# Uniqueness of the solution to some SDE

Consider the stochastic differential equation as follows:

$$X_t=X_0+t+\int_0^t\frac{dW_s}{1+m(s)},\quad \forall t\ge 0,~~~~~~~~~~~~~~~(\ast)$$

where $$X_0>0$$ is square integrable and $$m(t)=\mathbb P[\inf_{0\le s\le t}X_s>0]$$ for all $$t\ge 0$$. Can we prove that $$(\ast)$$ admits at most one solution $$(X,m)$$ (assuming its existence)?

• Are you assuming that $X_0$ and $(W_t)$ are independent? Dec 14, 2021 at 15:54
• @IosifPinelis Yes. They are independent Dec 14, 2021 at 15:56

$$\newcommand{\F}{\mathcal{F}}\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}$$According to a comment by the OP, $$X_0$$ and $$(W_t)$$ are independent. So, without loss of generality (wlog), $$X_0$$ is a real number $$x_0>0$$.

Let $$\F$$ denote the set of all nonincreasing functions from $$[0,\infty)$$ to $$[0,1]$$. Define the (nonlinear) operator $$F$$ from $$\F$$ to $$\F$$ as follows: for each $$f\in\F$$ and each real $$t\ge0$$, $$\begin{equation*} F(f)(t):=P(\inf_{s\in[0,t]}X^f_s>0), \tag{1} \end{equation*}$$ where $$\begin{equation*} X^f_t:=x_0+t+\int_0^t\frac{dW_s}{1+f(s)}. \tag{2} \end{equation*}$$ We have to show that the equation $$\begin{equation*} F(f)=f \tag{3} \end{equation*}$$ has at most one solution $$f$$ in $$\F$$.

Before doing this, let us prove, for completeness, the existence of a solution of (3). Here we just detail the argument in the comment by user GJC20 on this previous answer. Let $$f_0:=0$$ and $$f_n:=F(f_{n-1})$$ for all natural $$n$$. Then $$f_1=F(f_0)\ge0=f_0$$ and hence, by that previous answer, $$f_n\ge f_{n-1}$$ for all natural $$n$$. So, the uniformly bounded sequence $$(f_n)$$ converges (as $$n\to\infty$$) pointwise to some $$\bar f\in\F$$, which implies $$f_{n+1}=F(f_n)\to F(\bar f)$$ pointwise. So, $$F(\bar f)=\bar f$$, that is, $$\bar f$$ is a solution of (3).

To prove that (3) has at most one solution in $$\F$$, suppose that functions $$f_1$$ and $$f_2$$ in $$\F$$ are solutions of (3). Let $$\begin{equation*} t_0:=\sup\{t\in[0,\infty)\colon f_1(s)=f_2(s)\ \forall s\in[0,t)\}. \tag{4} \end{equation*}$$ Then wlog $$t_0<\infty$$. Take any $$h\in(0,1)$$. To obtain a contradiction, suppose that
$$\begin{equation*} \ep:=\|f_1-f_2\|>0, \tag{5} \end{equation*}$$ where $$\|g\|:=\sup\{|g(t)|\colon t\in[0,t_0+h]\}$$.

The crucial point is to use the same kind of time change as in the mentioned previous answer: The process $$\begin{equation*} \text{(X^{f_i}_t) equals the process (x_0+t+W_{\tau_i(t)}) in distribution,} \tag{*} \end{equation*}$$ where $$\begin{equation*} \tau_i(t):=\int_0^t\frac{ds}{(1+f_i(s))^2}; \tag{6} \end{equation*}$$ here and in what follows, $$i\in\{1,2\}$$.

Note that $$\frac1{(1+z)^2}$$ is $$2$$-Lipschitz in $$z\ge0$$. Therefore and in view of (6), (4), and (5), $$\begin{equation*} |\tau_1(t)-\tau_2(t)|\le\int_{t_0}^{\max(t_0,t)} ds\,\Big|\frac1{(1+f_1(s))^2}-\frac1{(1+f_2(s))^2}\Big| \le 2h\ep \tag{7} \end{equation*}$$ for all $$t\in[0,t_0+h]$$.

Since $$\frac14\le\frac1{(1+f_i)^2}\le1$$, the functions $$\tau_i$$ are Lipschitz-continuous and strictly increasing on $$[0,\infty)$$ from $$\tau_i(0)=0$$ to $$\tau_i(\infty-)=\infty$$, and the inverse functions $$\tau_i^{-1}$$ are defined, strictly increasing, and $$4$$-Lipschitz on $$[0,\infty)$$. It follows that for all $$u\in[0,\tau_2(t_0+h)]$$ $$\begin{equation*} |\tau_1^{-1}(u)-\tau_1^{-1}(u)|\le4\sup_{t\in[0,t_0+h]}|\tau_1(t)-\tau_2(t)|\le8h\ep, \tag{8} \end{equation*}$$ in view of (7).

