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)?

Any answer, comments or references are highly appreciated.

  • $\begingroup$ Are you assuming that $X_0$ and $(W_t)$ are independent? $\endgroup$ Dec 14, 2021 at 15:54
  • $\begingroup$ @IosifPinelis Yes. They are independent $\endgroup$
    – GJC20
    Dec 14, 2021 at 15:56

1 Answer 1


$\newcommand{\F}{\mathcal{F}}\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}$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.

  • 1
    $\begingroup$ Thank you GJC20 for your questions and comments, and @Iosif Pinelis for your answers. I am learning a lot by reading them. $\endgroup$
    – Nate River
    Dec 15, 2021 at 3:32
  • $\begingroup$ @NateRiver : Thank you for your comment. $\endgroup$ Dec 15, 2021 at 3:59
  • $\begingroup$ @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? $\endgroup$
    – GJC20
    Dec 15, 2021 at 9:06
  • $\begingroup$ @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 $D_1$. $\endgroup$ Dec 15, 2021 at 17:46
  • $\begingroup$ @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) $\endgroup$
    – GJC20
    Dec 15, 2021 at 18:44

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