Are elements of a sample i.i.d. realizations of the same random variable or realizations of different i.i.d. random variables?

If there is some sample $$X^n=(x_1,x_2,\dots,x_n)$$, do we consider the elements of this sample $$x_i$$ independent and identically-distributed realizations of the same random variable $$X$$ or are they all realizations of different independent and identically distributed random variables $$X_1,X_2,...,X_n$$ (observation $$x_1$$ is the realization of a random variable $$X_1$$, observation $$x_2$$ is the realization of a random variable $$X_2$$ etc.)?

I hope this makes sense.

• I do not see how this makes a difference to the distribution. But to the extent that $X_1$ is a random variable rather than an observation $x_1$ of a random variable, I suspect that the second description is more applicable Nov 29, 2019 at 12:57
• Just in case you don't get an answer that makes this concepts clear to you, I would suggest checking the difference between random variate (or realization) and random variable. Nov 29, 2019 at 13:30
• What exactly do you mean with "realization of random variable"? A random variable $X$ is not more than a function with specific properties. For every outcome $\omega$ in sample space $\Omega$ a value $X(\omega)$ "shows up". Is that the realization that you are talking about? If so then realize that there is only one such value if there is only one random variable $X$. Nov 29, 2019 at 13:42
• The key sentence is "The value of the random variable (that is, the function) $X$ at a point $\omega \in \Omega,$ i.e. $x=X(\omega )$ is called a realization of $X$." This would be for a single random variable. In your post $X^n=(x_1,x_2,...,x_n)$ is a realization of $n$ independent and identically distributed random variables, or the random vector $X=\begin{bmatrix}X_1,X_2,\cdots,X_n\end{bmatrix}^\top.$ Nov 29, 2019 at 13:53

You may think about $$(x_1,\ldots,x_n)$$ as a realization of $$n$$ independent copies of $$X$$. Basically, there is a probability space $$(\Omega,\mathcal{F},\mathsf{P})$$ in the background so that $$(x_1,\ldots,x_n)=(X_1(\omega)\ldots,X_n(\omega))$$ for some $$\omega\in\Omega$$, which is chosen randomly according to $$\mathsf{P}$$. Then the statement "independent and identically-distributed realizations of the same random variable" doesn't make sense. Although, sometimes $$(x_1,\ldots,x_n)$$ is referred to as a random sample from a particular distribution (e.g. $$F_X$$).

• thank you for your answer, very helpful. So if those copies of $X$ are separate entities for each $x_i$ in $(x_1,...,x_n)$, can you comment on why "sometimes $(x_1,...,x_n)$ is referred to as a random sample from a particular distribution"? I mean it makes sense: you have a distribution and you drag values from it. But on the other hand, a particular distribution describes probabilities of values for a particular random variable. Do those "copies of $X$" shrink to one entity when people talk about samples from a distribution? Nov 29, 2019 at 15:28
• "Drawing a random sample from a distribution" means obtaining a realization of i.i.d. random variables having that distribution. I think that the origin of that phrase may be related to the inverse transform sampling.
– user140541
Nov 29, 2019 at 15:35
• I got one question: Imagine that we have $n$ uncorrelated (not independent) copies of $X$. Now imagine we have one realization $(x_1,...,x_n)$ of these $n$ copies. Can we use this realization $(x_1,...,x_n)$ to estimate the mean of $X$? In other words, can we use samples from uncorrelated identically distributed random variables to estimate the mean (for example, using the sample mean)? Feb 6, 2020 at 20:04
• @Veljko Sure. The smaple average of pairwise uncorrelated r.v.s. converges to the mean of these r.v.s. (simple application of Chebyshev's inequality).
– user140541
Feb 6, 2020 at 20:28
• Not sure if you are still active, but I wondered if you could comment on the difference between your claim here and the claim made by the accepted answer on cross validated: math.stackexchange.com/questions/3455773/…. Thank you~
– S.C.
Mar 23 at 15:32