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The strong law of large numbers states that the sample average converges almost surely to the expected value $\overline{X}_n\ \xrightarrow{\text{a.s.}}\ \mu \qquad\mathrm{when}\ n \to \infty$ .

That is,$$\Pr\left( \lim_{n\to\infty}\overline{X}_n = \mu \right) = 1.$$

I want to ask whats the domain of the random variable $\overline{X}_n$, given that all $X_n$ have same domain $\Omega$?

For the coin tossing problem $\Omega =\{H,T\} $ and $X_n(H)=1 \quad and \quad X_n(T)=0 \quad\forall n $ so, if the domain of $\overline{X}_n$ was $\Omega$ the $\overline{X}_n(H)=1 \quad \overline{X}_n(T)=0 \quad \forall n$, so $\Pr\left({\omega \in \Omega : \overline{X}_n(\omega)=1/2}\right)=0$ , which is not what strong law says.

I want to know whats the domain of this random variable $\overline{X}_n$?

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  • $\begingroup$ Domain of $\overline X_n$ is $\Omega^n$. $\endgroup$ – A.S. Nov 11 '15 at 18:34
  • $\begingroup$ For the coin tossing problem, neither the domain $\Omega=\{H,T\}$ nor any domain $\Omega=\{H,T\}^n$ is suitable. To define a whole sequence $(X_n)$, hence every $\bar X_n$ necessary to even state that $\bar X_n\to\mu$, one option is to consider $\Omega=\{H,T\}^\mathbb N$ and to define each $X_n$ by $X_n(\omega)=\omega_n$ where $\omega=(\omega_n)_{n\in\mathbb N}$ (these are often called the canonical sample space and the canonical process on it). In any case, the domain of every $X_n$ and every $\bar X_n$ should be the same, and should be large enough to "receive" ... $\endgroup$ – Did Nov 11 '15 at 19:36
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    $\begingroup$ ... an i.i.d. sequence of Bernoulli random variables. Another choice is $\Omega=[0,1]$ with some suitable sigma-algebras and random variables $X_n$ (this one being called the standard space). $\endgroup$ – Did Nov 11 '15 at 19:40
  • $\begingroup$ So math.stackexchange.com/users/6179/did you mean is kind of lifted to the space $\Omega^\mathbb{N}$? $\endgroup$ – Mahbub Alam Nov 11 '15 at 19:46
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    $\begingroup$ *with some suitable sigma-algebra (singular). $\endgroup$ – Did Nov 11 '15 at 19:51
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Even if we could define the $\overline{X}_n: \Omega^n\to \{0,1\}$, we would end up with many random variables defined in different probability spaces. In that situation we would not be able to make any sense of the notion "probability that the sequence of $\overline{X}_n$ converges", because we wouldn't have a fixed probability measure to consider all of the outcomes of the $\{\overline X_k\}_{k=1}^\infty$ at once to assign a probability.

What we need is a probability measure defined on sequences of outcomes. More precisely, the sample space is the set of sequences with values in $\Omega = \{H,T\}$, that is, $\Omega^{\mathbb N}$.

In this setting, if $\omega = (\omega_1,\omega_2,\dots)\in \Omega^{\mathbb N}$, then

$$X_k(\omega) = \cases{1 & \text{if } \omega_k = H\\ 0 & \text{if } \omega_k = T}$$ and $$\overline{X}_n(\omega) = \frac{X_1(\omega)+\dots+X_n(\omega)}{n}.$$

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    $\begingroup$ Reaching a correct answer is much preferable to deleting a first version, of course. +1. $\endgroup$ – Did Nov 11 '15 at 20:11
  • $\begingroup$ @Did Agreed. Thank you! $\endgroup$ – dafinguzman Nov 11 '15 at 20:23

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