# A mean field SDE with hitting time

Let $$b\in \mathbb R$$ and $$\sigma>0$$ be given. For a fixed probability distribution $$\mu_0$$ on $$\mathbb R$$ s.t.

$$\int_{(0,\infty)}\mu_0(dx)=1,$$

consider the mean field SDE :

$$dX_t = \mathbf{1}_{\{X_t>0\}} \left[ bdt + \frac{\sigma}{1 + m_t {\bf 1}_{\{b>0\}}} dW_t \right],\quad \mbox{for all } t\ge 0,~~~~~~~~~~~~~~~~(\ast)$$

where $$X_0\sim \mu_0$$ is independent of the Brownian motion $$(W_t)_{t\ge 0}$$ and

$$m_t:=\int_{(0,\infty)}\mu_t(dx),\quad \mbox{for all } t\ge 0.$$

How can we show the existence and uniqueness of the (weak) solution to $$(\ast)$$?

Any answers, remarks or references are highly appreciated!

REMARK :

The case for $$b\le 0$$ is trivial. Indeed, $$(\ast)$$ reduces to $$dX_t = \mathbf{1}_{\{X_t>0\}} \big[ bdt + \sigma dW_t \big]$$ and the solution is given as $$X_t=Y_{t\wedge \tau}$$, where $$Y_t:=X_0+bt+\sigma W_t$$ and $$\tau:=\inf\{t\ge 0: Y_t\le 0\}$$. For the case $$b>0$$, $$(\ast)$$ turns to be

$$dX_t = \mathbf{1}_{\{X_t>0\}} \left[ bdt + \frac{\sigma}{1 + m_t} dW_t \right].$$

I do not find any literature on the existence of its solution.

REMARK 2 :

A heuristic argument is as follows : Let $$\ell:\mathbb R_+\to [0,1]$$ be some "nice" function. Consider the process

$$Y^{\ell}_t: = X_0+ bt+\int_0^t\frac{\sigma}{1+\ell(s)}dW_s.$$

Then we have $$(Y^{\ell}_{t\wedge \tau^{\ell}})_{t\ge 0}$$ is a solution to $$(\ast)$$ if $$\mathbb P[\tau^{\ell}>t]=m_t$$ for all $$t\ge 0$$, where $$\tau^{\ell}:=\inf\{t\ge 0: Y^{\ell}_t\le 0\}$$. Thus it remains to calculate the probability $$\mathbb P[\tau^{\ell}>t]$$ in terms of $$\ell$$. Is there any reference for this computation?

## 1 Answer

This is a rough idea. Fix an arbitrary $$T>0$$. For $$0\le t\le T$$ and $$x\ge 0$$, define $$u(t,x;T):=\mathbb P[\tau^{\ell}>T|Y^{\ell}_t=x]$$. One has by definition $$u(T,x;T)={\bf 1}_{\{x>0\}}$$ for all $$x\ge 0$$ and $$u(t,x;T)=0$$ for all $$0\le t\le T$$.

For $$0\le t and $$x>0$$, it follow from the tower property that

$$u(t,x;T)=\mathbb E\big[\mathbb E[{\bf 1}_{\{\tau^{\ell}>T\}}|Y^{\ell}_{s+h}]|Y^{\ell}_t=x\big]=\mathbb E\big[u(t+h,Y^{\ell}_{t+h};T)|Y^{\ell}_t=x\big], \quad \mbox{for all } 0

Admitting the regularity of $$u$$, one has by Ito's formula

$$\frac{1}{h}\mathbb E\left[\int_{t}^{t+h}\left(\partial_tu(s,Y^{\ell}_s;T) + b\partial_xu(s,Y^{\ell}_s;T) +\frac{\sigma^2}{2(1+\ell(s))^2}\partial_{xx}^2 u(s,Y^{\ell}_s;T)\right)ds~\Big|Y^{\ell}_t=x\right]~~=~~0.$$

Letting $$h\to 0$$, it yields

$$\partial_tu(t,x;T) + b\partial_xu(t,x;T) +\frac{\sigma^2}{2(1+\ell(t))^2}\partial_{xx}^2 u(t,x;T)=0.$$

Assume $$X_0=\bar x>0$$. Then one has coupled equations:

$$$$\partial_tu(t,x;T) + b\partial_xu(t,x;T) +\frac{\sigma^2}{2(1+\ell(t))^2}\partial_{xx}^2 u(t,x;T)=0,~ \forall 0\le t0 \\ u(T,x;T)={\bf 1}_{\{x>0\}},~ \forall x\ge 0 \\ u(t,0;T) = 0,~ \forall 0\le t\le T \\ u(0,\bar x,;T) = \ell(T),~ \forall T\ge 0.$$$$