Suppose we know $X_1, X_2, X_3, ..., X_n$ are identical and independent random variable, and $\overline{X}$ denote $\frac{X1+X2+...+Xn}n$. Is it necessarily true that $|X_1-\overline{X}|$ and $|X_2-\overline{X}|$ be independent random variable? I don't think it is true, because the fact that functions of independent random variables are still independent random variables cannot apply here as $X_1-\overline{X}$ is not just function of $X_1$, but also function of $X_2$.
If they are not independent random variables, that means that we cannot assert that because of law of large number $\frac{\sum|Xi-\overline{X}|}n$ converge in probability to $E(|X-\overline{X}|)$ right?
Indeed they are not independent and strong law cannot be applied to $\frac {|X_i - \bar {X}|} {n}$. For example take $no2$ and consider $|X_1- \bar{X}|= \frac {|X_1 -X_2|} 2$ and $|X_2- \bar{X}|= \frac {|X_2 -X_1|} 2$. Hence $|X_1 -\bar {X}|=|X_2 -\bar {X}|$ and independence fails unless $|X_1 -X_2|$ is a constant. There are many ways of showing that this can happen only if X_i is a constant random variable.