Given $v\_{ij} \in \{0,1\}$, $i \in \{1,2\}$, $j \in \{1,2,\ldots,n\}$. Let $X\_1, X\_2, \ldots, X\_n$ be random variables, $P[X\_i=1]=P[X\_i=0]=1/2$, $i \in \{1,\ldots, n\}$. By checking many examples, I think that the following is true: when $|v\_1|, |v\_2|$ are large enough,
\begin{align}
 \frac{1}{2} \sum\_{x\_1=0}^n \sum\_{x\_2=0}^n | P[ \sum\_{r=1}^n v\_{1r}X\_r = x\_1, \sum\_{r=1}^n v\_{2r}X\_r = x\_2 ] - P[ \sum\_{r=1}^n v\_{1r}X\_r = x\_1+1, \sum\_{r=1}^n v\_{2r}X\_r = x\_2+1 ] | < 1-\epsilon. 
\end{align}
Is there some method to prove this? Thank you very much.


For example, let $n=4$, and $v\_{1}=(v\_{11},\ldots,v\_{14})=(1,1,1,0), v\_2=(v\_{21},\ldots,v\_{24})=(0,0,1,1)$, then the following Python codes gives the result $0.3446349999999999$.



```
import numpy as np

N = 10**5  

XX = [np.random.randint(2, size=4) for n in np.arange(N)]

r=0
for x1 in range(0,5):
    for x2 in range(0,5):
        P1 = list(map(lambda X: (X[0]+X[1]+X[2]==x1)&(X[2]+X[3]==x2), XX))
        P2 = list(map(lambda X: (X[0]+X[1]+X[2]==x1+1)&(X[2]+X[3]==x2+1), XX))
        r=r+abs(np.mean(P1)-np.mean(P2))
r/2 

```

Let $n=4$, and $v\_{1}=(v\_{11},\ldots,v\_{14})=(0,1,1,1), v\_2=(v\_{21},\ldots,v\_{24})=(1,1,1,1)$, then the following Python codes gives the result $0.31340999999999997$.



```
import numpy as np

N = 10**5  

XX = [np.random.randint(2, size=4) for n in np.arange(N)]

r=0
for x1 in range(0,5): 
    for x2 in range(0,5):
        P1 = list(map(lambda X: (X[1]+X[2]+X[3]==x1)&(X[0]+X[1]+X[2]+X[3]==x2), XX))
        P2 = list(map(lambda X: (X[1]+X[2]+X[3]==x1+1)&(X[0]+X[1]+X[2]+X[3]==x2+1), XX))
        r=r+abs(np.mean(P1)-np.mean(P2))
r/2

```


Estimate an expression about probability about Bernoulli random variables
We shall assume that the $X\_i$'s are independent. The problem can be restated as follows: show that for some $h\in(0,1)$, all natural $n$, and all subsets $J$ and $K$ of the set $[n]:=\{1,\dots,n\}$
we have
\begin{equation\*}
S:=\sum\_{x,y}|P(X\_J=x,X\_K=y)-P(X\_J=x+1,X\_K=y+1)|\le2-h, \tag{1} 
\end{equation\*}
where $X\_J:=\sum\_{i\in J}X\_i$ and the sum in (1) is over all all integers $x,y$. Write
\begin{equation\*}
 S\le T+U, \tag{2}
\end{equation\*}
where
\begin{equation\*}
T:=\sum\_{x,y}|P(X\_J=x,X\_K=y)-P(X\_J=x+1,X\_K=y)|,
\end{equation\*}
\begin{align\*}
U&:=\sum\_{x,y}|P(X\_J=x+1,X\_K=y)-P(X\_J=x+1,X\_K=y+1)| \\ 
&=\sum\_{x,y}|P(X\_J=x,X\_K=y)-P(X\_J=x,X\_K=y+1)|. 
\end{align\*}


By the independence of the $X\_i$'s,
\begin{equation\*}
 P(X\_J=x,X\_K=y)=\sum\_z P(X\_{J\cap K}=z)P(X\_{J\setminus K}=x-z)P(X\_{K\setminus J}=y-z). 
\end{equation\*}
Hence,
\begin{align\*}
T&\le\sum\_zP(X\_{J\cap K}=z)\,\sum\_y P(X\_{K\setminus J}=y-z) \\ 
&\times\sum\_x|P(X\_{J\setminus K}=x-z)-P(X\_{J\setminus K}=x+1-z)| \\ 
&=\sum\_x|P(X\_{J\setminus K}=x)-P(X\_{J\setminus K}=x+1)|=:D\_{|J\setminus K|}, 
\end{align\*}
where $|\cdot|$ denotes the cardinality.
Similarly, $U\le D\_{|K\setminus J|}$, so that, by (2)
\begin{equation\*}
 S\le D\_{|J\setminus K|}+D\_{|K\setminus J|}. \tag{3}
\end{equation\*}
Note that $D\_0=1$ and, by [this answer](https://mathoverflow.net/a/307506/36721), for $k\ge1$ we have
\begin{align\*}
 D\_k=\frac1{2^k}\,\Big(2\binom k{m+1}-1\Big)\le\frac34, 
\end{align\*}
where $m:=\lfloor (k-1)/2\rfloor$. So, by (3),
\begin{equation\*}
 S\le 1+3/4=7/4 
\end{equation\*}
if $J\ne K$.


In the remaining case when $J=K$,
\begin{equation\*}
S=\sum\_x|P(X\_J=x)-P(X\_J=x+1)|=D\_{|J|}\le1. 
\end{equation\*}


Thus, in all cases (1) holds with $h=2-\max(7/4,1)=1/4>0$, as desired.


