Maximum occupancy balls in bins with limited independence

Consider the probability that $k$ particular balls go into the same bin. By $k$-wise independence, this is $n^{-k}$ for each bin, or $n^{-k+1}$ when we sum over all bins. On the other hand, if we choose a random $k$-tuple of indices, the probability that these are all sent to the same bin is at least the probability that all $k$ are sent to the bin with the highest load $X_n$, so if we condition on $X_n$, the probability is at least ${X_n \choose k} / {n \choose k}$.

Unfortunately, $x \choose k$ is not always convex as a function of $x$, so we have to modify it slightly to use convexity. For $k \in \mathbb{N}, x\in \mathbb{R}$, define ${x \choose k}^+ = {\begin{cases} {x \choose k}, x \gt k-1 \newline 0, x \le k-1\end{cases}} = \frac{1}{k!}\prod_{i=0}^{k-1} \max (0,x-i).$ Since each factor $\max (0,x-i)$ is convex and nonnegative, the scaled product ${x \choose k}^+$ is convex.

By Jensen's inequality, the probability $k$ balls go to the same bin is at least ${E[X_n] \choose k}^+ / {n \choose k}$. So, $k$-wise independence implies

$$\begin{eqnarray}{E[X_n] \choose k}^+ &\le & {n \choose k}n^{-k+1} \newline &\le & \frac{n}{k!} \newline \prod_{i=0}^{k-1} \max(0,X_n-i) &\le & n \newline \max(0,E[X_n]-k+1)^k &\le & n \newline E[X_n] &\le & \sqrt[k]{n} + k-1. \end{eqnarray}$$

I think https://arxiv.org/abs/1502.05729 might be at least a partial answer to my question. It shows "a $k$-independent family of functions that imply [heaviest loaded bin] size is $\Omega(n^{1/k})$".