Is the Feynman's path integral a density? The Feynman-Kac path integral formula is used to solve parabolic equations related to stochastic processes. Considering the probabilistic expression, the solution is indeed not a density. However, after undergoing an exponential transform, it does have the form of a density (once normalized). Does there exist a stochastic process, the density of which is governed by the transformed solution?
Consider a one dimensional process $dx_s = b(x_s) ds + \nabla u(x_s) + \sigma dw_s$ where $0< \sigma$ and $w_s$ is the standard Brownian motion and we denote $x_s \sim p(s, x)$. The density evolution is governed by the Fokker-Planck PDE as $$\partial_s p = -\nabla ((b - \nabla u) p) + \frac{1}{2} \sigma^2 \Delta p,$$ where $x_0 \sim p(0,x)$, and the Feynman-Kac formula $$\psi(t,x) = \mathbb{E} \left[\int_t^T e^{-\int_t^s V(x_\tau) d \tau} \psi(T, x_T) ds \bigg| x_t = x \right]$$ solves the final value problem $$-\partial_s \psi = - V \psi + \nabla \psi \cdot b + \frac{1}{2} \sigma^2 \Delta \psi,$$ where $\psi(T, x) = e^{-f(x)}$. We note that the variable transform $$\psi = e^{-u}$$ attributed to Schrödinger, when applied to the final value problem $$- \partial_s u = V - \frac{1}{2}|\nabla u|^2 + \nabla u \cdot b + \frac{1}{2} \sigma^2 \Delta u,$$ where $u(T, x) = f(x)$, yields the PDE representation of the previous final value problem. The question then, is whether the normalized transformed variable $\frac{\psi (s, \cdot)}{\int \psi (s, \cdot) d(\cdot)}$, is the density of a backward-in-time process directly related to the forward-in-time process $dx_s = b(x_s) ds + \sigma dw_s$.
 A: I preface my answer by saying that this is really a non-answer and more of a long comment or just a rewrite of the relevant parts of the article that you cited.
So I will start by reformulating the question slightly differently, and, for simplicity, in 1D. 
So we have a system $x_t$ and a control $u_t$, with the system evolving as:
$$\mathrm{d}x_t = (b(t, x_t) + au_t)\mathrm{d}t + \sigma\mathrm{d}W_t$$
So our control is linear. We want to find the optimal control, and we restrict ourselves to cost rate functions that are quadratic in the control. So the cost going from time $t$ given system state $x_t$ to the final $t_f$:
$$C(t, x_t, u(t \to t_f)) = \left\langle D(x_{t_f}) + \int_t^{t_f}\left(\frac{1}{2}cu_t^2 + V(t, x_t)\right)\mathrm{d}t\right\rangle$$
and so the control is to minimize this, i.e.
$$J(t, x_t) = \min_{u(t \to t_f)} C(t, x_t, u(t \to t_f))$$
With some Ito expansions, and thanks to the quadratic form of the cost, we can find analytical solution to the minimization:
$$-\partial_t J = -\frac{a^2}{2c}(\partial_x J)^2 + V + b(\partial_x J) + \frac{1}{2}\sigma^2(\partial_x^2 J)$$
So you solve this backwards starting with the boundary condition $J(t_f, x) = D(x)$. Now we do the change of variables, defining $J = -\lambda \log \psi$, where $\lambda = \frac{\sigma^2c}{a^2}$ and we end up with the (still) backwards PDE:
$$\partial_t\psi = -\left(-\frac{V}{\lambda} + b\partial_x + \frac{\sigma^2}{2}\partial_x^2\right)\psi = -H\psi$$
We can easily cast this into the forward PDE (a shameless plug for another answer of mine):
$$\partial_t\rho = H^\dagger\rho = \left(-\frac{V}{\lambda} - b\partial_x + \frac{\sigma^2}{2}\partial_x^2\right)\rho$$
with the initial condition $\rho(0,x) = \delta(x-x_0)$. So this is diffusion (and it looks like the original system's evolution without control), but some of the probability disappears at a rate $\frac{V(t, x_t)}{\lambda}$. So already based on this, I would guess that no, you cannot remove probability in a state dependent fashion and then rescale the distribution to add it back and have the resulting equation satisfy the diffusion equation. What you could do, of course, is set up an SDE $\mathrm{d}x_t = b(t, x_t)\mathrm{d}t + \sigma\mathrm{d}W_t$ that can randomly end in the interval $[t, t+\mathrm{d}t)$ with probability $\frac{V}{\lambda}\mathrm{d}t$, but that's not really a standard form diffusion SDE. 
