Strange deduction about relation of median and mean

Question

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 $(1)$?

Answer 1

Let $m$ denote the median. By the triangle inequality $\bigl|\Bbb E F(X)-m\bigr|\leq \Bbb E(|F(X)-m|)$. But this expectation is $$ E(|F(X)-m|) = \int_0^\infty \Bbb P(|F(X) - m| \geq \lambda)\,d\lambda \leq KC\int_0^\infty e^{-c\lambda^2}\,d\lambda = O(1). $$

Answer 2

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 $|R(X)|\leqslant CK\sqrt{\pi/(4c)}$.