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Does the Central Limit Theorem require the number of random variables to increase to a sufficiently large number or the number of samples of each random variable to increase to a sufficiently large number?

If the number of random variables is 1 for the chi-square statistic (as an example), then the degrees of freedom is 1, and the distribution would not represent a normal distribution, no matter how many samples. However, if the number of random variables is large, it seems we would still require more than a few samples for the chi-square distribution to resemble a normal distribution.

Note: I am starting to self learn statistics, and may need some contextual info in answers.

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  • $\begingroup$ Do you mean a situation where you write something like $X_1, X_2,X_3,...\overset{iid}{\sim} f(x)$? $\endgroup$ – Dave Jul 22 '20 at 3:11
  • $\begingroup$ Yes, that's correct $\endgroup$ – imagineerThat Jul 22 '20 at 4:27
  • $\begingroup$ Of possible interest: stats.stackexchange.com/questions/473455/… $\endgroup$ – Dave Oct 15 '20 at 21:13
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Your question shows some confusion! A sample is usually represented by a sequence of random variables iid $X_1, X_2, \dotsc, X_n, \dotsc$, where each random variable is a function from some sample space $\omega \mapsto X(\omega)$, the argument $\omega$ often omitted from the notation. So a realization is only one value, always. So $n$ in the CLT refers to the number of observations, each observation represented with one random variable.

So in your chisquare example, $n$ has nothing to do with the number of degrees of freedom $\nu$. For the iid CLT, the only assumption is that expectation and variance exists as finite numbers. Since that is the case for the chisquare distribution, the CLT applies and when the number of observations $n \to\infty$, the standardized mean converges to a normal distribution.

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