# Max decoupling inequality

Let $$X_1,\ldots,X_n$$ be $$\{0,1\}$$-valued random variables drawn from some joint distribution. Let $$\tilde X_1,\ldots,\tilde X_n$$ be their independent version: $$\mathbb{E}X_i=\mathbb{E}\tilde X_i$$ for each $$1\le i\le n$$ and the $$\tilde X_i$$ are mutually independent (as well as being independent of the $$X_i$$'s). I would like to compare $$M:=\mathbb{E}\max_{1\le i\le n}X_i$$ and $$\tilde M:=\mathbb{E}\max_{1\le i\le n}\tilde X_i$$. It's obvious that $$M$$ can be made arbitrarily small, while $$\tilde M$$ is arbitrarily close to $$1$$, so in general, there can be no bound on $$\tilde M/M$$. But it seems that in the reverse situation, one should be able to say something. Is some bound of the form $$M \le c\tilde M$$ known, where $$c$$ is an absolute constant? If such a thing is impossible (what's the counterexample?), what's the best dependence on $$n$$ in the bound?

Yes, this inequality holds with $$c:=\frac e{e-1}.$$
Indeed, let $$A_i:=\{X_i=1\}$$. Then $$M=P\Big(\bigcup_i A_i\Big)\le\min(s,1)\le c(1-e^{-s}),$$ where $$s:=\sum_i P(A_i)$$. On the other hand, $$\tilde M=1-\prod_i(1-P(A_i)) \ge1-\prod_i\exp(-P(A_i))=1-e^{-s}.$$ So, $$M\le c\tilde M$$.