Maximal deviation from mean of a bounded random variable Is there a non-probabilistic Hoeffding like inequality which tells me the deviation between a bounded random variable and its expectation?
Let $X$ be a random vector such that $||X|| \leq c$. I am looking for
$$||X-\mathbb{E}[X]|| \leq ?\quad ( \text{maybe }c)$$
if such an inequality exists.
I think $\mathbb{E}[X]$ lies in the ball with radius $c/2$ and therefore $||X-\mathbb{E}[X]|| \leq c$, but I don't know how to proof it.

$X$ lies in the ball around $0$ with radius $c$ and so does the expectation of $X$ by Jensen's inequality, since the norm is a convex function. The distance between $X$ and $\mathbb{E}[X]$ is therefore bounded almost surely by $2c$. This upper bound is tight. To see this, take your probability space to be $[0,1]$ with the uniform distribution and define the random variable $X_n$ by
$$X_n(\omega)=\begin{cases}
c,  & \text{if }\omega\in[0,1/n) \\
-c, & \text{if }\omega\in[1/n,1] 
\end{cases}.$$
For each $n$ and $\omega\in[0,1/n)$, one has $$\big| X_n-\mathbb{E}[X_n]\big|=\big|c-1/n c-(1-1/n)(-c)\big|=|2(1-1/n)c|,$$
and this expression gets arbitrarily close to $2c$.
