Let $X$ and $Y$ be distributions with statistical distance (total variation distance) at most $d$. What is the best upper bound you can give on the statistical distance between $k$ independent copies of $X$ and $k$ independent copies of $Y$?
I can show using a "hybrid argument" that it is at most $k\cdot d$, and I am looking for something better (maybe not always, but at least in some cases).
For example if $d=1/2$ and $k=2$ this bound is $1$ which is meaningless and it cannot be reached by actual distributions because $X_1,X_2$ and $Y_1,Y_2$ have statistical distance $1$ iff their supports are disjoint which would imply that the supports of $X$ and $Y$ are also disjoint in contradiction to $X$ and $Y$ having statistical distance at most $1/2$. The best example I could find was $X\equiv1$ and $Y$ being a uniformly distributed bit so the distance between $X$ and $Y$ is $1/2$ and the distance between $X_1,X_2$ and $Y_1,Y_2$ is $3/4$.
For general $k,d$ I'd like to show that the distance between $k$ samples of $X$ and $k$ samples of $Y$ is at most $1-\left(1-d\right)^k$. (For sanity check: $\left(1-d\right)^k\geq 1-d\cdot k$ for $k\geq 1,d\in\left[0,1\right]$ can be easily shown using basic calculus).
Can you do better (give a counter example), or can you prove that it is the best?