 Solution

Independently retain each event with probability $t$, using a fresh Bernoulli mark for every index. Its retained occurrence is the original principal up-set together with its mark being one, so the first proof applies in the enlarged independent product space. The new parameters are ${\mu }_t=t\mu$ and ${\Delta }_t=t^2\Delta$: distinct retained events have independent marks, and pairs of disjoint original supports remain independent.

If no original event occurs, no retained event occurs. ThusWhen $\Delta \ge \mu >0$, choose $t=\mu /\Delta$ to minimize this quadratic exponent. It givesIf $\mu =0$, the appropriate trivial bound is one, rather than an undefined zero-over-zero exponent. The argument is the Janson lower-tail bound by independent thinning.