Strange deduction about relation of median and mean On his blog T. Tao's proves the following concentration inequality, due to Talagrand.
Let $K>0$, and let $X_{1},..., X_{n}$ be iid complex random variables all bounded by $K$. Let $F:\mathbb{C}^n\rightarrow \mathbb{R}$ be a $1$-Lipshitz convex function (for this identify $\mathbb{C}^n$ with $\mathbb{R}^{2n}$ ). Then for any $\lambda$ one has:
$$ \mathbb{P}(|F(X)-MF(X)|\geq\lambda K)\leq C \exp(-c\lambda ^{2})  \tag{1}$$
and
$$ \mathbb{P}(|F(X)-\mathbb{E}F(X)|\geq\lambda K)\leq C \exp(-c\lambda ^{2})  \tag{2}.$$
He claims that it is sufficient to prove  $(1)$, because $(1)$ imples in turn that:
$$\mathbb{E}F(X)=MF(X)+\mathcal{O}(1)$$
which then gives $(2)$. 
WHY IS $$\mathbb{E}F(X)=MF(X)+\mathcal{O}(1)$$ true, and in particular why can it be deduced from $(1)$?

For every nonnegative random variable $Y$,
$$
E[Y]=\int_0^\infty P[Y\geqslant y]\,\mathrm dy.
$$
Applying this to $Y=|F(X)-MF(X)|$ and using (1), one gets
$$
E[|F(X)-MF(X)|]\leqslant\int_0^\infty C\mathrm e^{-cy^2/K^2}\,\mathrm dy=CK/\sqrt{c}\int_0^\infty\mathrm e^{-y^2}\mathrm dy.
$$
Thus, $E[F(X)]=MF(X)+R(X)$, where $|R(X)|\leqslant CK\sqrt{\pi/(4c)}$.