By (*), for all real $$t\ge0$$, $$\begin{equation*} F(f_1)(t)-F(f_2)(t)=D_1(t)+D_2(t), \tag{9} \end{equation*}$$ where \begin{equation*} \begin{aligned} D_1(t)&:=P(\inf_{u\in[0,\tau_1(t)]}(x_0+\tau_1^{-1}(u)+W_u)>0) \\ &-P(\inf_{u\in[0,\tau_2(t)]}(x_0+\tau_1^{-1}(u)+W_u)>0) \end{aligned} \end{equation*} and \begin{equation*} \begin{aligned} D_2(t)&:=P(\inf_{u\in[0,\tau_2(t)]}(x_0+\tau_1^{-1}(u)+W_u)>0) \\ &-P(\inf_{u\in[0,\tau_2(t)]}(x_0+\tau_2^{-1}(u)+W_u)>0). \end{aligned} \end{equation*}

Using e.g. (as an overkill) Lemma 8, p. 407 (or its Russian original, Lemma 8, p. 423) and (8), we get $$\begin{equation*} \|D_2\|\le C\Big(1+\frac1{\sqrt{t_0+h}}\Big)h\ep\le2C\sqrt h\,\ep, \tag{10} \end{equation*}$$ where $$C$$ is some universal positive real constant.

Let $$\tau_{\min}(t):=\min(\tau_1(t),\tau_2(t))$$ and $$\tau_{\max}(t):=\max(\tau_1(t),\tau_2(t))$$. Take any $$t\in[0,t_0+h]$$ and then let $$u_1:=\tau_{\min}(t)$$, $$\de:=\tau_{\max}(t)-\tau_{\min}(t)$$, $$x_1:=x_0+\tau_1^{-1}(u_1)$$, and $$G:=1-\Phi$$, where $$\Phi$$ is the standard normal cdf. Let $$G\big(\frac{x_1}{\sqrt{u_1}}\big):=0$$ if $$u_1=0$$. Then \begin{equation*} \begin{aligned} &|D_1(t)| \\ &=P(\inf_{u\in[0,\tau_{\min}(t)]}(x_0+\tau_1^{-1}(u)+W_u)>0) \\ &-P(\inf_{u\in[0,\tau_{\max}(t)]}(x_0+\tau_1^{-1}(u)+W_u)>0) \\ &=P(\inf_{u\in[0,\tau_{\min}(t)]}(x_0+\tau_1^{-1}(u)+W_u)>0, \\ &\inf_{u\in(\tau_{\min}(t),\tau_{\max}(t)]}(x_0+\tau_1^{-1}(u)+W_u)\le0) \\ &\le P(\inf_{u\in[0,u_1]}W_u>-x_1,\inf_{u\in(u_1,u_1+\de]}W_u\le-x_1) \\ &=P(\inf_{u\in[0,u_1+\de]}W_u\le-x_1) -P(\inf_{u\in[0,u_1]}W_u\le-x_1) \\ &=2G\Big(\frac{x_1}{\sqrt{u_1+\de}}\Big)-2G\Big(\frac{x_1}{\sqrt{u_1}}\Big) \\ &\le\frac\de{x_1^2}\le\frac\de{x_0^2}\le\frac{2h\ep}{x_0^2}. \end{aligned} \tag{11} \end{equation*} The first inequality in (11) follows because the function $$\tau_1^{-1}$$ is increasing and $$x_0+\tau_1^{-1}(u_1)=x_1$$. The fourth, last equality in (11) follows by the reflection principle.
The second inequality there follows because $$\frac{\partial}{\partial u}G\big(\frac x{\sqrt u}\big)\le\frac1{2x^2}$$ all real $$x>0$$ and $$u>0$$. The last inequality in (11) follows by (7), since $$\de=\tau_{\max}(t)-\tau_{\min}(t)$$.

Collecting (5), (9), (11), and (10), for any real $$h\in(0,1)$$ such that $$\frac{2h}{x_0^2}+2C\sqrt h<1$$, we have $$\begin{equation*} \ep=\|f_1-f_2\|=\|F(f_1)-F(f_2)\|\le\|D_1\|+\|D_2\|\le\Big(\frac{2h}{x_0^2}+2C\sqrt h\Big)\ep<\ep, \end{equation*}$$ which is the mentioned desired contradiction with (5).

So, $$\ep=0$$, which means that $$f_1(s)=f_2(s)\ \forall s\in[0,t_0+h)$$, which contradicts (4). This final contradiction shows that $$t_0=\infty$$ in (4), and we are done.

• Thank you GJC20 for your questions and comments, and @Iosif Pinelis for your answers. I am learning a lot by reading them. Dec 15, 2021 at 3:32
• @NateRiver : Thank you for your comment. Dec 15, 2021 at 3:59
• @IosifPinelis Thank you so much for the answer which I find amazing. I just have one question concerning $t_0$. Can we show $t_0>0$ here? If not, I think it is better to use this definition $t_0:=\{t\in [0,\infty): f_1(s)=f_2(s),~ \forall s\in [0,t_0]\}$. Do you think so? Dec 15, 2021 at 9:06
• @GJC20 : Thank you for your comment. I think the definition $t_0:=\sup\{t\in[0,\infty)\colon f_1(s)=f_2(s)\ \forall s\in[0,t)\}$ is OK, as well as the definition $t_0:=\sup\{t\in[0,\infty)\colon f_1(s)=f_2(s)\ \forall s\in[0,t]\}$ would be. If there is no $t>0$ such that $f_1(s)=f_2(s)\ \forall s\in[0,t)$, then $t_0=0$ according to either one of these two definitions, because $f_1(s)=f_2(s)\ \forall s\in[0,0)$. Also, I have corrected the bounding of $D_1$. Dec 15, 2021 at 17:46
• @IosifPinelis Thanks again for your time. This question is related to one ongoing project. I will email you as soon as I finish (if you don't mind) Dec 15, 2021 at 18:44