# What is the probability space of a random variable according to central limit theorem?

According to Central Limit Theorem:

$$\Omega$$ = {$$\omega_1$$, $$\omega_2$$,..., $$\omega_n$$}, $$Event$$, $$p$$, $$X$$ : $$\Omega$$ $$\rightarrow$$ $${\rm I\!R}$$

$$n$$ elements $$e_1$$,...,$$e_n$$ are drawn independently from $$\Omega$$.

Therefore, mean of X is:

$$\mu_s$$ = $$\frac{X(e_1) +,...,+X(e_n)}{n}$$

Then $$\mu_s$$ is a random variable whose probability distribution approaches with increasing $$n$$ the normal distribution with mean $$\mu$$ and standard deviation $$\frac{\sigma}{√n}$$

Although the theorem states that $$\mu_s$$ is a random variable, a random variable is a function defined in the context of a probability space.

What is that probability space here?

How can I specify

$$\Omega$$', $$Event$$', $$p$$' : $$Event$$' $$\rightarrow$$ [0, 1]

and define $$\mu_s$$ : $$\Omega$$' → $${\rm I\!R}$$ as a function in terms of the given $$\Omega$$, $$Event$$, $$p$$, and $$X$$?

• A close read of the theorem you are quoting indicates the $\mu_s$ do not have to be random variables on a common probability space, because the conclusion only concerns convergence in distribution. Thus, you need only discover what the probability space of one particular $\mu_s$ might be. I know this is answered in another thread, but it might be hard to find. Nevertheless, many related threads with informative answers can be found by searching for sample probability space. – whuber Jun 17 '19 at 12:23
• Inductive application of the construction for $n=2$ will answer your question: see stats.stackexchange.com/questions/364739. – whuber Jun 17 '19 at 12:24
• Also posted at math.stackexchange.com/questions/3265203/…. – StubbornAtom Jun 17 '19 at 13:49

The issue here isn't really about the central limit theorem at all, but about defining a random variable that is a function of multiple independent realisations of some underlying random variable. (I have taken the liberty of removing the central limit theorem tag from your post.) In your question, you have defined a probability space for a single random variable $$X: \Omega \rightarrow \mathbb{R}$$, but you then want to use a vector of independent realisations of this random variable. This would require you to define a large probability space with outcomes corresponding to this new random vector.

Defining a probability space for a vector of IID values: Suppose we want our probability space to be able to accommodate a data vector with $$n \in \mathbb{N}$$ independent data points. Given the initial probability space $$(\Omega, \mathscr{G}, \mathbb{P})$$, we can define the new sample space and sigma field:

\begin{aligned} \Omega_* &\equiv \Omega \times \cdots \times \Omega. \\[6pt] \mathscr{G}_* &\equiv \{ \mathcal{E}_1 \times \cdots \times \mathcal{E}_n | \mathcal{E}_i \in \mathscr{G} \} \\[6pt] \end{aligned}

We then define the corresponding probability measure $$\mathbb{P}_*$$ by:

$$\mathbb{P}_*(\mathcal{E}_*) = \prod_{i=1}^n \mathbb{P}(\mathcal{E}_i) \quad \quad \quad \text{for all } \mathcal{E}_* = \mathcal{E}_1 \times \cdots \times \mathcal{E}_n \in \mathscr{G}_*.$$

We now have a probability space $$(\Omega_*, \mathscr{G}_*, \mathbb{P}_*)$$ that can accommodate an $$n$$-tuple of (independent) events in the original probability space. Now that we have this, we define the random vector $$\mathbf{X} = (X_1,...,X_n)$$ by:

$$\mathbf{X}(\omega_*) = (X(\omega_1),...,X(\omega_n)) \quad \quad \quad \text{for all } \omega_* = (\omega_1,...,\omega_n) \in \Omega_*.$$

This now gives us a random vector $$\mathbf{X}: \Omega_* \rightarrow \mathbb{R}^n$$ in the larger probability space. Assuming that $$X$$ was measureable in the original probability space, it can be shown that $$\mathbf{X}$$ is a measureable function in this new probability space, and is thus a valid random variable.

This is sufficient to handle $$n$$ data points. Usually we would go further than this, and define our probability space to be able to accommodate this data vector for all possible values $$n \in \mathbb{N}$$ (rather than a pre-specified fixed value). Thus, we would usually extend the above definition to take $$n=\infty$$, which would allow us to define a sequence of independent data values within a stipulated probability space. (This extension is not necessary for the present question, but it is important to mention it.)

Defining the sample mean: Once we have defined the data vector properly, it is then trivial to define the sample mean as a valid random variable in the new probability space. (I will depart from the notation in your question and use the standard notation $$\bar{X}$$ to denote the sample mean.) We define $$\bar{X}: \Omega_* \rightarrow \mathbb{R}$$ by:

$$\bar{X}(\omega_*) \equiv \frac{1}{n} \sum_{i=1}^n X(\omega_i) \quad \quad \quad \text{for all } \omega_* = (\omega_1,...,\omega_n) \in \Omega_*.$$

As you can see, the sample mean is a random variable defined in the probability space $$(\Omega_*, \mathscr{G}_*, \mathbb{P}_*)$$, which is the same probability space as the data vector. Again, so long as the underlying random variables are measureable, this function is also measureable, and thus a well-defined random variable.

You can now look at the probability distribution of the sample mean, and you can apply the central limit theorem if this is applicable. The central limit theorem does not come into any of the above setup, and it is something you apply after you have well-defined random variables.

This is what I have so far... But still a bit lost.

Considering a coin toss experiment:

$$\Omega$$ = {$$H$$,$$T$$}, $$H$$$$T$$, $$P(\Omega) = 1$$

$$p$$:$$Event$$ $$\rightarrow$$ [0,1], $$p(H)$$ = $$\alpha$$

$$H: n\alpha$$

$$T: n(1-\alpha)$$

The elementary results for $$n=2$$ are:

$$\Omega' = \{h,t\}$$

$$p' : Event' \rightarrow [0,1]$$

Since the probability space is $$(\Omega, Event, p)$$...

$$X(H) = h$$

$$X(T) = t$$

...are both random variables. Therefore:

$$p'(h) = P(H)$$

$$p'(t) = P(T)$$

$$E(X) = \mu$$

$$\mu_s = \frac{X(e_1) + X(e_2)}{2}$$