Under what conditions is mean square displacement $\text{MSD}(t)=4Dt+v^2t^2$ a valid asymmetric random walk model? I am reading the paper Actin dynamics drive microvillar motility and clustering during brush border assembly by Meenderink et al. (2019). In this paper, the authors fit the mean square displacement (MSD) of "particle" tracking data to what they describe as an "active movement model" given by
$$\text{MSD}(t)=4Dt+v^2t^2$$
where $D$ represents a diffusion coefficient, and $v$ represents velocity. I cannot find the details of where this result comes from, but it looks like the $4Dt$ term is what you usually get for an unbiased random walk in two dimensions, and the other term represents active movement.
However, in going over the mathematical details more carefully, e.g. in these slides (specifically slide 7), it looks like the MSD for an asymmetric random walk$^*$ should have an additional term that is linear in $t$:
$$\text{MSD}(t)=(4D-v^2\Delta t)\,t+v^2t^2$$
This result makes more sense once I see it; for example, if $q=1$ for both horizontal and vertical movement, then the linear term drops out, and we just get $\text{MSD}(t)=v^2t^2$.
So then, is the equation used in the paper incorrect, or is there a different type of random walk model that results in the used MSD equation? If it is incorrect, is there a way to still understand the model in a different light: such as $4D-v^2\Delta t$ being an "effective" diffusion constant that one can obtain from the data, or conditions where $v^2\Delta t$ is negligible compared to $4D$ (I am wary of this last point, since both $D$ and $v^2\Delta t$ have the same dependencies on $\ell$ and $\Delta t$, so the only difference would be the dependency on $q$)?

$^*$ The asymmetric random walk is modeled on a lattice with spatial separation $\ell$, where for every unit of time $\Delta t$ the "particle" has a probability of $0\leq q\leq1$ to move to the right and a probability of $1-q$ to move towards the left. For this system, you can define a diffusion coefficient $D=\ell^2/2\Delta t$ and velocity $v=(2q-1)\ell/\Delta t$. This can easily be generalized to more than one dimension, where there is an asymmetric walk in each direction.
 A: This is just diffusion with a drift.
Starting with the SDE
$$\mathrm{d}x_t = \mu\,\mathrm{d}t + \sigma\,\mathrm{d}W_t$$
which assuming constant $\mu$, $\sigma$ has the solution
$$x_t - x_0 \sim \mathcal{N}(\mu t, \sigma\sigma^\mathrm{T} t)$$
assuming further that $\sigma$ is uniform and $\mu$ only applies to one dimension (say z), i.e. $\mu = (0, 0, v)$, we then have
$$\mathbb{E}(\|x_t - x_0\|^2) = \mathrm{Var}(\|x_t - x_0\|^2) + \mathbb{E}(\|x_t - x_0\|)^2 = n\sigma^2t + v^2t^2$$
where $n$ is the dimensionality.
Now the diffusion constant is defined via the relation of SDEs and the Fokker-Planck equation, for the original SDE could have equivalently been written as a PDE for its PDF $p$:
$$\partial_tp=-\nabla\cdot(\mu p) + \frac{1}{2}\nabla^2(\sigma\sigma^\mathrm{T}p)$$
where we then define $D = \frac{1}{2}\sigma\sigma^\mathrm{T}$ to make the equation to look nicer, and in so doing define the diffusion tensor (or constant as the case may be for uniform diffusion).
So with the final definition of $\mathrm{MSD}(t) = \mathbb{E}(\|x_t - x_0\|^2)$ and all that mathematical background out of the way, we finally arrive at:
$$\mathrm{MSD}(t) = 2nDt + v^2t^2$$
which, in 2D, is the equation you have.
And just to finish up with a short comment; The asymmetric diffusion on a lattice model as per your formulation, as you found out, is effectively trading between variance and drift - with q at either of the extremes, the model becomes deterministic. A discrete model more in line with my post would be one where the jumps are symmetric but that's all superposed on a deterministic, constant, move to one of the directions (perhaps that move could be smaller or larger than the random jump sizes).
