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.



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 $t$.

  • $\begingroup$ Thanks for your reply. I agree with your first comment that the equation does not provide any probability of absorbing states, see the added remark. The second statement of your answer seems to me to support the equation itself. $\endgroup$ – kenneth Nov 7 '17 at 2:40
  • $\begingroup$ Yes, I agree with you, the second statement is supposed to be the answer to your question, in the case $b=0, \sigma =1$. The first statement was just to make things clear. If you want a reference, you might be interested in Section 40.3 of "Stochastic Processes" by Richard Bass. $\endgroup$ – Kore-N Nov 7 '17 at 12:40
  • $\begingroup$ I've checked Bass's book, but it does not seem to directly answer this question. Especially, why the boundary shall be set zero is not explained. $\endgroup$ – kenneth Nov 8 '17 at 12:00
  • $\begingroup$ In my opinion it addresses exactly your issue. Indeed the $p^0$ he finds is the $m$ in your notation. Since he has an explicit description of $p^0$ in terms of a series expansion you can directly see that the boundary conditions are verified (and maybe you can generalize this series expansion to other situations). You also have the stochastic representation as $m(t,y) = \mathbb{P}(Y_t = y, t < T),$ which tells you that up to a normalizing constant $m(t, \cdot)$ is the transition probability conditioned on the hitting time being larger than $t$ $\endgroup$ – Kore-N Nov 8 '17 at 12:53
  • $\begingroup$ The same author addresses similar problems also in "Diffusions and Elliptic Operators". $\endgroup$ – Kore-N Nov 8 '17 at 12:55

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 $Y$.


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.

  • $\begingroup$ Thanks for your reply, let me think about it and reply you later. $\endgroup$ – kenneth Nov 7 '17 at 2:38
  • $\begingroup$ Thanks for your idea. I could not be completely convinced by this argument. At least I believe the equation itself is correct in $(0, \infty) \times (0, 1)$ for its local property of the diffusion. $\endgroup$ – kenneth Nov 8 '17 at 11:58
  • $\begingroup$ @kenneth A local solution of the PDE in m does have a stochastic representation, which is given in my answer. It, however, has no immediate relation to the representation you expect. $\endgroup$ – Nawaf Bou-Rabee Nov 8 '17 at 13:02
  • $\begingroup$ Indeed your suggestion is quite fresh to me. But I do not see how this argument implies that the PDE solution $m$ is not the desired density. For instance, if we consider the solution $m$ of the first Fokker-Planck equation for the density of $X$ in the above, one can write $m(t, y) = \mathbb E_y [ m_0(Z_t) ]$ by Feymann-Kac formula. However, I do not see how $m$ relates to the density of $X_t$ from this representation, although it's the fact. $\endgroup$ – kenneth Nov 8 '17 at 13:36
  • $\begingroup$ @kenneth I added some background material. I hope it helps. $\endgroup$ – Nawaf Bou-Rabee Nov 8 '17 at 15:08

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