I have read in a paper about the following result:


Let $V$ be a separable Hilbert space and $(\Omega,A\_{\Omega},P)$ a probability space. Suppose that $Y\_1,Y\_2,...$ is a sequence of independent $V$-valued random variables. If $E\left(\Vert Y\_i\Vert\_{V}^m\right)\leq \frac{1}{2} m! B^2 L^{m-2}$ $\forall m\geq 2$, then, $\forall n\in N$ and $\epsilon>0$


$$ P\big(\big\Vert \frac{1}{n} \sum\_{i=1}^n Y\_i\big\Vert\_{V}>\epsilon\big)\leq 2 \exp\Big(-\frac{n\epsilon^2}{B^2+L\varepsilon+B\sqrt{B^2+2L\epsilon}}\Big).$$


There was no proof given but it was mentioned that the result is well known. I think I found some literature that shows the result if we would assume that $E(Y\_i)=0$ for all $i\in N$ holds (see [Theorem 3.3 of Pinelis](https://arxiv.org/pdf/1208.2200.pdf)). But I don't know much about martingales so maybe it wouldn't follow. Does anyone know if the result is actually true and if $E(Y\_i)=0$ is a necessary condition?



Concentration inequality for Hilbert space valued random variables
The condition $EY\_i=0$ cannot be dropped.


Indeed, if e.g. the $Y\_i$'s are iid with $\mu=EY\_i\ne0$ and $n\to\infty$, then, by the law of large numbers, the left-hand side of the inequality in question will go to $1$ for each $\epsilon\in(0,\|\mu\|\_V)$, whereas the right-hand side of the inequality will go to $0$.


