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?


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.

  • $\begingroup$ in your example, $X^n$ would all have the same distribution as the $X$ you described, hence their expecation is same as that of $X$. $\endgroup$
    – bdeonovic
    Oct 8, 2016 at 15:36

2 Answers 2


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}


Sample can be drawn from population, not from random variable. "Sample of $n$ random variables" is a simplified way of saying that we have a sample drawn from the population, that we assume to be $n$ identically distributed random variables. So such sample behaves like $n$ random variables. It's ambiguous because it mixes terminology used in probability and statistics. The same with simulation, where samples are drawn from common distribution. In both cases sample is the data you have. Samples are considered as random variables because random processes lead to drawing them. They are identically distributed since they come from common distribution. For dealing with samples we have statistics, while statistics use abstract, mathematical description of the it's problems in terms of probability theory, so the terminology is mixed. Random variables are functions assigning probabilities to events that can be encountered in your samples.

  • $\begingroup$ What about in the Monte Carlo simulation context. There, the samples are not from a population. They are from random number generators. $\endgroup$
    – sk1ll3r
    Oct 8, 2016 at 14:37
  • $\begingroup$ @sk1ll3r still it's sample, drawn from a common distribution. $\endgroup$
    – Tim
    Oct 8, 2016 at 14:48
  • $\begingroup$ So would I treat it as an element from $\Omega_2$ or a function from $\Omega_1$ to $\Omega_2$? $\endgroup$
    – sk1ll3r
    Oct 8, 2016 at 15:00
  • $\begingroup$ @sk1ll3r as bdeonovic said, it just an ordinary random variable, nothing more then this. $\endgroup$
    – Tim
    Oct 8, 2016 at 20:01
  • 1
    $\begingroup$ @LewisKelsey. It confuses me and I don't know if the concepts are mixed, but for example in the definition " We say that $X_1, X_2,...,X_n$ is a random sample of size n from a population if the $X_i$ are i.i.d. and their common distribution is that of the population." And then, a sample mean is the sum of $X_1, X_2,...,X_n$ divided by n. Why to get the sample mean are suming the $X_i$s?, are they not supposed to be functions? or are the realization or outcome of the $X_i$ random variable (and why the same name?) $\endgroup$
    – jessica
    Nov 6, 2020 at 2:31

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