Hello,


Let $x\in[0;1]$ and $(B\_i)\_i$ be events defined by $P(B\_i)\leq x, \forall i$.
Furthermore, this inequality is independent of the other events $B\_i$ but the events are not necessarily independent.


I want to upperbound the probability of $A\_k = (B\_1\cup B\_2)\cap (B\_2\cup B\_3)\cap\cdots\cap (B\_k\cup B\_{k+1})$. The first terms give (I simplified each $A\_i$):


$$P(A\_1)\leq 2x$$


$$P(A\_2)=P((B\_1\cap B\_3) \cup B\_2)\leq x^2 + x$$


$$P(A\_3)=P((B\_2\cap B\_3)\cup(B\_2\cap B\_4)\cup(B\_1\cap B\_3))\leq 3x^2$$


What's the upperbound for any $k$ ?


Thank you very much



Apparently simple probability
Let $m:=\big\lfloor\frac{k+1}{2} \big \rfloor$. Then, assuming $\mathbb{P}(B \_ i)\le x$, 
$$\mathbb{P}\big((B \_ 1\cap B \_ 2) \cup(B \_ 2\cap B \_ 3) \dots \cup(B\_k\cap B\_{k+1})\big)\le mx\wedge 1\, ,$$
which follows immediately by the inclusion 
$$(B \_ 1\cap B \_ 2) \cup(B \_ 2\cap B \_ 3)\cup \dots \cup(B\_k\cap B\_{k+1}) \subset B \_ 2 \cup B \_ 4\cup \dots \cup B\_ {2m} \, .$$
The equality is realized in any atomless measure space, taking a sequence of measurable sets $(B \_ i) \_ {i\ge1}$ such that $B\_{2i-1}=B\_{2i}$ and $\mathbb{P}(B \_ {2i})=x\wedge \frac{1}{m}$, for all $1\le i\le m$, $B\_{2i}\cap B\_{2j}=\emptyset$ for $i\neq j$,  

and $B \_ i=\emptyset$ for all $i > m$. 


