Let $(\Omega,\mathcal F, \mathbb P)$ be a probability space on which a Brownian motion $W$ is defined, and $\mathcal U$ be the set of progressively measurable (w.r.t. the Brownian filtration) processes $p=(p_t)_{t\ge 0}$ taking values in $\mathbb R_+$ s.t.

$$\mathbb E\left[\int_0^T p_t dt\right]<\infty,\quad \forall T\ge 0. $$

For each $n\ge 1$, denote by $\mathcal U_n\subset \mathcal U$ the subset of $p=(p_t)_{t\ge 0}$ taking values in $[1/n,n]$. Consider the stochastic control problems: for $t\in [0,1]$ and $x\in [0,1]$,

\begin{eqnarray} v_n(t,x):=\sup_{p\in \mathcal U_n}\mathbb E\left[\int_t^{\min(1,\tau^{p,t,x})}\big(1+\log(p_s)\big)ds\right] \\ v(t,x):=\sup_{p\in \mathcal U}\mathbb E\left[\int_t^{\min(1,\tau^{p,t,x})}\big(1+\log(p_s)\big)ds\right], \end{eqnarray}

where $\tau^{p,t,x}:=\{s\ge t: X^{p,t,x}_s\notin (0,1)\}$ and $dX^{p,t,x}_s=\sqrt{2p_s}dW_s$ for all $s\ge t$ with $X^{p,t,x}_t:=x$.

I have two questions :

1. Does the pointwise convergence $v_n\to v$ hold?
2. Does there exist $N$ large enough s.t. $v_n=v_N$ for all $n\ge N$?

PS : If $p$ is a constant control, i.e. $p\equiv u$ for some $u>0$, then one has (by Prob 8.14 of Karatzas and Shreve)

$$\mathbb E[\tau^{p,t,x}-t]=\frac{x(1-x)}{2u}.$$

As $u\to 0+$,

$$\mathbb E\left[\int_t^{\min(1,\tau^{p,t,x})}\big(1+\log(u)\big)du\right]\approx (1-t)\big(1+\log(u)\big)\to -\infty;$$

and as $u\to\infty$

$$\mathbb E\left[\int_t^{\min(1,\tau^{p,t,x})}\big(1+\log(u)\big)du\right]\approx \mathbb E[\tau^{p,t,x}-t]\big(1+\log(u)\big) = \frac{x(1-x)\big(1+\log(u)\big)}{2u}\to 0.$$

This illustrates in some sens when the diffusion coefficient is too small or too large, it can never be optimal (a straightforward computation shows that $u=2$ yields a strictly positive expectation).

Let $(\Omega,\mathcal F, \mathbb P)$ be a probability space on which a Brownian motion $W$ is defined, and $\mathcal U$ be the set of progressively measurable (w.r.t. the Brownian filtration) processes $p=(p_t)_{t\ge 0}$ taking values in $\mathbb R_+$ s.t.

$$\mathbb E\left[\int_0^T p_t dt\right]<\infty,\quad \forall T\ge 0. $$

For each $n\ge 1$, denote by $\mathcal U_n\subset \mathcal U$ the subset of $p=(p_t)_{t\ge 0}$ taking values in $[1/n,\infty)$. Define, for $p\in\mathcal U$, $t\in [0,1]$ and $x\in [0,1]$,

\begin{eqnarray} J(t,x,p):=\mathbb E\left[\int_t^{\min(1,\tau^{p,t,x})}\big(1+\log(p_s)\big)ds\right] \end{eqnarray}

where $\tau^{p,t,x}:=\{s\ge t: X^{p,t,x}_s\notin (0,1)\}$ and $dX^{p,t,x}_s=\sqrt{2p_s}dW_s$ for all $s\ge t$ with $X^{p,t,x}_t:=x$. Can we prove the existence of some $N$ s.t. for any $p\in\mathcal U$, there exists $q\in\mathcal U_N$ satisfying $J(t,x,p)\le J(t,x,q)$?

Let $(\Omega,\mathcal F, \mathbb P)$ be a probability space on which a Brownian motion $W$ is defined, and $\mathcal U$ be the set of progressively measurable (w.r.t. the Brownian filtration) processes $p=(p_t)_{t\ge 0}$ taking values in $\mathbb R_+$ s.t.

$$\mathbb E\left[\int_0^T p_t dt\right]<\infty,\quad \forall T\ge 0. $$

For each $n\ge 1$, denote by $\mathcal U_n\subset \mathcal U$ the subset of $p=(p_t)_{t\ge 0}$ taking values in $[1/n,n]$. Consider the stochastic control problems: for $t\in [0,1]$ and $x\in [0,1]$,

\begin{eqnarray} v_n(t,x):=\sup_{p\in \mathcal U_n}\mathbb E\left[\int_t^{\min(1,\tau^{p,t,x})}\big(1+\log(p_s)\big)ds\right] \\ v(t,x):=\sup_{p\in \mathcal U}\mathbb E\left[\int_t^{\min(1,\tau^{p,t,x})}\big(1+\log(p_s)\big)ds\right], \end{eqnarray}

where $\tau^{p,t,x}:=\{s\ge t: X^{p,t,x}_s\notin (0,1)\}$ and $dX^{p,t,x}_s=\sqrt{2p_s}dW_s$ for all $s\ge t$ with $X^{p,t,x}_t:=x$.

