When is the sum of uncorrelated (not necessarily with the same distribution) r.v.'s bounded in Probabilty? Let $v_{i},\;i=1,\cdots,N$ be such that $E\left(v_{i}\right)=0$, $E(v_{i}^{2})=1$  and $E\left(v_{i}v_{j}\right)=0\;for\;i\neq j$. So $v_{i}$'s are mean zero with unitary variance, uncorrelated and not necessarily with identical distributions.
Is it true that
$\frac{1}{\sqrt{N}}\sum_{i=1}^{N}\left(v_{i}^{2}-1\right)=O_{p}(1)$  (bounded in probability, not necessarily asymptotic normal)?
What would be the minimum set of assumptions in order to get this result?
 A: So if you define $X_n = \frac{1}{\sqrt{n}} \sum_{i=1}^n (V_i^2-1)$, then you want to show for each $\epsilon>0$ there is a constant $M$ such that $P[|X_n|>M]\leq \epsilon$ for all $n \in \{1, 2, 3, ...\}$. 

No.  Try $\{V_i\}_{i=1}^{\infty}$ independent with $$V_i = \left\{ \begin{array}{ll}
0 &\mbox{ with prob $1 - \frac{1}{4^i}$} \\
4^{i/2}  & \mbox{ with prob $\frac{1/2}{4^i}$}\\
-4^{i/2} & \mbox{ with prob $\frac{1/2}{4^i}$}
\end{array}
\right.$$
Define event $\mathcal{A} =\cap_{i=1}^{\infty} \{V_i=0\}$.  Then $P[\mathcal{A}]>0$, and if $\mathcal{A}$ happens then $\frac{1}{\sqrt{n}}\sum_{i=1}^n(V_i^2-1) = -\sqrt{n}$ for all $n \in \{1, 2, 3, ...\}$. 

On the other hand, if you assume $\{V_i^2\}_{i=1}^{\infty}$ are pairwise uncorrelated, so $E[V_i^2V_j^2]=E[V_i^2]E[V_j^2]$ whenever $i \neq j$, and if you assume there is a constant $D$ such that $Var(V_i^2)\leq D$ for all $i$, then it is true.  That is because: 
$$X_n = \frac{1}{\sqrt{n}} \sum_{i=1}^n (V_i^2-1) \implies E[X_n]=0, Var(X_n)=\frac{1}{n}\sum_{i=1}^nVar(V_i^2-1) \leq D$$
and so by the Chebyshev inequality we have for all $M>0$ and all $n\in\{1, 2, 3, ...\}$: 
$$ P[|X_n|\geq M] \leq \frac{Var(X_n)}{M^2} \leq \frac{D}{M^2}$$
