Let $X_1, \ldots, X_n$ be independent Bernoulli random variables. Then $Pr[X_i=1]=Pr[X_i=0]=1/2$. Let $X = (X_1, \ldots, X_n)$ and $v \in \{0,1\}^n$, $Y=v \cdot X$, $Z=Y-1$. Let \begin{align} \mu_1(x) = Pr[Y=x], \ \mu_2(x)=Pr[Z=x], \ x \in [n]=\{1, \ldots, n\}. \end{align} Are there some method to estimate the total variation distance? \begin{align} d_{TV}(\mu_1, \mu_2) & = \frac{1}{2} \sum_{x \in [n]} | \mu_1(x) - \mu_2(x) | \\ & = \frac{1}{2} \sum_{x \in [n]} | Pr[Y=x] - Pr[Z=x] |. \end{align} Without loss of generality, we many assume that $v=(1,\ldots, 1,0,\ldots,0)$, where the number of $1$'s is $k$ ($0 < k \leq n$). Then \begin{align} d_{TV}(\mu_1, \mu_2) & = \frac{1}{2} \sum_{x \in [n]} | Pr[Y=x] - Pr[Z=x] | \\ & = \frac{1}{2} \sum_{x \in [n]} | Pr[X_1+\cdots+X_k=x] - Pr[X_1+\cdots + X_k=x+1] |. \end{align}

Assume that $n$ is sufficient large. Do we have $d_{TV}(\mu_1, \mu_2)< 1-\epsilon$ for some $0 \leq \epsilon \leq 1$? Thank you very much.