I have seen two definitions of universal hash functions in the literature.
For any $i \geqslant 2$ let $[i]=\{1,\ldots,i\}$.
Definition 1:
A family $\mathcal H$ of hash functions from $[n]$ to $[m]$ is a $k$-universal family of hash functions if for any distinct $x_1,\ldots,x_k\in[n]$ and any possibly identical $v_1,\ldots,v_k\in[m]$, $$ \Pr\limits_{h\in\mathcal H} [h(x_1)=v_1\wedge \cdots \wedge h(x_k) = v_k] \leqslant 1/m^k \enspace . $$
Definition 2:
A family $\mathcal H$ of hash functions from $[n]$ to $[m]$ is a $k$-universal family of hash functions if for any distinct $x_1,\ldots,x_k\in[n]$, $$ \Pr\limits_{h\in\mathcal H} [h(x_1)=\cdots=h(x_k) ] \leqslant 1/m^{k-1} \enspace . $$
In both definitions if equality holds then it is called strongly $k$-universal or alternatively $k$-wise independent hash family.
Clearly the first definition implies the second. But is the other direction true?
It is easy to see that this is not the case for $n\leqslant m$.
This is a counter example when $[n]\subseteq [m]$:
Consider the family $\mathcal H$ composed of only one element, the identity function: $x \mapsto x$. It satisfies the second definition since for all $x \neq y \in [n], \Pr_{h\in\mathcal H} [h(x) = h(y)] = 0 \leqslant 1/m^{k-1}$. However, it does not satisfy the first definition since for all $x,y\in[n], \Pr_{h\in\mathcal H}[h(x) = x \wedge h(y) = y] = 1 > 1/m^k$.
This answer is inspired by this other answer.
However, hash functions are usually employed in applications where $n>m$. So my question is whether this is also the case for $n>m$.
