Generalization of the Bollobas theorem in extremal set theory The Bollobas'1965 theorem is the following:
If $A_1,...,A_n$ and $B_1,...,B_n$ are two sequences of subsets of $X=\{1,...,r\}$ such that $A_i\cap B_j = \emptyset$ if and only if $i=j$, then
$$\sum_{i=1}^n\binom{|A_i|+|B_i|}{|A_i|}^{-1}\leq 1.$$
I am interested in a generalization of this theorem where for each set $A_i$, there are more than one set $B_j$ for which we have $A_i\cap B_j = \emptyset$.
More formally, the assumptions of my problem can be expressed as:
Let $A_1,...,A_n$ and $B_1,...,B_n$ be two sequences of subsets of $X=\{1,...,r\}$. For each $i=1,...,n$, we have a subset $J_i=\{i,i+1,...,i+t\}$(mod n) with $|J_i|=t+1$. $A_i\cap B_j = \emptyset$ if and only if $j\in J_i$.
If $t=0$, we obtain the basic version of the Bollobas theorem.

How can I adapt the result of the Bollobas theorem taking into account these new assumptions for $t\geq 1$?

 A: Just for being more specific than in the comments above.
Consider a random map $\pi$ from the ground set $\cup A_i\cup B_i$ to, say, $(0,1)$. 
Define sets of indices 
$$
F(i)=\{j:\max_{A_i} \pi<\min_{B_j} \pi\}\subset J_i.
$$ 
Note that for any two indices $i_1,i_2$ either $F(i_1)\subset F(i_2)$ or $F(i_2)\subset F(i_1)$, depending on which of two numbers $\max \pi(A_1)$, $\max \pi(A_2)$ is greater. That is, the family of sets $F(i)$ is nested. In particular, all non-empty sets $F(i)$ have a common element $j_0$, thus $F(i)$ may be non-empty only for $i\in\{j_0,j_0-1,\dots,j_0-t\}$. Therefore at most $t+1$ sets may be non-empty. Denote 
$$
p(i)={\rm prob}\, (F(i)\ne \emptyset)=\mathbb{E} {\mathbf 1}_{F(i)\ne \emptyset}.
$$
Then we have 
$$
\sum p(i)=\mathbb{E} |\{i:F(i)\ne \emptyset\}|\leqslant t+1.
$$
For $t=0$ this becomes Bollobas bound. In general case $p(i)$ does not have such a nice formula, but there are some lower bounds like 
$$
p(i)\geqslant \max_{j\in J_i} \frac1{\binom{|A_i|+|B_j|}{|A_i|}}.
$$
Maybe other arguments may be applied in your specific situation.
