Let $X = \sum_{i=1}^n X_i$ and $Y = \sum_{i=1}^n Y_i$ be two Poisson Binomial distributions (PBDs) with respective parameter vectors $p = (p_1, p_2, \dots, p_n)$ and $q = (q_1, q_2, \dots, q_n)$. Note that $X_i$ and $Y_i$ are each independent Bernoulli RVs with parameters $p_i, q_i$. Suppose that $p$ and $q$ are $\epsilon$ apart entrywise (or for any other norm). I am interested in bounding the total variation (TV) distance $d_{TV}(X, Y)$ in terms of $\epsilon$ and $n$.
The top answer on this post is able to bound the TV of the associated Bernoulli product measures by $\epsilon$, but I do not see how this would extend to a bound on the TV of the PBDs $d_{TV}(X, Y)$.
From the definition of TV, I believe it suffices to attempt to bound $|\Pr(X = k) - \Pr(Y = k)|$ for all $k$. To do this, I have tried using the DFT representation of the PMF of PBDs (see this for a nice treatment: $$ \Pr(X = k) = \frac{1}{n+1} \sum_{j=0}^n \exp(-iwkj)x_j $$ for $w = 2\pi / (n+1)$ and $x_j = \prod_{k=1}^n (1 - p_k + p_k \exp(iwj)$. However, it doesn't appear that we can get this difference into terms which we can bound by $\epsilon$ (due to the many multiplicative factors).
I have also tried seeing if the learnability results for PBDs (e.g. Theorem~1) imply that the parameters $p$ and $q$ must be close, but I am not sure that is true / helps us here.
I would also appreciate any pointers, this seems like it should be morally true but perhaps there is a counterexample.
A sufficient thing to show would also be $d_{TV}(\sum_{i=1}^n X_i, \sum_{i=1}^n Y_i) \leq \sum_{i=1}^n d_{TV}(x_i, y_i)$, which I believe is true by the triangle inequality for TV?