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I toss 2 dice to get their sum; now, this means I have a sample space from 2 to 12. Now, when I repeat this experiment 10 times, do I have 10 different random variables all of which are from that sample space?

The problem that I have is terminology. Some online sources suggest that in the above situation, I have 10 independent random variables, all of which are random variables from the 2–12 sample space. To me it seems that I have just one random variable, but used 10 times.

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  • $\begingroup$ You can say the latter, but somehow you must operationalize your words mathematically. What do you intend "used 10 times" to mean, as a mathematical model? There are various good ways to answer that--and one of them is to construct ten random variables on a common sample space. $\endgroup$
    – whuber
    Commented Feb 4, 2021 at 18:18
  • $\begingroup$ I'm kind of newbie to Statistics. I'm studying central limit theorem for sums. Some sources take this from distribution point of view and say "We take a sample of size 10"; the ones that look at this from random variable point of view, say "We take 10 random variables." So, I am assuming that these 2 approach are the same; and by 10 random variables, they meant the same as "a sample of size 10." $\endgroup$ Commented Feb 4, 2021 at 18:38
  • $\begingroup$ The problem with the "10 random variables" approach is that it's difficult (although not impossible) to define what you might mean by their sum (or average or whatever). When the random variables are all defined on the same space, there is no problem at all. $\endgroup$
    – whuber
    Commented Feb 4, 2021 at 18:46
  • $\begingroup$ Now, I get it. Thank you. $\endgroup$ Commented Feb 4, 2021 at 18:59

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This is a modelling decision, and both could be correct. A single random variable has a distribution, so one thinks of it as allowing for multiple realisations. Therefore you are right that one can say you have 10 realisations of the same random variable.

However, one may also be interested in computing the probability of joint events, for example rolling a number below 4 in the first 5 repetitions and then a number higher then 4 in all the following ones. This requires a model that treats the repetitions separately, and they would normally, in this situation, be modelled as independent.

So both can be correct, depending on your modelling aim.

Note that a model is something you set up in order to do analyses. It is not a "true" feature of the situation, and therefore you can set up your model, including what you treat as random variable, in different ways.

(Just to avoid misunderstandings, there can be models that are wrong for the situation at hand in the sense of being incompatible with the values you observe or even with the possibilities or impossibilities of what can happen in that situation, but that's not the issue here. Both the two candidates you propose look fine, and in fact cannot be distinguished by data, so this is a decision of the modeller.)

PS having seen a comment by the thread opener: The central limit theorem involves distribution statements about functions of vectors of random variables, so should normally be formalised in terms of random variables $X_1,\ldots,X_n,\ n\to\infty$ (often, although not in the most general sense, independently and identically distributed), rather than in terms of repetitions of the same random variable, although one could say, "let $X$ be a random variable distributed according to $P$, and let $X_1,\ldots,X_n$ be independent replicates of $X$", which should mean the same thing.

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  • $\begingroup$ Very insightful; thank you. Just one other question; why do they have to be considered independent random variables (rather than realizations of the same random variable) when the subject is application of central limit theorem? $\endgroup$ Commented Feb 4, 2021 at 18:52
  • $\begingroup$ In order to prove the CLT, you need to make statements about the distribution of the whole vector of results, which means that you need an assumption about how the different repetitions relate to each other. It works with independence, in fact you could even assume something weaker (but complicated) but in any case you need some kind of assumption of this kind, which requires all the replicates to be modelled separately, $\endgroup$ Commented Feb 4, 2021 at 18:55

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