Solved – a sample of a random variable

Question

Random variable $X$ is defined as a measurable function from one $\sigma$-algebra $(\Omega_1, \mathcal F_1)$ with the underlying measure $P$ to another $\sigma$-algebra $(\Omega_2, \mathcal F_2)$.

How do we talk about a sample $X^n$ of this random variable? Do we treat it as an element from $\Omega_2$? Or as the same measurable function as $X$?

Where can I read more about this?

Example:

In Monte Carlo estimation, we prove the unbiasedness of the estimator by considering the samples $(X^n)_{n = 1}^N$ to be the functions. If an expectation of a random variable $X$ is defined as

\begin{align}
\mathbb E[X] = \int_{\Omega_1} X(\omega_1) \,\mathrm dP(\omega_1)
\end{align}

and assuming that $X^n$ are functions and $X^n = X$, we can proceed as follows:

\begin{align}
\mathbb E\left[\frac{1}{N} \sum_{n = 1}^N f(X^n)\right]
&= \frac{1}{N} \sum_{n = 1}^N \mathbb E[f(X^n)] \\
&= \frac{1}{N} \sum_{n = 1}^N \mathbb E[f(X)] \\
&= \mathbb E[f(X)].
\end{align}

If $X^n$ was just an element from $\Omega_2$, we couldn't have written the last set of equations.

Answer 1

A sample $(X^1,\ldots,X^N)$ is a measurable function from $\Omega_1$ to $\Omega_2^N$. A realisation of this sample is the value taken by the function at $\omega\in\Omega_1$, $(x^1,\ldots,x^N)=(X^1(\omega),\ldots,X^N(\omega))$.

When stating

assuming that $X^n$ are functions and $X^n=X$

The functions $X^n$ are all different functions, which means that the images $X^1(\omega),\ldots,X^N(\omega)$ may be different for a given $\omega$. When the sample is iid (independent and identically distributed), the functions $X^n$ are different with two further properties

1. identical distribution, meaning that $\mathbb{P}(X^1\in A)=\cdots=\mathbb{P}(X^N\in A)$ for all measurable sets $A$ in $\mathcal{F}_2$;
2. independence, meaning that $\mathbb{P}(X^1\in A^1,\ldots,X^N\in A^N)=\mathbb{P}(X^1\in A^1)\cdots\mathbb{P}(X^N\in A^N)$ for all measurable sets $A^1,\ldots,A^N$ in $\mathcal{F}_2$

Your definition

\begin{align} \mathbb E[X] = \int_{\Omega_1} X(\omega_1) \,\mathrm d\omega_1 \end{align}

is incorrect: it should be

\begin{align} \mathbb E[X] = \int_{\Omega_1} X(\omega_1) \,\mathrm dP(\omega_1) \end{align}