For $v, w \in \{0,1\}^n$, denote $v w = (v\_1 w\_1, \ldots, v\_n w\_n)$ and $|v|=\sum\_{i} v\_i$. 


Let $v\_1, v\_2 \in \{0,1\}^n$ and
\begin{align\*}
f(x\_1, x\_2) = \sum\_{d=0}^{|v\_1 v\_2|} \frac{1}{2^{|v\_1|+|v\_2|-|v\_1 v\_2|}} {|v\_1| - |v\_1 v\_2| \choose x\_1 - d} {|v\_2| - |v\_1 v\_2| \choose x\_2 - d}. 
\end{align\*}


How to find the asymptotic formulas for $f(x\_1, x\_2)$ ($|v\_1|, |v\_2| \to \infty$)? 


Is it possible to prove that there is a constant $\epsilon$, $0<\epsilon<1$, such that 
\begin{align}
\sum\_{x\_1=0}^{|v\_1|} \sum\_{x\_2=0}^{|v\_2|} f(x\_1, x\_2) < 1 - \epsilon?
\end{align} 


These questions relate to [the question](https://mathoverflow.net/questions/307584/how-to-compute-the-asymptotic-of-a-summation-which-involves-binomial-coefficient).


Any help would be greatly appreciated!



How to estimate a summation?
Put $a=|v\_1|$, $b=|v\_2|$, $c=|v\_1v\_2|$. Then we have
$$
\sum\_{i=0}^a\sum\_{j=0}^b\sum\_{k=0}^c\binom{a-c}{i-k}\binom{b-c}{j-k}= \sum\_{k=0}^c\left(\sum\_{i=0}^a\binom{a-c}{i-k}\right)\left(\sum\_{j=0}^b\binom{b-c}{j-k}\right).
$$
As $i$ runs from $0$ to $a$, $i-k$ runs from $-k$ to $a-k$. Since $0\leq k\leq c$, this range covers the interval from 0 to $a-c$, thus $\sum\_{i=0}^a\binom{a-c}{i-k} = \sum\_{i=0}^{a-c}\binom{a-c}{i}=2^{a-c}$. Hence
$$
2^{a+b-c}\sum\_{i=0}^a\sum\_{j=0}^b f(i,j) = \sum\_{k=0}^c 2^{a-c}\cdot 2^{b-c} = (c+1)2^{a+b-2c},
$$
that is,
$$
\sum\_{i=0}^a\sum\_{j=0}^b f(i,j) = \frac{c+1}{2^c}.
$$
Hence the answer to the last question is yes if and only if $|v\_1v\_2|>1$. A pointwise estimate for $f$ is more difficult, as it depends on the relative size of $a$, $b$ and $c$.


