F}:Q_{ij}\supset Q_{ik}\;\forall( j\leq k)$ and $\lim \inf Q_{in} = s_i\in S$

Note that $P'$ can be extended to $\sigma(S)$ via additivity.

Question 1: Under what conditions does $P'$ exist?

Question 2: What additional conditions are needed for the above to be true if $|S| = \infty$?


My thinking so far

The use of the limit in $P'$ requires a metric space $(\Omega,d)$, such that $P(B_r(s) :=\{\omega \in \Omega: d(s,\omega)<r \})>0 \;\forall(s\in S, r>0)$, which allows the above limit to define a probability measure on $\sigma(S)$. However, I'm not sure if this condition is sufficient, necessary or both?

I initially developed the above notions by working with the simple probability space $([0,1],\mathcal{B}([0,1]),P(A\subset \mathcal{B}([0,1]) = \lambda(A))$. If $|S|=M$, then we have a set of $M$ distinct points in $[0,1]$. We can define $Q_{in} := [(s_i - \frac{s_i}{n}),(s_i+\frac{1-s_i}{n})] \;\forall i:s_i \in S$. This results in a possible candidate for $P'$:

$P'(s_i) = \lim \limits_{n\rightarrow \infty} \frac{\lambda([(s_i - \frac{s_i}{n}),(s_i+\frac{1-s_i}{n})])}{ \lambda \left(\bigcup\limits_{j \in \{1...|S|\}}[(s_i - \frac{s_j}{n}),(s_j+\frac{1-s_j}{n})]\right) }$.

The numerator is easy to calculate for all $n$; however, the $Q_{in}$ are not initially disjoint so the denominator is complicated at the beginning of the sequence. However, since the elements of $S$ are countable and distinct, $\exists n_0: Q_{in}\cap Q_{jn} = \emptyset \; \forall (i\neq j, n>n_0)$. Therefore, we can restrict analysis of the above limit to $n>n_0$ without loss of generality. The benefit of doing so is that the union of the $Q_{in}$ becomes a disjoint union, and we can get a simple formula for the denominator. Specifically,

$P'(s_i) = \lim \limits_{n\rightarrow \infty} \frac{\lambda([(s_i - \frac{s_i}{n_0+n}),(s_i+\frac{1-s_i}{n_0+n})])}{ \lambda \left(\biguplus\limits_{j \in \{1...|S|\}} [(s_i - \frac{s_j}{n_0+n}),(s_j+\frac{1-s_j}{n_0+n})]\right) } = \frac{1/(n_0+n)}{M/(n_0+n)} = \frac{1}{M}\; \forall s_i \in S$

Thus, in this very simple example, $P'$ converges at $n_0<\infty$.

It seems straightforward to extend this to a countable set on the domain of a non-uniform distribution function F (e.g., gaussian). The denominator is guaranteed to be $\leq 1$ since F is a distribution; therefore, the denominator will always be a subset of the domain of F and the numerator, being a subset of the union in the denominator, will always be $\leq$ the denominator. Therefore, it seems like this is quite general, since all random variables map to the real numbers (hence we have a metric space).

I am not sure if I am missing some, possibly pathological case, where you cannot define $P'$ from $P$

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  • $\begingroup$ Clearly, you need to relate somehow the sets $Q_{in}$ for different $i$ between them, otherwise the limit could be whatever. As you suggest, a metric would be a useful tool to relate different $Q_{in}$. Or, as it vaguely appears in the example on $[0,1]$, simply an other measure (there, it's the Lebesgue). So I think it could be useful to think in terms of the Radon Nikodym theorem: start from 2 measure, and get a function (which you can see as a measure after normalization). $\endgroup$ – Pietro Majer Jun 13 '14 at 17:02
  • $\begingroup$ @PietroMajer Thanks for the input. In my simple examples, the underlying balls in $\mathbb{R}$ were simply intervals in the sequence of Q's, with the lebesgue measure serving the role of the probability measure (since I was using the unit interval for simplicity). Are you saying that as long as the probability measure possesses a radon-nikodym derivative, then the limit will exist (which seems to make sense). $\endgroup$ – user42192 Jun 13 '14 at 17:12
  • $\begingroup$ Yes: for the Lebesgue measure, the ratio $m(B(x,r))/\lambda (B(x,r))$ converges a.e. to $dm/d\lambda$ by the Lebesgue Differentiation Theorem (and as $L^1$-limit for even simpler reasons). For measures in metric spaces the analog is also true under mild assumptions. Check any textbook on GMT, e.g. Leon Simon's "Lectures on Geometric Measure Theory". $\endgroup$ – Pietro Majer Jun 13 '14 at 17:48
  • $\begingroup$ @PietroMajer Thanks, I'll take a look at GMT to see what those conditions are...haven't gone beyond basic measure theory and metric spaces myself. $\endgroup$ – user42192 Jun 13 '14 at 18:04
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The condition you have is not sufficient. Although the question has some wording issues, I can answer it as I think you intended it.

