# Concentration around mean from concentration around median

I got stuck solving exercise 11.3 from the book Concentration of Measure for the Analysis of Randomized Algorithms. The setting is:

Consider again the situation of Section $$7.2,$$ the number of non-empty bins when $$m$$ balls are thrown independently and uniformly at random into $$n$$ bins.

It is shown that $$\operatorname{Pr}[|f-\mathrm{M}[f]|>t] \leq 2 e^{-t^{2} / n}$$.

Then I want to show

Exercise $$11.3 .$$ Check that in this case, concentration around $$\mathrm{M}[f]$$ can be used to deduce similar concentration around $$\mathrm{E}[f]$$

$$M[X]$$ is the median of $$X$$. How do I do this? I do not know how to start.

The tail bound shows that $$f-M[f]$$ is subgaussian with subgaussian norm $$O(\sqrt{n})$$. Centering a subgaussian random variable loses the subgaussian norm by a factor of $$2$$. The centered version of $$f-M[f]$$ is $$(f - M[f]) - \mathbb{E}(f-M[f]) = f - \mathbb{E}f$$ hence $$f-\mathbb{E}f$$ is subgaussian with subgaussian norm $$O(\sqrt{n})$$ and thus $$\Pr\{ |f-\mathbb{E} f| > t\} \leq 2e^{-ct^2/n}$$ for some absolute constant $$c$$.
Alternatively, you can just plainly integrate the tail bound to show that $$|\mathbb{E}f - M[f]| = O(\sqrt n)$$, this gives you a similar concentration around $$\mathbb{E}f$$, too.