# 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