Update: Thanks to GJC20's answer on the existence and uniqueness. Let me reformulate my questions 3/4 as follows: There exists a unique non-increasing and continuously differentiable function $f:\mathbb R \to [0,1]$ s.t.

$$X_t=X_0+W_t-1+f(t) \mbox{ and } f(t)= \mathbb E \left[ \exp\left(-\frac 1 \epsilon \int_0^t X_s^-ds\right)\right],\quad \forall t\ge 0.$$

What is a characterization for $f$?


$$X_t=X_0+ W_t-1 + \mathbb E \left[ \exp\left(-\frac 1 \epsilon \int_0^t X_s^-ds\right)\right],\quad \forall t\ge 0,\quad\quad\quad(\ast)$$

where $X_0>0$ is a random variable as nice as possible, $(W_t)_{t\ge 0}$ is an independent Brownian motion, $\epsilon>0$ is fixed and can be as small as possible.

  1. Has this equation been studied (I don't even know how to call such equation)?
  2. Do we have the existence/uniqueness result?
  3. If $X_t$ has a density function, denoted by $f(t,x)$, is there any analytic equation for $f$?
  4. Let $\tau:=\inf\{t\ge 0: X_t \le 0\}$. If $X_t{\bf 1}_{\{\tau>t\}}$ has a density function, denoted by $g(t,x)$, is there any analytic equation for $g$?

Many thanks for the answer, comments and references.

PS : Here I adopt the notation $a^-:=\max(-a, 0)$ for all $a\in \mathbb R$.

  • $\begingroup$ To clarify, am I reading correctly that you have a deterministic function $\{Y_t\}_t$ such that $X$ satisfies $X_t-X_0=W_t-1+Y_t$? $\endgroup$ Feb 21, 2022 at 17:17
  • $\begingroup$ Yes. An equivalent formulation is to find out $Y_t$ s.t. $X_t-X_0=W_t-1+Y_t$ and $Y_t=\mathbb E[\exp(-\int_0^t X^-_sds/\epsilon)]$ $\endgroup$
    – user420828
    Feb 21, 2022 at 17:47

1 Answer 1


Updates : This is an answer to the simplest question above, existence and uniqueness.

If the existence holds, then any solution must be a strong solution. Let $X, Y$ be two arbitrary solution, it appears that $|X_t-Y_t|$ is deterministic and satisfies further

$$\mathbb E|X_t-Y_t| = |X_t-Y_t| \le \mathbb E\left[\frac{1}{\epsilon}\int_0^t |X^-_s-Y^-_s|ds\right] \le \frac{1}{\epsilon}\int_0^t \mathbb E[|X_s-Y_s|]ds,\quad \forall t\ge 0,$$

which yields the uniqueness by Gronwall's inequality.

In the following, let us show the existence. To do so, we adopt the argument of fixed point. Let $C$ be the space of continous functions on $\mathbb R_+$ and $C_{\preceq}\subset C$ be the subset of non-increasing functions $f$ s.t. $f(0)=1$ and $\inf_{t\ge 0} f(t)\ge 0$. Define further the operator $\Gamma: C_{\preceq}\to C_{\preceq}$ by

$$\Gamma[f](t):=\mathbb E\left[\exp\left(-\frac{1}{\epsilon}\int_0^t\big(X^f_s\big)^-ds\right)\right],\quad \forall t\ge 0,$$

where $X^f_t:=X_0+W_t-1+f(t)$. Clearly $\Gamma[f]\in C_{\preceq}$ for every $f\in C_{\preceq}$. Then it suffices to show $\Gamma$ has a fixed point in $C_{\preceq}$. First, we observe that $\Gamma$ is monotone in the following sense: if $f\preceq g$, then $\Gamma[f]\preceq \Gamma[g]$. Here we say $f\preceq g$ iff $f(t)\le g(t)$ for all $t\ge 0$. Next, set $f_0\equiv 1\in C_{\preceq}$ and define $f_{n+1}:=\Gamma[f_n]$ for every $n\ge 0$. By induction, one has that $n\mapsto f_n(t)$ is non-increasing for all $t\ge 0$. Thus the pointwise $f(t):=\lim_{n\to\infty}f_n(t)$ exists. Finally, using the equality

$$f_{n+1}(t)=\mathbb E\left[\exp\left(-\frac{1}{\epsilon}\int_0^t\big(X_0+W_s-1+f_n(s)\big)^-ds\right)\right]$$

and the dominated convergence theorem, we may conclude the desired existence.

  • $\begingroup$ Really nice arguments. Thanks for the answer. Do you know the analytic characterization by chance? $\endgroup$
    – user420828
    Feb 21, 2022 at 18:07
  • $\begingroup$ @Philo18 My pleasure. Unfortunately I don't know any references on your questions 3 & 4 $\endgroup$
    – GJC20
    Feb 21, 2022 at 18:08

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