# Intuition behind the no convergence of the variance of sum of random variables

$$Var[\bar{X}] = \sigma^2/n$$ $$Var [\sum{X}_i] = n\sigma^2$$ $$lim_{n \to \infty} Var[\bar{X}] = 0$$ 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 $$X_i in my$$\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.

• What is the limit of a constant times a number ($n$) that goes to $\infty$? – jbowman Dec 26 '18 at 3:16
• I don't understand your question? just if the sample mean is constant why the sample sum isn't – Youssef Dec 26 '18 at 3:21
• The sample sum equals the sample mean times $n$. If the sample mean is constant, what happens to the sample sum as $n \to \infty$? – jbowman Dec 26 '18 at 4:33
• 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 – 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 $$n$$.
• 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. – Youssef Feb 4 '19 at 16:20