$$Var[\bar{X}] = \sigma^2/n $$ $$Var [\sum{X}_i] = n\sigma^2$$ $$lim_{n \to \infty} Var[\bar{X}] = 0 $$ wich means at $\infty$ we will always get the same $\bar{X}$ after every simulation. I understand this such as if I have an observation $X_i >E[X]$ I'm sure that I also have $X_i <E[X]$ in my$ \infty $ observations that will compensate it. is this correct?

What I don't understand is if $\bar{X}$ is constant in $ \infty $ why $\sum{X}_i = n\bar{X}$ isn't? That is to say why $$lim_{n \to \infty} Var[\sum{X}_i] \neq 0$$ I'm not looking for mathematical proof but the idea.

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    $\begingroup$ What is the limit of a constant times a number ($n$) that goes to $\infty$? $\endgroup$ – jbowman Dec 26 '18 at 3:16
  • $\begingroup$ I don't understand your question? just if the sample mean is constant why the sample sum isn't $\endgroup$ – Youssef Dec 26 '18 at 3:21
  • $\begingroup$ The sample sum equals the sample mean times $n$. If the sample mean is constant, what happens to the sample sum as $n \to \infty$? $\endgroup$ – jbowman Dec 26 '18 at 4:33
  • $\begingroup$ The sample sum tends to infinity but its variance is not 0 . The sum is infinity and variable, at the same time the sum divided by n is a constant. I find this counter-intuitive $\endgroup$ – Youssef Dec 26 '18 at 7:11

You asked for some intuition. Well, even when the law of large numbers hold, so that $\bar{X}_n \to \mu$ when $n \to \infty$, in practice you will never have $n=\infty$, so the variance of $\bar{X}_n$ is $\sigma^2/n$ which will be positive, maybe tiny but not zero (as long as $\sigma^2 >0$.)

If you multiply $\bar{X}_n$ with $\sqrt{n}$ so to get a constant variance not depending on $n$, you can think of that as looking at the small deviations from $\mu$ through a magnification lens, so as to see the details. Multiplying that with $\sqrt{n}$ another time to get the sample sum, you are looking at the first magnification lens through another magnification lens, and so now the image will be increasing in size with $n$.

  • $\begingroup$ Thank you, But if you magnify something that is decreasing (the variance of $\bar X$) it's true that it will give something bigger ( The variance of $S_n$) but magnifying won't make the two variances go in different directions ( one decreases and the other increases as n increases). This is why I don't find this explanation satisfying. Please explain me more if you think I still don't get your idea. $\endgroup$ – Youssef Feb 4 '19 at 16:20

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