I have two questions :

1. Does the pointwise convergence $v_n\to v$ hold?
2. Does there exist $N$ large enough s.t. $v_n=v_N$ for all $n\ge N$?

PS : If $p$ is a constant control, i.e. $p\equiv u$ for some $u>0$, then one has (by Prob 8.14 of Karatzas and Shreve)

$$\mathbb E[\tau^{p,t,x}-t]=\frac{x(1-x)}{2u}.$$

As $u\to 0+$,

$$\mathbb E\left[\int_t^{\min(1,\tau^{p,t,x})}\big(1+\log(u)\big)du\right]\approx (1-t)\big(1+\log(u)\big)\to -\infty;$$

and as $u\to\infty$

$$\mathbb E\left[\int_t^{\min(1,\tau^{p,t,x})}\big(1+\log(u)\big)du\right]\approx \mathbb E[\tau^{p,t,x}-t]\big(1+\log(u)\big) = \frac{x(1-x)\big(1+\log(u)\big)}{u^2}\to 0.$$

This illustrates in some sens when the diffusion coefficient is too small or too large, it can never be optimal (a straightforward computation shows that $u=2$ yields a strictly positive expectation).

Let $(\Omega,\mathcal F, \mathbb P)$ be a probability space on which a Brownian motion $W$ is defined, and $\mathcal U$ be the set of progressively measurable (w.r.t. the Brownian filtration) processes $p=(p_t)_{t\ge 0}$ taking values in $\mathbb R_+$ s.t.

$$\mathbb E\left[\int_0^T p_t dt\right]<\infty,\quad \forall T\ge 0. $$

For each $n\ge 1$, denote by $\mathcal U_n\subset \mathcal U$ the subset of $p=(p_t)_{t\ge 0}$ taking values in $[1/n,n]$. Consider the stochastic control problems: for $t\in [0,1]$ and $x\in [0,1]$,

\begin{eqnarray} v_n(t,x):=\sup_{p\in \mathcal U_n}\mathbb E\left[\int_t^{\min(1,\tau^{p,t,x})}\big(1+\log(p_s)\big)ds\right] \\ v(t,x):=\sup_{p\in \mathcal U}\mathbb E\left[\int_t^{\min(1,\tau^{p,t,x})}\big(1+\log(p_s)\big)ds\right], \end{eqnarray}

where $\tau^{p,t,x}:=\{s\ge t: X^{p,t,x}_s\notin (0,1)\}$ and $dX^{p,t,x}_s=\sqrt{2p_s}dW_s$ for all $s\ge t$ with $X^{p,t,x}_t:=x$.

I have two questions :

1. Does the pointwise convergence $v_n\to v$ hold?
2. Does there exist $N$ large enough s.t. $v_n=v_N$ for all $n\ge N$?

PS : If $p$ is a constant control, i.e. $p\equiv u$ for some $u>0$, then one has (by Prob 8.14 of Karatzas and Shreve)

$$\mathbb E[\tau^{p,t,x}-t]=\frac{x(1-x)}{2u}.$$

As $u\to 0+$,

$$\mathbb E\left[\int_t^{\min(1,\tau^{p,t,x})}\big(1+\log(u)\big)du\right]\approx (1-t)\big(1+\log(u)\big)\to -\infty;$$

and as $u\to\infty$

$$\mathbb E\left[\int_t^{\min(1,\tau^{p,t,x})}\big(1+\log(u)\big)du\right]\approx \mathbb E[\tau^{p,t,x}-t]\big(1+\log(u)\big) = \frac{x(1-x)\big(1+\log(u)\big)}{2u}\to 0.$$

This illustrates in some sens when the diffusion coefficient is too small or too large, it can never be optimal (a straightforward computation shows that $u=2$ yields a strictly positive expectation).

