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I consider a standard normal random variable $X$ and a Vitali set $V$. $P(X\in V)$ can not be computed as $V$ is not measurable.

Now I consider the outcome of the following experiment $E_S$ : $N_S$ is the number of experiments $X_i$ (all independent and equals to $X$, and $i\in\mathbb N$) such that $X_i\in S$.

  1. If $P(X\in S)=0$ then $P(N_S=0)=1$
  2. If $P(X\in S)>0$ then $P(N_S=\infty)=1$
  3. What happens for $S=V$ ? I think that $P(N_V=\infty)=0$ and $P(N_V=0)<1$. Am I right and can we obtain some more precise results ?

Thank you for your answers !

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    $P(X\in V)$ is not defined. Why do you think this number should be "almost" zero, whatever that means?2012-07-21
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    Because when you define $V$, you can force $V\subset [a,b]$, for any $a$ and $b$. Hence, $V$ can be as "small" as you want...2012-07-21
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    This confuses measure and cardinality: there is a bijection between $(0,\varepsilon)$ and $\mathbb R$ but the measure of $(0,\varepsilon)$ can be made as small as desired while the probability measure of $\mathbb R$ is $1$.2012-07-21
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    We agree, and if $A\subset B$ then $\mu(A)\le\mu(B)$, if defined, no ?2012-07-21
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    Thanks for the support, even if it is not an answer. I agree with you, but the probability point of view of this question is very practical : Can I hope $N_V$ to be $1$ or $3$ ? For sure I think this is linked to how is defined $V$, but I like the question that can be linked to some finite integer point of view :)2012-07-21
  • 0
    Perhaps you can explain with more details in an answer why this is the wrong random variable model ?2012-07-23
  • 0
    In English, "experience" and "experiment" are two different words. The one you want here is "experiment".2012-08-01
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    thanks for the english lesson :)2012-08-01

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It is a bit surprising to see that this post is still alive... but here we go. The question is:

What happens for $S=V$?

In a nutshell, and as was already explained in the comments, what happens is that nothing guarantees that the sets $[N_V=\infty]$ or $[N_V=0]$ are measurable, hence neither $P(N_V=\infty)$ nor $P(N_V=0)$ is defined. Thus, asking if these probabilities are $0$ or $\lt1$ or whatever has no sense.

Let us recall why (the function) $N_S$ is measurable when (the subset) $S$ is measurable. One writes $$ N_S=\sum_{i=1}^{+\infty}\mathbf 1_{A^S_i},\qquad A^S_i=[X_i\in S], $$ and, by the definition of the measurability of $X_i$ and $S$, each set $A^S_i$ is measurable hence each function $\mathbf 1_{A^S_i}$ is measurable and, by measurability of pointwise limits, the function $N_S$ is measurable.

When $V$ is not measurable, the reasoning above breaks down at the moment when one needs each $A^V_i$ to be measurable. For example, $$ [N_V=0]=\bigcap_{i=0}^{+\infty}(\Omega\setminus A_i^V)=\bigcap_{i=0}^{+\infty}[X_i\notin V], $$ and none of the subsets $[X_i\notin V]$ is measurable, a priori. If you find a way to prove that these subsets are in fact measurable, or only that their whole intersection is measurable (something which could happen without every $[X_i\notin V]$ being measurable), please go ahead. Otherwise...

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    This is exactly the point of this question, how do you show that the intersection is not measurable ? The usual proof for $V$ doesn't seem to work on that case...2012-08-01
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    **As I said, this is impossible**, since for example the sigma-algebra on $\Omega$ could be $2^\Omega$, in which case every function defined on $\Omega$ would be measurable. But the point of the answer is that **YOU** have to prove that these subsets are measurable before writing things like $P(N_V=0)$.2012-08-01
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    I'm sorry, but I don't ask question if I know the answer, and I can (and it happens all the time) make mistakes. So if your answer is that **I** have to do it myself, I don't think it's a very good answer. Or is there some misunderstanding ? I wrote $P(N_V=0)$ because that was what I was talking about, and if you can give me a proof why it is or it is not measurable, I'll be happy :)2012-08-01
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    Please read: I never wrote that you ought not to make mistakes (this would be stupid, we all do mistakes) nor that $[N_V=0]$ is not measurable. On the contrary, I repeated ad nauseam that it is **impossible** to prove that $[N_V=0]$ is not measurable but that one ought to prove that (some still unmentioned aspects of the setting make that) $[N_V=0]$ is in fact measurable **before** writing things like $P(N_V=0)$. I understand this is bad news for you because you want nevertheless to manipulate the undefined number $P(N_V=0)$... Well, go ahead--but do not think you are still doing mathematics.2012-08-01
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    Ok, I got your point, but could you please give me some details about "the unmentioned aspects of the setting" ? What do I need to make it possible ? Perhaps just a link to some explanations ?2012-08-01