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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.

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    $\begingroup$ 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 $\endgroup$
    – Henry
    Nov 29, 2019 at 12:57
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    $\begingroup$ 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. $\endgroup$ Nov 29, 2019 at 13:30
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    $\begingroup$ 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 $X$. $\endgroup$
    – drhab
    Nov 29, 2019 at 13:42
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    $\begingroup$ 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 $X=\begin{bmatrix}X_1,X_2,\cdots,X_n\end{bmatrix}^\top.$ $\endgroup$ Nov 29, 2019 at 13:53
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    $\begingroup$ @AntoniParellada thank you for your helpful comments $\endgroup$ Nov 29, 2019 at 15:36

1 Answer 1

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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. $F_X$).

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  • $\begingroup$ 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? $\endgroup$ Nov 29, 2019 at 15:28
  • $\begingroup$ "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. $\endgroup$
    – user140541
    Nov 29, 2019 at 15:35
  • $\begingroup$ 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)? $\endgroup$
    – Veljko
    Feb 6, 2020 at 20:04
  • $\begingroup$ @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). $\endgroup$
    – user140541
    Feb 6, 2020 at 20:28
  • $\begingroup$ 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~ $\endgroup$
    – S.C.
    Mar 23 at 15:32

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