Let $(\Omega,\mathcal F, \mathbb P)$ be a probability space on which a Brownian motion $W$ is defined, and $\mathcal U$ be the set of progressively measurable (w.r.t. the Brownian filtration) processes $p=(p_t)_{t\ge 0}$ taking values in $\mathbb R_+$ s.t.

$$\mathbb E\left[\int_0^T p_t dt\right]<\infty,\quad \forall T\ge 0. $$

For each $n\ge 1$, denote by $\mathcal U_n\subset \mathcal U$ the subset of $p=(p_t)_{t\ge 0}$ taking values in $[1/n,n]$. Consider the stochastic control problems: for $t\in [0,1]$ and $x\in [0,1]$,

\begin{eqnarray} v_n(t,x):=\sup_{p\in \mathcal U_n}\mathbb E\left[\int_t^{\min(1,\tau^{p,t,x})}\big(1+\log(p_s)\big)du\right] \\ v(t,x):=\sup_{p\in \mathcal U}\mathbb E\left[\int_t^{\min(1,\tau^{p,t,x})}\big(1+\log(p_s)\big)du\right], \end{eqnarray}

where $\tau^{p,t,x}:=\{s\ge t: X^{p,t,x}_s\notin (0,1)\}$ and $dX^{p,t,x}_s=\sqrt{2p_s}dW_s$ for all $s\ge t$ with $X^{p,t,x}_t:=x$.

I have two questions :

1. Does the pointwise convergence $v_n\to v$ hold?
2. Does there exist $N$ large enough s.t. $v_n=v_N$ for all $n\ge N$?

PS : If $p$ is a constant control, i.e. $p\equiv u$ for some $u>0$, then one has (by Prob 8.14 of Karatzas and Shreve)

$$\mathbb E[\tau^{p,t,x}-t]=\frac{x(1-x)}{2u}.$$

As $u\to 0+$,

$$\mathbb E\left[\int_t^{\min(1,\tau^{p,t,x})}\big(1+\log(u)\big)du\right]\approx (1-t)\big(1+\log(u)\big)\to -\infty;$$

and as $u\to\infty$

$$\mathbb E\left[\int_t^{\min(1,\tau^{p,t,x})}\big(1+\log(u)\big)du\right]\approx \mathbb E[\tau^{p,t,x}-t]\big(1+\log(u)\big) = \frac{x(1-x)\big(1+\log(u)\big)}{u^2}\to 0.$$

This illustrates in some sens when the diffusion coefficient is too small or too large, it can never be optimal (a straightforward computation shows that $u=2$ yields a strictly positive expectation).

Let $(\Omega,\mathcal F, \mathbb P)$ be a probability space on which a Brownian motion $W$ is defined, and $\mathcal U$ be the set of progressively measurable (w.r.t. the Brownian filtration) processes $p=(p_t)_{t\ge 0}$ taking values in $\mathbb R_+$ s.t.

$$\mathbb E\left[\int_0^T p_t dt\right]<\infty,\quad \forall T\ge 0. $$

For each $n\ge 1$, denote by $\mathcal U_n\subset \mathcal U$ the subset of $p=(p_t)_{t\ge 0}$ taking values in $[1/n,n]$. Consider the stochastic control problems: for $t\in [0,1]$ and $x\in [0,1]$,

\begin{eqnarray} v_n(t,x):=\sup_{p\in \mathcal U_n}\mathbb E\left[\int_t^{\min(1,\tau^{p,t,x})}\big(1+\log(p_s)\big)ds\right] \\ v(t,x):=\sup_{p\in \mathcal U}\mathbb E\left[\int_t^{\min(1,\tau^{p,t,x})}\big(1+\log(p_s)\big)ds\right], \end{eqnarray}

where $\tau^{p,t,x}:=\{s\ge t: X^{p,t,x}_s\notin (0,1)\}$ and $dX^{p,t,x}_s=\sqrt{2p_s}dW_s$ for all $s\ge t$ with $X^{p,t,x}_t:=x$.

I have two questions :

1. Does the pointwise convergence $v_n\to v$ hold?
2. Does there exist $N$ large enough s.t. $v_n=v_N$ for all $n\ge N$?

PS : If $p$ is a constant control, i.e. $p\equiv u$ for some $u>0$, then one has (by Prob 8.14 of Karatzas and Shreve)

$$\mathbb E[\tau^{p,t,x}-t]=\frac{x(1-x)}{2u}.$$

As $u\to 0+$,

$$\mathbb E\left[\int_t^{\min(1,\tau^{p,t,x})}\big(1+\log(u)\big)du\right]\approx (1-t)\big(1+\log(u)\big)\to -\infty;$$

and as $u\to\infty$

$$\mathbb E\left[\int_t^{\min(1,\tau^{p,t,x})}\big(1+\log(u)\big)du\right]\approx \mathbb E[\tau^{p,t,x}-t]\big(1+\log(u)\big) = \frac{x(1-x)\big(1+\log(u)\big)}{u^2}\to 0.$$

This illustrates in some sens when the diffusion coefficient is too small or too large, it can never be optimal (a straightforward computation shows that $u=2$ yields a strictly positive expectation).