Statistical distance between two coin-flipping distributions I have two distributions that I suspect are close in statistical (variational) distance i.e. $\sum_{x} \| D_1(x) - D_2(x) \|$ is small.
The first distribution is $N$ coin-flips of iid coins with probability $p$ of coming up heads.
For the second distribution, I first pick $F$ locations uniformly randomly out of the $N$ and set them to heads. For the remaining locations, I flip iid coins with bias $p' = \frac{pN - F}{N- F}$. (The rationale for the adjusted probability is so that the expected number of heads match up in both situations.)
How close are these distributions in statistical distance? Is there an easy bound one can place? Here I'm thinking of $p = N^{-1/10}$ and $F \le N^{1/100}$.
My progress
Let's note that both distributions are permutation symmetric so the statistical distance between the two distributions is equivalent to the statistical distance between random variables $X$ and $Y$ where
$X \sim \textrm{Binomial}(N, p)$ and $Y \sim F + \textrm{Binomial}(N-F, p')$.
The intuition as to why these random variables should have similar distribution is that as $N \rightarrow \infty$, the distributions approach normal distributions which are both centered at $pN$ but $X$ concentrates faster than $Y$.
 A: Hopefully, I didn't make too many mistakes in the asymptotic bounds in the end... but this should work and give something (not necessarily the best bound, but something).

This may be overkill, but let
$$
\sigma_1^2 := Np(1-p), \qquad \sigma^2_2 := (N-F)p'(1-p'), \qquad \mu_1=\mu_2 := Np
$$
Then, using the triangle inequality and comparison between Poisson Binomial Distributions (of which both your distributions are a special case: one being binomial, and the other being a PBD with only two distinct values of $p_i$, namely $1$ and $p'$) and "translated Poisson distributions" from Lemmas 1 and 2 of [DDS12], we get
$$\begin{align*}
\mathrm{TV}(X,Y) &\leq \mathrm{TV}(X,_X)+\mathrm{TV}(Y,_Y) +\mathrm{TV}(_X,_Y) \\
&\leq
 \frac{2+\sqrt{Np^3(1-p)}}{\sigma^2_1}+\frac{2+\sqrt{(N-F)p'^3(1-p')}}{\sigma^2_2}+\frac{1+|\sigma^2_1-\sigma^2_2|}{\sigma^2_1}
\end{align*}$$
denoting by $_X$ and $_Y$ the two translated Poisson with parameters $\mu_1,\sigma_1$ and $\mu_2,\sigma_2$, respectively (see [DDS12] for the definition).
In the setting of parameters you consider ($F \ll N$, $p \ll 1/F$, etc.), the first two terms are
$$
\frac{2+\sqrt{Np^3(1-p)}}{\sigma^2_1}+\frac{2+\sqrt{(N-F)p'^3(1-p')}}{\sigma^2_2} \leq \frac{2+\sqrt{Np^3}}{\sigma^2_1}+\frac{2+\sqrt{N p^3}}{\sigma^2_2} = O\left(\sqrt{\frac{p}{N}}\right)
$$
while the third is
$$
\frac{1+|\sigma^2_1-\sigma^2_2|}{\sigma^2_1} \leq O\left( \frac{F}{Np}\right)
$$
(not being too careful with the constants, gratefully using the $O(\cdot)$ notation, but that should be correct), So you should get some bound of the form
$$
\mathrm{TV}(X,Y) = O\left(\sqrt{\frac{p}{N}}+\frac{F}{Np}\right) = O\left(\frac{1}{N^{11/20}}\right)
$$
the last bound for the specific setting $p=1/N^{1/10}$ and $F\leq N^{1/100}$ you mentioned.

[DDS12] Learning Poisson Binomial Distributions. Daskalakis, Diakonikolas, and Servedio, STOC'12. https://arxiv.org/abs/1107.2702
