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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 Oct 15 '15 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 Oct 15 '15 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 ? $\endgroup$ – pikachuchameleon Oct 15 '15 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 $\omega_1,\omega_2,...,\omega_n$ ? $\endgroup$ – pikachuchameleon Oct 15 '15 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 Oct 15 '15 at 23:15

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