For an iid sample $X_1, \dots, X_n$, is the rank statistic distribution free? I.e. for each $k$ s.t $1 \leq k \leq n$, will the distribution of the rank of $X_k$ not depend on the distribution of $X_i$?
I think yes. Suppose $X_i$ has a cdf $F$ which admits an inverse $F^{-1}$, then $F^{-1}(X_1), \dots, F^{-1}(X_n)$ will still be iid but with the uniform distribution over $[0,1]$. Since $F$ is invertible, both $F$ and $F^{-1}$ are strictly increasing, so the rank of $X_k$ and the rank of $F^{-1}(X_k)$ will be the same. So we can assume the distribution of $X_i$ to be the uniform distribution over $[0,1]$, and thus the rank of $X_k$ doesn't depend on the distribution of $X_i$.
If $F$ is not invertible, the ties might change when going from $X_1, \dots, X_n$ to $F^{-1}(X_1), \dots, F^{-1}(X_n)$. We cannot say the distribution of the rank statistic doesn't depend on the distribution of $X_i$?
Thanks and regards!