Suppose $(\mathbb{R},\mathcal{B},P)$ is our probability triple, where $P([0,\frac{1}{2^n}])=4^{-\lceil\frac{n}{2}\rceil}$. In this case, there can be no limit defining $P'(\{0\})$ if the sets you are using in your limit are balls size 2^{-n}$.

A simpler counterexample (that you may be more or less happy with):

$P$ has probability density function $f(x)=\left\{\begin{array}\,e^{-x}&x>0\\0&x<0\end{array}\right.$. In this case, depending on whether the sets tending down to zero are to the left or the right, you will get different answers for $P'(\{0\})$

Note: You can always define some $P'$ based on some specific sequence $(Q_{in})_{i=1,n=1}^{i=|S|,n=\infty}$ using your definition; it just isn't unique or useful. The way to define it in general is to first choose any sequence $Q_{in}$ with $P(Q_{in})>0$ for all $i$ and $n$, and use Bolzano-Weierstrauss to find a convergent subsequence of the sequence of $|S|$-tuplets $P(Q_{in})$, and name this sequence (Q^*_{in}). Then define $Q^{*,1}_{in}=Q^*_{in}$ and define P'_1:{{s_1}}\rightarrow [0,1] by $P'_1({s_1})=1$.

Inductively define $P'_k:\{\{s_i\}|i\le k\}\rightarrow [0,1]$ by choosing an $s_i$ with $P'_{k-1}(s_i)\ne 0$ and a subsequence $Q^{*,k}_{in}$ of $Q^{*,k-1}_{in}$ s.t. $\frac{P(Q^{*,k}_{in})}{P(Q^{*,k}_{kn})}$ has a limit in $[0,\infty]$. Then we can define $P'_k(s_i)$ just by keeping everything in the appropriate ratios. $P'_{|S|}$ will be a probability distribution defined by your limit from the sequence of sets $(Q^{*,|S|}_{in})_{i=1,n=1}^{i=|S|,n=\infty}$

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  • $\begingroup$ Thanks for the help...and pathological case! It appears that P must be continuous for this to work out, as both your examples employ discontinuities. For my future reference, what about my post was confusing. Is there something that I did not mention? I tried to show my thinking in the second part, but I explicitly stated the two main issues in the front part of my post. $\endgroup$ – user42192 Jun 13 '14 at 17:07
  • $\begingroup$ If $P$ has a continuous probability density function this is indeed well-defined. Note that in certain cases $P$ can blow up to infinity at some of the $s_i$ and we may still have a sensible way to define $P'$; it can't have discontinuities that have multiple limits. The confusing part was that you didn't state the conditions on $Q_{in}$ relating to the metric space in a clear and explicit manner, and you asked the question before even discussing the metric space. You also asked about existence but not uniqueness. which I think you were probably more interested in. $\endgroup$ – Zorgoth Jun 13 '14 at 17:43
  • $\begingroup$ Thanks again...I think I have what I was looking for..I'll limit it to P's with Radon-Nikoydm derivatives and that possess the "mild conditions" as hinted by Pierto...time to look up some Geometric Measure Theory :-) Thanks to both of you. $\endgroup$ – user42192 Jun 13 '14 at 18:03

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