Symmetry Argument in the Proof of McDiarmid’s Inequality

Question

I am wondering why in the proof the author says it suffices to show the one-sided estimate. Is this because of the Lipschitz hypothesis for $X_1,..., X_n$?

In (2.15), the $\sigma^2$ on the left hand side should be $\sigma$ instead.

Answer 1

Suppose you can prove $$P(F(X)-E[F(X)]\ge \lambda \sigma^2)\le C\exp(-c\lambda^2).\tag1$$ Since $-F$ satisfies the same hypotheses of the theorem, you also have $$P(-F(X)-E[-F(X)]\ge \lambda \sigma^2)=P(F(X)-E[F(X)]\le -\lambda\sigma^2)\le C\exp(-c\lambda^2).\tag2$$ Adding $(1)$ to $(2)$ then gives $$ P(|F(X)-E[F(X)]|\ge \lambda \sigma^2)\le 2C\exp(-c\lambda^2). $$