uniform integrability of a squared sum of iid variables I'm trying to prove that if $X_i$ are independent, identically distributed random variables such that $E X_i = 0$ and $E X_i^2 < \infty$ then the sequence $\frac{(\sum_{i=1}^{n} X_i)^2}{n}$ is uniformly integrable. Actually I've been told that even something stronger holds, namely that 
$$\frac{\max_{k\leq n}(\sum_{i=1}^{k} X_i)^2}{n}$$
is uniformly integrable. Can someone please give me a hint or a reference to some proof? I've been told that the Hoffmann-Jorgensenn inequality might come in handy, but I suppose that's just for the generalization with $\max$. I know this would be trivial if we had $\frac{\sum_{i=1}^{n} X_i^2}{n}$, but the problem is that the whole sum is squared, not each variable separately. 
Thank you very much for your help.
 A: I think there are two ways. 


*

*Assume you know that $S_n/\sqrt{n}$ converges in distribution to $N(0,1)$ (that is, you know that central limit theorem holds). Then $P(S_n^2>nR)\to P(N^2>R)$ for all $R>0$. 
Conclude using a $2\varepsilon$-argument and the fact that for a non-negative random variable $Y$, 
$$E(Y\chi_{\{Y>R\}})=R\cdot P(Y>R)+\int_R^{+\infty}P(Y>t)dt$$

*Using a truncation argument and a fourth-moments inequality. Such a method gives the wanted uniform integrability for max after having used Kolmogorov inequality for example. 
A: ok, here's what I got so far:
Davide - I didn't really know how to follow your truncating idea - I tried to do it, but I can't figure how to use fourth moment inequality in a sensible way, but instead I tried using the first approach, although slightly modified so that I'd get uniform integrability with max, I'd be grateful if someone took a look at it:
I'm using Kolmogorow's inequality 
$$ P (\frac{1}{n}\max_{k \leq n} S_k^2 \geq M) =  P(\max_{k \leq n} |S_k| \geq \sqrt{Mn}) \leq \frac{1}{M} \sigma^2$$
and the following Hoffmann-Jorgensen inequality - if $X_i$'s are independent then for all nonnegative s, r, t we have
$$P(\max_{k \leq n} |S_k| > s + r + t) \leq P(\max_{k \leq n} |X_k| > s) + 2P(\max_{k \leq n} |S_k| > t)P(\max_{k \leq n} |S_n - S_k| > r/2)
$$
taking s = t = r/2 and by stationarity (same distribution) $S_n - S_k \sim S_j$ we get
$$P(\max_{k \leq n} |S_k| > 4s) \leq P(\max_{k \leq n} |X_k| > s) + 2[P(\max_{k \leq n} |S_k| > s)]^2
$$
so by using
$$P (\frac{1}{n}\max_{k \leq n} S_k^2 \geq s) =  P(\max_{k \leq n} |S_k| \geq \sqrt{sn})$$
we obtain
$$ P (\frac{1}{n}\max_{k \leq n} S_k^2 \geq s) \leq P(\max_{k \leq n} |X_k| > \sqrt{sn}) + 2[P(\max_{k \leq n} |S_k| > \sqrt{sn})]^2
$$
by using Kolmogorov on the second summand
$$P (\frac{1}{n}\max_{k \leq n} S_k^2 \geq s) \leq P(\max_{k \leq n} |X_k| > \sqrt{sn}) + \frac{2}{s^2} \sigma^4
$$
which is cool because the part with $1/s^2$ is nice and integrable, so we could use the formula for expectation suggested by Davide and this ingredient would be less than $\epsilon$ for R sufficiently large, now (if I didn't make a mistake along the way) we just need a sensible estimate on $P(\max_{k \leq n} |X_k| > \sqrt{sn})$ which shouldn't be that hard. I thought about writing something like
$$P(\max_{k \leq n} |X_k| > \sqrt{sn}) = 1 - [P(|X_1| \leq \sqrt{sn})]^n$$
$$P(|X_1| \geq \sqrt{sn}) \leq \frac{\sigma^2}{ns}$$ so
$$P(\max_{k \leq n} |X_k| > \sqrt{sn}) \leq 1 - (1 - \frac{\sigma^2}{ns})^n$$
and I guess we could work with it, but I was hoping for something nicer. Does anyone have an idea?
