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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 $\mathbb P(X_0 \in dy) = m_0(y)$, then it's standard that the density $m(t, y) := \mathbb P(X_t \in dy)$ 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 $m(t, y) := \mathbb P(Y_t \in dy)$ for $y\in (0,1)$ solve the following equation? $$\partial_t m = G^* m \hbox{ on } (0, \infty) \times (0, 1), \quad m(0, y) = m_0(y), \quad m(t, 0) = m(t, 1) = 0.$$

[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.

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.

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 $\mathbb P(X_0 \in dy) = m_0(y)$, then it's standard that the density $m(t, y) := \mathbb P(X_t \in dy)$ 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 $m(t, y) := \mathbb P(Y_t \in dy)$ for $y\in (0,1)$ solve the following equation? $$\partial_t m = G^* m \hbox{ on } (0, \infty) \times (0, 1), \quad m(0, y) = m_0(y), \quad m(t, 0) = m(t, 1) = 0.$$

[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 $m(t, 1)$.

Thanks.

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 $\mathbb P(X_0 \in dy) = m_0(y)$, then it's standard that the density $m(t, y) := \mathbb P(X_t \in dy)$ 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 $m(t, y) := \mathbb P(Y_t \in dy)$ for $y\in (0,1)$ solve the following equation? $$\partial_t m = G^* m \hbox{ on } (0, \infty) \times (0, 1), \quad m(0, y) = m_0(y), \quad m(t, 0) = m(t, 1) = 0.$$

[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.

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 $\mathbb P(X_0 \in dy) = m_0(y)$, then it's standard that the density $m(t, y) := \mathbb P(X_t \in dy)$ 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 $m(t, y) := \mathbb P(Y_t \in dy)$ for $y\in (0,1)$ solve the following equation? $$\partial_t m = G^* m \hbox{ on } (0, \infty) \times (0, 1), \quad m(0, y) = m_0(y), \quad m(t, 0) = m(t, 1) = 0.$$

Thanks.

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 $\mathbb P(X_0 \in dy) = m_0(y)$, then it's standard that the density $m(t, y) := \mathbb P(X_t \in dy)$ 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 $m(t, y) := \mathbb P(Y_t \in dy)$ for $y\in (0,1)$ solve the following equation? $$\partial_t m = G^* m \hbox{ on } (0, \infty) \times (0, 1), \quad m(0, y) = m_0(y), \quad m(t, 0) = m(t, 1) = 0.$$

[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 $m(t, 1)$.

Thanks.