Timeline for Central limit theorem on distributions with support other than $\mathbb{R}$ [duplicate]
Current License: CC BY-SA 4.0
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| Aug 20, 2019 at 0:45 | comment | added | Dave | @Ben Your answer in the linked thread nailed it. | |
| Aug 20, 2019 at 0:45 | history | duplicates list edited | whuber♦ | duplicates list edited from How can the central limit theorem hold for distributions which have limits on the random variable? to What intuitive explanation is there for the central limit theorem?, How can the central limit theorem hold for distributions which have limits on the random variable? | |
| Aug 20, 2019 at 0:44 | history | closed |
Ben Sextus Empiricus whuber♦ |
Duplicate of How can the central limit theorem hold for distributions which have limits on the random variable? | |
| Aug 20, 2019 at 0:44 | comment | added | whuber♦ | My answer at stats.stackexchange.com/a/3904/919 makes a point of explaining and illustrating this with a more extreme example: that of a Bernoulli distribution. You need to remember that the CLT concerns the standardized distribution of the mean (or sum). BTW, there is no such thing as "getting to" infinity. | |
| Aug 20, 2019 at 0:40 | review | Close votes | |||
| Aug 20, 2019 at 0:45 | |||||
| Aug 20, 2019 at 0:27 | comment | added | Sextus Empiricus | The variable $\bar{X}$ does indeed not become negative when every $X_i>0$. But, it's not the variable $\bar{X}$ that's approaching the normal distribution, and instead, it is some transformed version of it that approaches the normal distribution. | |
| Aug 20, 2019 at 0:27 | comment | added | BruceET | Two approaches to answer: (1) The definition of convergence dist'n is satisfied, and it if doesn't match with your intuitive views about support, then your intuitive views don't matter. (2) Focus on the shape of the distribution over regions where most of the probability is. The dist'n of the mean of $n$ iid exponentials is gamma with shape parameter $n$. If you plot the gamma dist'n for $\bar X_{100}$ and superimpose the normal dist'n with same $\mu$ and $\sigma,$ you won't be able to distinguish between them. And 0 is so many SD's from the mean that the probability below 0 is essentially 0. | |
| Aug 19, 2019 at 23:57 | history | asked | Dave | CC BY-SA 4.0 |