From what I understand, Liouville's theorem is about the probability density $\rho$ of an ensemble existing in a differential volume in phase space $d\mathbf{r}d\mathbf{p}$.

So the statement for [Liouville's theorem](https://en.wikipedia.org/wiki/Liouville%27s_theorem_(Hamiltonian)) is
\begin{align\*}
\frac{d}{dt} \rho (\mathbf{r}, \mathbf{p}, t) &= \frac{\partial \rho}{\partial t} + \sum\_{i=1}^N \frac{\partial \rho}{\partial q\_i}\cdot \frac{dq\_i}{dt}+\frac{\partial \rho}{\partial p\_i} \cdot \frac{dp\_i}{dt} \quad ...(1)\\ &=0
\end{align\*}
which arises out of the continuity equation coupled with Hamilton's relations.

All good so far.

Now we have [Boltzmann's transport equation](https://en.wikipedia.org/wiki/Boltzmann_equation), which is given by

$$\frac{\partial f}{\partial t} + \frac{\mathbf{p}}{m}\cdot \nabla F + \mathbf{F}\cdot \nabla \_p f = \left( \frac{df}{dt} \right) \_{\rm coll}$$

where $f$ "probability density function: $f(\mathbf{r}, \mathbf{p}, t)$".

If I replace the $\frac{dq}{dt}$ by $\frac{\mathbf{p}}{m}$ and $\frac{dp}{dt}$ by $\mathbf{F}$ in equation $(1)$, I recover the same equation as the Boltzmann transport equation.

My question is: What is the difference between $\rho$ and $f$? Why are there no collisions in an ensemble as seen by Liouville, but we consider collisions when thinking about Boltzmann's transport equation?