Fokker-Planck equation for a truncated process

Question

Let $X_t = x + bt + \sigma W_t$ be an arithmetic Brownian motion, where $x$ is a random variable independent to $W$, and $\sigma>0$. Suppose the initial distribution is given by $\mathbb P(X_0 \in dy) = m_0(y)dy$, then it's standard that the density $$m(t, y) := \frac{\mathbb P(X_t \in dy)}{dy}$$ can be solved from the following Fokker-Planck equation: $$\partial_t m = G^* m, \quad m(0, y) = m_0(y),$$ where $$G^* f(t, y) = (- b \partial_y f + \frac 1 2 \sigma^2 \partial_{yy} f)(t, y).$$

Now, Let $T = \inf\{t>0, X_t \notin (0, 1)\}$ be the first exit time of $X$ from $(0, 1)$, and $Y_t$ is the process $X_t$ truncated by $T$, i.e. $$Y_t = X_{t \wedge T}.$$

[Q]. Does the density $m(t, y)$ for $y\in (0,1)$ solve the following equation? $$\partial_t m = G^* m \hbox{ on } (0, \infty) \times (0, 1), \quad m(0, y) = m_0(y), \quad m(t, 0) = m(t, 1) = 0.$$ If yes, how does one argue the zero boundairy condition, i.e. $$\lim_{y\to 0} m(t, y) = \lim_{y\to 1} m(t, y) = 0?$$

[Remark]. The density $m(t, y)$ of $Y_t$ is only required on the open interval $(0, 1)$. In other words, the above Parabolic equation with Cauchy-Dirichlet boundary does not and can not intend to provide any probability of $Y_t$ at the absorbing boundary. By local property of the diffusion the equation itself shall be correct in the domain $(0, \infty) \times (0, 1)$, and the main concern is on the setting of Dirichlet data of $m(t, 0)$ and $m(t, 1)$.

It seems to me a fundamental question and there supposed to be an existing literature.

Thanks.

Answer 1

The answer to this question is no.

Indeed it holds that $$P(Y_t = 1) = P\big( \max\limits_{s\le t} X_s \ge 1\big) > 0.$$ So the density has an atom in $1$ and similarly in $0$.

Let us derive some heuristics. You are right in implicitly saying that the Brownian Motion (plus a possible drift and variance) with absorbing boundary has as a generator the operator $\mathcal{G}$ with Dirichlet boundary conditions. And from this indeed it follows that the probability density solves the adjoint equation, related to $\mathcal{G}^*$ in the interior of the domain. But since you are testing only against functions with Dirichlet boundary conditions you cannot deduce anything about $p(t,0)$ or $p(t,1).$

On the other hand, one can ask whether the only bad things are the atoms. I.e., whether the absolutely continuous part w.r.t. to the Lebesgue measure solves the forward Kolmogorov equation with Dirichlet boundary.

In this case, if my calculations are correct, the answer is positive, at least for $\sigma =1$ and $b = 0$. Indeed we would be interested in the limit $$\lim_{\epsilon \to 0} m(t, 1-\epsilon)$$ Forgetting the lower boundary, which does not play a role in this, and replacing $1$ with $R$ for clarity, we can compute: $$ m(t, R-\epsilon) \le \lim_{h \to 0 } \frac{1}{h}\mathbb{P}(X_t \in ( R{-}\epsilon{-}h, R{-}\epsilon), M_t < R) $$ where $M_t$ is the running maximum of $X_t$. Now the joint distribution of $X_t$ and $M_t$ is well known through the reflection principle. If my calculations are correct the last term can be computed as $$ \lim_{h \to 0 } \frac{1}{h}\mathbb{P}(X_t \in ( R{-}\epsilon{-}h, R{-}\epsilon), M_t < R)= \varphi_t(R-\epsilon) - \varphi_t(R+\epsilon) $$ where $\varphi_t$ is the density of a $N(0,t)$ random variable. Taking the limit over $\epsilon$ gives the result. In the case of drift and diffusion one has to use a more general formula for the distribution density of the process and it's running maximum (unless one comes up with an intelligent trick). Such a formula can be found for example here but I was to lazy to do the calculations.

EDIT: in the spirit of finding some heuristics, I believe that one can think of the second case as follows: the probability of being near to the boundary without the maximum ever hitting the boundary vanishes, I would guess due to the fluctuations of the B.M.. It would be like asking for the probability of assuming the maximum exactly at time $t$.

Answer 2

Let me offer another viewpoint. By a Feynman-Kac formula, a solution to $$ \partial_t m(t, y) = -b \partial_y m(t,y) + \frac{1}{2} \sigma^2 \partial_y^2 m(t,y) \;, ~~ m(0,y) = m_0(y)\;, ~~ m(t,0) = m(t,1) = 0\;, $$ admits the following stochastic representation $$ m(t, y) = \mathbb{E}_y \{ 1_{\{T \ge t\}} m_0(Z_t) \} $$ where $\mathbb{E}_y$ is an expected value over the process which satisfies $$ Z_t = y - b t + \sigma W_t $$ and $T$ is the first exit time of $Z$ from $(0,1)$. Alternatively, one can write $m(t,y)$ as $$ m(t, y) = \mathbb{E}_y \{ m_0(Z_t) \mid T \ge t \} \mathbb{P}_y(T \ge t) $$ Due to the truncation of $X$, there does not seem to be a simple relation between this $m(t,y)$ and the transition density of $Y$.

Add

In this addendum, I briefly explain how $m(t,y)$ is related to the transition density of $Z$ in the unbounded case where, as @kenneth states, $$ m(t, y) = \mathbb{E}_y m_0(Z_t) $$ satisfies $$ \partial_t m(t, y) = -b \partial_y m(t,y) + \frac{1}{2} \sigma^2 \partial_y^2 m(t,y) \;, ~~ m(0,y) = m_0(y)\;. $$ To obtain the corresponding Fokker-Planck equation for the transition density $p(t,y,\xi)$ of the process $Z$, one "differentiates under the integral sign" and integrates by parts to get \begin{align*} \partial_t \mathbb{E}_y m_0(Z_t) &= \int_{\mathbb{R}} m_0(\xi) \partial_t p(t,y,\xi) d\xi \\ &= \int_{\mathbb{R}} \left( - b m_0'(\xi) + \frac{1}{2} \sigma^2 m_0''(\xi) \right) p(t,y,\xi) d\xi \\ &= \int_{\mathbb{R}} m_0(\xi) \left( b \partial_{\xi} p(t,y,\xi) + \frac{1}{2} \sigma^2 \partial_{\xi}^2 p(t,y,\xi) \right) d\xi \end{align*} Hence, we get $$ \int_{\mathbb{R}} m_0(\xi) \left( - \partial_t p(t,y,\xi) + b \partial_{\xi} p(t,y,\xi) + \frac{1}{2} \sigma^2 \partial_{\xi}^2 p(t,y,\xi) \right) d\xi = 0 $$ If this holds for all $m_0 \in C_0^2(\mathbb{R})$, then one gets the Fokker-Planck equation $$ \partial_t p(t,y,\xi) = b \partial_{\xi} p(t,y,\xi) + \frac{1}{2} \sigma^2 \partial_{\xi}^2 p(t,y,\xi) $$ If I'm not mistaken, this procedure can't be repeated in the bounded case under consideration.