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Consider the drifted Brownian motion $X_t=1+\lambda(t)+W_t$, where $\lambda: \mathbb R\to [0,\infty)$ with $1\le \lambda'(t)\le 2$ and $(W_t)_{\ge 0}$ denotes a Brownian motion. Define the hitting time

$$\tau:=\inf\{t\ge 0: X_t\le 0\}$$

and further the function $p(t):=\mathbb P[\tau>t]$ for all $t\ge 0$. Can we show that $t\mapsto p(t)$ is Lipschitz?

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  • $\begingroup$ What happens if you try to use Girsanov to remove the drift? $\endgroup$
    – Nate River
    Jun 30 at 6:44
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    $\begingroup$ @NateRiver Using Girsanov theorem, I only obtain the Holder continuity $\endgroup$
    – GJC20
    Jun 30 at 10:30
  • $\begingroup$ Isn't this easier to state without the hitting time, e.g. $p(t) = \mathbb{P}[\forall u\le t,X_u\ge0]$$? $\endgroup$
    – Matt F.
    Jul 5 at 19:54
  • $\begingroup$ @MattF. I tried it by rewritting $p(t)-p(t+\Delta t)=\mathbb P[\inf_{s\le t}X_s>0, X_t+\inf_{t\le u\le t+\Delta t}(X_s-X_t)\le 0]$, but then I don't know how to proceed $\endgroup$
    – GJC20
    Jul 5 at 20:48
  • $\begingroup$ @MattF. What I mean is that, unless using Girsanov, I don't know how to deal with the above probability $\endgroup$
    – GJC20
    Jul 5 at 20:48
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I claim this is not a complete answer. I wonder whether it can be improved to obtain the Lipschitz continuity.

Define

$$\left(\frac{d\mathbb Q}{d\mathbb P}\right)_t := \exp\left(-\int_0^t \lambda'(s)dW_s-\frac{1}{2}\int_0^t(\lambda'(s))^2ds\right).$$

Then $B_t:=W_t+\lambda(t)$ is a Brownian motion under $\mathbb Q$. Thus,

$$p(t) = \mathbb E^{\mathbb Q}\left[\left(\frac{d\mathbb P}{d\mathbb Q}\right)_{t}{\bf 1}_{\{\inf_{0\le r\le t}(1+\lambda(r)+W_{r})>0\}}\right] = \mathbb E^{\mathbb Q}\left[\left(\frac{d\mathbb P}{d\mathbb Q}\right)_{t}{\bf 1}_{\{\inf_{0\le r\le t}(1+B_{r})>0\}}\right].$$

For every $0\le t<t+\Delta t$, it holds

$$0\le p(t)-p(t+\Delta t) = \mathbb E^{\mathbb Q}\left[\left(\frac{d\mathbb P}{d\mathbb Q}\right)_{t+\Delta t}\left({\bf 1}_{\{\inf_{0\le r\le t}(1+B_{r})>0\}}-{\bf 1}_{\{\inf_{0\le r\le t+\Delta t}(1+B_{r})>0\}}\right)\right].$$

Using Holder inequality for $a, b>1$ with $1/a+1/b=1$, one has

$$p(t)-p(t+\Delta t)\le \mathbb E^{\mathbb Q}\left[\left(\frac{d\mathbb P}{d\mathbb Q}\right)_{t+\Delta t}^a\right]^{1/a} \left(\mathbb Q\left[\inf_{0\le r\le t}B_{r}>-1\right]-\mathbb Q\left[\inf_{0\le r\le t+\Delta t}B_{r}>-1\right]\right)^{1/b}.$$

Let $\Phi:\mathbb R\to (0,1)$ be the CDF defined by

$$\Phi(x):=\int_{-\infty}^x\frac{1}{\sqrt{2\pi}}e^{-z^2/2}dz.$$

Then we finally obtain

$$p(t)-p(t+\Delta t)\le \mathbb E^{\mathbb Q}\left[\left(\frac{d\mathbb P}{d\mathbb Q}\right)_{t+\Delta t}^a\right]^{1/a}2^{1/b} \left( \Phi\left(\frac{1}{\sqrt{t}}\right)-\Phi\left(\frac{1}{\sqrt{t+\Delta t}}\right)\right)^{1/b}.$$

It follows from some tedious computation, we have

$$ \mathbb E^{\mathbb Q}\left[\left(\frac{d\mathbb P}{d\mathbb Q}\right)_{t+\Delta t}^a\right]^{1/a} \le \exp(2(a-1)(t+\Delta t))$$

and

$$2^{1/b} \left( \Phi\left(\frac{1}{\sqrt{t}}\right)-\Phi\left(\frac{1}{\sqrt{t+\Delta t}}\right)\right)^{1/b} \le \left(\frac{2}{\pi t^3}\right)^{1/2b} \exp\left(-\frac{1}{2bt}\right) \Delta t^{1/b}.$$

We conclude thus the (local) $1/b-$Holder continuity. Is there any chance to have the Lipschitz continuity?

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