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We say that $X_1,X_2,...X_n$ are i.i.d random variables if they have identical distribution and are mutually independent, given probability space $(\Omega, \mathcal{F},\mathbb{P})$.

My doubt is that if we say we have a realization $(x_1,x_2,...,x_n)=(X_1(\omega),X_2(\omega),...,X_n(\omega))$, since $X_1,...,X_n$ are identical as random variables, shouldn't all values of realization be equal since their input $\omega$ is the same, i.e $x_1=x_2=...=x_n$ ?

What exactly does identical mean ? Does it mean that they all are identical measurable functions from $(\Omega, \mathcal{F},\mathbb{P})$ to Borel $\sigma$-algebra ?

Or just that they have same distribution but they are different when seen as measurable functions ?

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    $\begingroup$ Yes, to assert that $(X_1,\ldots,X_n)$ are i.i.d. is to assume that 1. they are all defined on a common probability space $(\Omega,\mathcal F,P)$, 2. they all have the same distribution, meaning that, for every $B$, $P(X_k\in B)$ does not depend on $k$, and 3. they are independent, meaning that for every $(B_k)$, $P(X_1\in B_1,\ldots,X_n\in B_n)=P(X_1\in B_1)\ldots P(X_n\in B_n)$. $\endgroup$
    – Did
    Commented Oct 15, 2015 at 16:54
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    $\begingroup$ "since $X_1,...,X_n$ are identical as random variables, shouldn't all values of realization be equal since their input ω is the same" This is where you go astray in your analysis: the $X_k$s are not identical as random variables, only their distributions coincide. Example: on $\Omega=[0,1]$ with the Borel sigma-algebra and the Lebesgue measure, $X$ and $Y$ defined by $X(ω)=ω$ and $Y(ω)=1-ω$ for every $ω$ in $\Omega$ are not equal (right?) but they have the same distribution (uniform on $[0,1]$). (But beware that these $X$ and $Y$ are not independent either...) $\endgroup$
    – Did
    Commented Oct 15, 2015 at 16:57
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    $\begingroup$ @Did : Then can we say that, $X_1(.),X_2(.),...$ are different when viewed as measurable functions in the view that $X_1(\omega)$ may not be equal to $X_2(\omega)$ for all $\omega \in \Omega$ but they just happen to have same probability distribution, right ? $\endgroup$ Commented Oct 15, 2015 at 20:16
  • $\begingroup$ @Did: Suppose we have a realization $(x_1,x_2,...,x_n)$, then should we write it as $X_1(\omega_1),...,X_n(\omega_n)$ for different $\omega_1,\omega_2,...,\omega_n$ ? $\endgroup$ Commented Oct 15, 2015 at 20:18
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    $\begingroup$ Once again: there is no $\omega_1$, ..., $\omega_n$. An observed sequence $(x_1,\ldots,x_n)$ is always $(X_1(\omega),\ldots,X_n(\omega))$ for some common argument $\omega$. $\endgroup$
    – Did
    Commented Oct 15, 2015 at 23:15

1 Answer 1

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I know that the comments by @Did are informative. But, I thought I would add a more mathematically rigorous answer here:


Preliminary Concepts

We know that a random variable is a function, $X: \Omega \rightarrow \Omega'$. For the element space $\Omega'$, we have a corresponding probability space that we associate in this context, namely the $(\Omega', \mathcal{F}', \mathbb{P}_X)$ space. Here $\mathcal{F}'$ is a choice of the statistician who is doing all the modeling. In most cases when $\Omega'$ is $\mathbb{R}$, we use $\mathcal{F}' = \mathcal{B}(\mathbb{R})$ - the Borel $\sigma$-algebra. On the other hand, the choice for $\mathbb{P}_X$ is relatively fixed. It's defined as follows \begin{align} \mathbb{P}_X &= \mathbb{P} \circ X^{-1} \\ where, \ \ \ \ \ X^{-1}(A') &= \{\omega: X(\omega) \in A'\}\ \ \ \ \forall A' \in \mathcal{F'} \end{align}

The $P_X$ measure is called the pushforward measure or the distribution or the law of $X$, in the literature.

Generated $\sigma$-algebra: The set $\sigma(X) = X^{-1}(\mathcal{F'}) = \{X^{-1}(A'): A' \in \mathcal{F}'\}$ is referred to as the $\sigma$-algebra generated by the random variable $X$.


Main Answer

A family $(X_i)_{i \in I}$ of random variables defined on the same probability space $(\Omega, \mathcal{F}, \mathbb{P})$ is called identically distributed if $\mathbb{P}_{X_i} = \mathbb{P}_{X_j}$ $\forall i,j \in I$

A pair of measurable sets $A,B\in\mathcal{F}$ is called independent if $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)$.
A pair of measurable families $\mathcal{A},\mathcal{B}\subset\mathcal{F}$ is called independent if this holds for every pair $A\in\mathcal{A}$, $B\in\mathcal{B}$.

A family $(X_i)_{i \in I}$ of random variables defined on the same probability space $(\Omega, \mathcal{F}, \mathbb{P})$ is called independent if the family $(\sigma(X_i)_{i \in I}$ of $\sigma$-algebras is independent.

A family of I.I.D random variables satisfies both of the above conditions.

As you can see they are identical not in their mappings from $\Omega$ to $\Omega'$ but in their pushforward measures. It's not necessary to have the same mapping to have the same pushforward measure - An example of this was aptly given by @Did in the comments below the question.

Hope this helps.

Edit: Reference - Probability Theory by Achim Klenke

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