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I've been thinking about the concept of random sample for hours, and I don't really understand it.

Imagine I toss a fair coin three times, and I want to study the number of heads. I define $\Omega=\{HHH,HHC,HCH,CHH,HCC,CHC,CCH,CCC\}$ and $X:\Omega\rightarrow\mathbb{R}$ as $X(\omega)=\text{number of heads in $\omega$}$. The distribution for $X$ is a $\text{binomial}(3,0.5)$.

The definition of random sample is: we say that $\{X_1,\ldots,X_n\}$ is a random sample if the random variables $X_1,\ldots,X_n$ are independent and identically distributed. In this case, imagine that $X_1,\ldots,X_n\sim\text{binomial}(3,0.5)$.

My doubt is: we have $X_1,\ldots,X_n:\Omega\rightarrow\mathbb{R}^n$, and each $X_i$ represents the number of heads. Then, don't we have $X_1(\omega)=X_2(\omega)=\ldots=X_n(\omega)\,$?

For example, in the law of large numbers, why don't we write $$\frac{X(\omega_1)+\ldots+X(\omega_n)}{n}\stackrel{n}{\longrightarrow} E[X]\,?$$

What I don't really understand is why $X_1(\omega),\ldots,X_n(\omega)$ represent different observations of the same phenomenon, and not $X(\omega_1),\ldots,X(\omega_n)$.

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    Think of it as $X:\Omega^n\to\mathbb R^n$ with $X_i=p_i\circ X$ where $p_i:\mathbb R^n\to\mathbb R$ denotes the projection. Space $\Omega^n$ is equipped with product probability measure $P^n$ where $P$ is the uniform probability on $\Omega$. That will result in iid binomially distributed $X_i$. – drhab Sep 11 '16 at 12:11
  • @drhab Then, for $\omega=(\omega_1,\ldots,\omega_n)\in\Omega^n$, one has $X_i(\omega)=\text{number of heads in $\omega_i$}$, right? If I define $Y:\Omega\rightarrow\mathbb{R}$ as $Y(\eta)=\text{number of heads in $\eta$}$, $\eta\in\Omega$, then $X_i(\omega)=Y(\omega_i)$, right? –  Sep 11 '16 at 15:42
  • Yes, that is correct. If e.g. $\omega_i=HCH$ then $X_i(\omega)=2$. I see now that you edited your comment. This is an answer to the original. – drhab Sep 11 '16 at 15:46
  • Also the second part of your comment is correct. If e.g. $\omega=(CHC, HCH,CCH,CHC,\dots)$ then $Y(\omega_2)=Y(HCH)=2$ and $X_2(\omega)=p_2\circ X(\omega)=p_2(1,2,1,1,\dots)=2$. Actually like this we have $X=(X_1,\cdots,X_n)$ where the $X_i$ are iid random variables equipped with binomial distribution, as you describe in your question. – drhab Sep 11 '16 at 16:01
  • So in order to construct that situation you should start with $\Omega={HHH,\dots, CCC}^n$ instead of $\Omega={HHH,\dots, CCC}$. – drhab Sep 11 '16 at 16:08
  • I asked approximately the same question here and got several glib answers and at least one that was quite satisfactory. Roughly @drhab's comment here is the answer, but perhaps a little too condensed to be useful to someone who doesn't already understand it. :( – John Hughes Sep 11 '16 at 12:32

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