# Is it possible to improve the order of convergence of averages of random variables if they are not identically distributed?

Let $X_n$ be a sequence of independent random variables (but not necessarily identically distributed) taking values in $[-1,1]$ that have the following property:

1) The average $A_n := \frac{(X_1+ \ldots + X_n)}{n}$ converges in probability to $0$, i.e. for every $\epsilon >0$, the probability that $|A_n| > \epsilon$ converges to $0$.

2) For some $p$, the distribution of $n^p A_n$ converges to something non degenerate.

3) The distribution of $X_n$ converges to something non degenerate (ideally the uniform distribution, but not necessarily).

$\textbf{Question 1:}$ Is it possible for property 1) and 2) to hold and $p$ to be something bigger than $\frac{1}{2}$?

$\textbf{Question 2:}$ Is it possible for property 1), 2) and 3) to hold and $p$ to be something bigger than $\frac{1}{2}$?

$\textbf{Remark:}$ Note that if $X_n$ were identically distributed, with expectation value $0$, then 1), 2)and 3) hold with $p =\frac{1}{2}$. Property 1) is Weak Law of Large Numbers and 2) is Central Limit Theorem. In fact something stronger than 1) holds, namely almost sure convergence.

I am wondering if a better order of convergence is possible if we do not demand that the random variables are identically distributed (and we do not require almost sure convergence). Note that I am also not requiring that the expectation value of $X_n$ is actually $0$ for each $n$. I am simply requiring that the average $A_n$ converges in probability to $0$.

$\textbf{Remark$1$:}$ The answer to Question $1$ seems to be yes; take $X_n$ to be infinite summable almost surely.

$\textbf{Remark$2$:}$ The answer to Question $2$ also seems to be yes. The example suggested is as follows; let $Y_n$ be i.i.d. random variables. Define $$X_n := Y_n+ n^{-\alpha}$$ for some $\alpha \in (0, \frac{1}{2})$. Then it seems that the distribution of $n^{1-\alpha} A_n$ will converge to a Dirac Delta not centered around the origin. I am wondering if someone can explain the reason behind this (or point out some references).

• Just make the $X_n$ summable almost surely, then this property holds with $p=1$... – Martin Hairer Jan 8 '15 at 8:43
• @Martin: Thank you for the example; I guess I was tacitly making another assumption and did not state it. Could you take a look at the revised question now? – Ritwik Jan 8 '15 at 9:27
• If by "non-degenerate" in 2) you just mean "not a Dirac at the origin", then you can take $X_n = n^{-\alpha} + Y_n$ for i.i.d. $Y_n$ and you get your result with $p = 1-\alpha$ provided that $\alpha \in (0,1/2)$. In this case, the limit of $n^p A_p$ will be deterministic though (but non-zero). – Martin Hairer Jan 8 '15 at 11:45
• @Martin: This example is very interesting (it is the sort of example I was really looking for). But can you explain a bit more as to what the limit of n^p A_n is going to be? Why is it deterministic? – Ritwik Jan 8 '15 at 12:01