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Mathematical description of a random sample: which one is it and why?

  1. $X_1(\omega), X_2(\omega), ..., X_n(\omega)$, where $X_1, ..., X_n$ are different but i.i.d. random variables.

  2. $X(\omega_1), X(\omega_2), ..., X(\omega_n)$, where $X$ is a (single) random variable.

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    Option 1. $ $ $ $2011-06-25
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    What about the "why" part?2011-06-25
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    Because the whole purpose of probability theory is to avoid actually manipulating little-omegas. I know that in some first courses one wastes an incredible amount of time specifying what is Omega... This is nonsense, all the quantities one needs are in fact "at the other end of the arrow". The best option is to establish once and for all that suitable spaces Omega exist that are sufficient for the families of random variables one has in mind, and then to proceed happily. .../...2011-06-25
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    .../... Leo Breiman must have explained this much better, somewhere. (OK, the structure of Omega may become an interesting question again, much later--but not before one begins to tackle questions of a much higher level of sophistication.)2011-06-25
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    @Leo: A choice other than $X_1, X_2, \dots, Xn$ makes it difficult to talk about the *distribution* of various statistics.2011-06-25
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    Isn't 1 and 2 equivalent?2011-06-25
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    @Leo: $X+X$ is not the same random variable as $X_1+X_2$. They have in general very different distributions.2011-06-25
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    For given $X_1, X_2$ and $\omega$ I can choose $X$, $\omega_1$ and $\omega_2$ in such a way that $X(\omega_1) = X_1(\omega)$ and $X(\omega_2) = X_1(\omega)$.2011-06-25
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    Edit to the previous comment: $X(\omega_2) = X_2(\omega)$.2011-06-25
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    @Leo, what is there in the comments above which you do not understand and/or which makes these comments not fully answer your question? Since you now offer a bounty, I guess this is so but I cannot figure why...2011-07-30
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    @Didier, consider this example: I throw a die 10 times to get a sample of 10 integer numbers. Why should I consider this as 1 and not 2? If I throw 10 dice simultaneously to get 10 numbers, then I'm OK to consider this as 1.2011-07-30
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    What would be the image set of X if you model this experiment as 2.?2011-07-30
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    $X: \{\omega_1, \omega_2, ..., \omega_6 \} \to \{1, 2, ..., 6\}$2011-07-30
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    And then I have $X(t_1), X(t_2), ..., X(t_n)$ where $t_i \in \{\omega_1, ..., \omega_6\}$2011-07-30
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    Then, for $X(t_i)$ to be random one needs to pick $t_i$ randomly in the set $\{\omega_1,\ldots,\omega_6\}$ (otherwise $X(t_i)$ is a number and not a random variable). Question: how do you choose $t_i$? The answer seems to be that you choose $t_i$ **at random**, in other words $t_i$ is a map from an unspecified probability space $S$ to $\{\omega_1,\ldots,\omega_6\}$... in other words, you simply encoded each random variable $X_i$ of option 1. as a random variable $X\circ t_i$ defined on $S$. In the end, option 2. does not exist. (Unrelated: please use the @ sign to notify your comments.)2011-07-31

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