# Has this "stochastic differential equation" been studied?

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 $$f$$?

Consider

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

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 $$a\in \mathbb R$$.

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

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.

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