# Lipschitz continuity of $\mathbb P[\tau>t]$ with respect to $t$

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?

• What happens if you try to use Girsanov to remove the drift? Jun 30, 2021 at 6:44
• @NateRiver Using Girsanov theorem, I only obtain the Holder continuity Jun 30, 2021 at 10:30
• Isn't this easier to state without the hitting time, e.g. $p(t) = \mathbb{P}[\forall u\le t,X_u\ge0]$$? Jul 5, 2021 at 19:54 • @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 Jul 5, 2021 at 20:48
• @MattF. What I mean is that, unless using Girsanov, I don't know how to deal with the above probability Jul 5, 2021 at 20:48

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, 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?