# Confusion regarding the weak law of large numbers

I can intuitively understand that as I take more samples from a random variable $X$ (gaussian distribution), the mean would approach $E(X)$.

But what I don't get is if I look at it mathematically. Say I took two samples $x_1$, $x_2$. The mean would be $(x_1+x_2)/2$. But variances add, which is contrary to the previous idea that the variance for the mean should get smaller as I increase the number of samples.

I've asked this question elsewhere and from what I understand variances in this case do not add. Somebody talked about there being a difference between "i.i.d. and independence". But I don't get it, either something is independent or it is not. The way I see it acting like $x_1$ and $x_2$ are dependent would be committing the gamblers fallacy.

Or is there a difference between a) the process of first sampling $x_1$, then $x_2$ and b) viewing the sampling of $x_1$ and $x_2$ as a single process?

I tried following this proof this for the sum of variances and using twice the same random variable $X$ but that resulted in nonsense.

In short: What's the deal with independence in this context and how does it influence $var(x_1 +x_2)$?

• "Variances add". So the variance of $X_1+X_2$ is the sum of the variances. But we're not talking about $X_1+X_2$, we're talking about $(X_1+X_2)/2$. – David C. Ullrich Feb 9 '16 at 13:37

$V(aX)=a^2V(X)$. Hence, if $X_1,\dots X_n$ are i.i.d.: $$V\left(\frac{1}{n}\sum_{i=1}^nX_i\right)=\frac{1}{n^2}V\left(\sum_{i=1}^nX_i\right)=\frac{1}{n^2}nV(X_1)=\frac{V(X_1)}{n}$$

So the variance of the sample mean is indeed decreasing as the sample size increases.