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This question can be seen as a continuation of Lipschitz continuity of $\mathbb P[\tau>t]$ with respect to $t$

Consider the martingale given as

$$X_t=1+\int_0^t a(s,X_s)dW_s,\quad \forall t\ge 0.$$

Denote $\tau:=\inf\{t\ge 0: X_t\le0\}$. My question is whether $t\mapsto \mathbb P[\tau>t]$ is Holder continuous, i.e. $\exists C>0, \alpha\in (0,1]$ s.t.

$$0\le \mathbb P[\tau>t]-\mathbb P[\tau>t+\Delta t]\le C\Delta t^{\alpha}.$$

Any solution, references or comments are appreciated. Here we assume that $0<\underline a \le \inf_{(t,x)} a(t,x)\le \sup_{(t,x)} a(t,x)\le \overline a$, $t\mapsto a(t,x)$ is continuous and $|a(t,x)-a(t,y)|\le L|x-y|$ for some $L>0$.

PS : My idea is as follows :

\begin{eqnarray} \mathbb P[\tau>t]-\mathbb P[\tau>t+\Delta t] &=& \mathbb P\left[\inf_{0\le s\le t}X_s>0, X_t+\inf_{t\le u\le t+\Delta t}\int_t^ua(s,X_s)dW_s\le 0\right] \\ &=& \mathbb P\left[\inf_{0\le s\le t}X_s>0, X_t>0, X_t+\inf_{t\le u\le t+\Delta t}\int_t^ua(s,X_s)dW_s\le 0\right] \\ &\le &\mathbb P\left[X_t>0, X_t+\inf_{t\le u\le t+\Delta t}\int_t^ua(s,X_s)dW_s\le 0\right] \\ &=& \int_{(0,\infty)}\mathbb P\left[x+\inf_{t\le u\le t+\Delta t}\int_t^ua(s,X_s)dW_s\le 0\Big|X_t=x\right]\mathbb P[X_t\in dx]. \end{eqnarray}

Therefore, it suffices to estimate the conditional probability

$$\mathbb P\left[x+\inf_{t\le u\le t+\Delta t}\int_t^ua(s,X_s)dW_s\le 0\Big|X_t=x\right]$$

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  • $\begingroup$ Do you have something missing in the last line? $\endgroup$
    – Nate River
    Oct 6, 2021 at 2:47
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    $\begingroup$ @NateRiver Absolutely right. I forgot $\le 0$ $\endgroup$
    – GJC20
    Oct 6, 2021 at 5:25

1 Answer 1

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For $h:=\Delta t>0$, you had $$P(\tau>t)-P(\tau>t+h)=\int_{(0,\infty)} P(M\le-x|X_t=x)\,P(X_t\in dx),$$ where $$M:=\inf_{t\le u\le t+h}J_u,\quad J_u:=\int_t^u a(s,X_s)\,dW_s.$$ By Doob's martingale inequality, $$P(M\le-x|X_t=x)\le x^{-2}\,E(J_{t+h}^2|X_t=x) \le x^{-2}\,ha_2^2,$$ where $a_2:=\overline a$ and $a_1:=\underline a$.

The crucial point is that the pdf $p_t$ of $X_t$ for $t>0$ is bounded so that $$p_t(x)\le\frac c{\sqrt t}\,e^{-bx^2/t}\le\frac c{\sqrt t}$$ for some positive real constants $c,b$ depending only on $a_1,a_2,L$ and for all real $x$. So, $$ \begin{align} P(\tau>t)-P(\tau>t+h)&\le \int_{(0,\infty)}\min(1,x^{-2}\,ha_2^2)\,\frac c{\sqrt t}\,dx \\ &= \sqrt h\int_{(0,\infty)}\min(1,u^{-2}\,a_2^2)\,\frac c{\sqrt t}\,du \\ &=\frac C{\sqrt t}\,\sqrt h, \end{align} $$ where $C>0$ is a real constant depending only on $a_1,a_2,L$.


To get now a uniform Hölder continuity, we can reason as follows: $$P(\tau>t)-P(\tau>t+h)\le1-P(\tau>t+h)=P(\tau\le t+h)\le a_2^2(t+h)/1^2,$$ again by Doob's martingale inequality. So, $$P(\tau>t)-P(\tau>t+h)\le\min\Big(1,\frac C{\sqrt t}\,\sqrt h,a_2^2(t+h)\Big)\le C_1h^{1/3},$$ where $C_1>0$ is a real constant depending only on $a_1,a_2,L$ (to verify the latter inequality, consider separately the three cases when (i) $h\ge1$, (ii) $t\le h^{1/3}<1$, or (iii) $t>h^{1/3}$).

Using here an exponential inequality (see e.g. Theorem 3.1) instead of Doob's one, one can improve the factor $h^{1/3}$ to $h^{1/2}\ln\frac1h.$

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  • $\begingroup$ Thank you so kindly for your answer. Indeed I saw your post (mathoverflow.net/questions/405624/…) which I believe is relevant to my question. This question is related to my current project (on mean field games). I'm happy to acknowledge your help in our paper if you don't mind $\endgroup$
    – GJC20
    Oct 7, 2021 at 22:07
  • $\begingroup$ You have taken $\alpha=1/2$ in your reasoning, and we see $t\mapsto \mathbb P[\tau>t]$ is not uniformly $1/2-$Holder continuous (nearby zero). However, if we choose $\alpha=1/4$, then we may expect the uniform $1/4-$Holder continuity. To do so, it suffices to use an alternative estimate for the probability $\mathbb P[M\le -x|X_t=x]=\mathbb P[-M\ge x|X_t=x]=\mathbb P[(-M)^{\alpha}\ge x^{\alpha}|X_t=x]$. If you don't mind, could you please modify your arguments to obtain the uniform $\alpha-$Holder continuity? $\endgroup$
    – GJC20
    Oct 7, 2021 at 22:33
  • $\begingroup$ @GJC20 : I have added a piece on a uniform Hölder continuity. $\endgroup$ Oct 8, 2021 at 0:09
  • $\begingroup$ Your solution is really amazing. Thanks infinitely for your help $\endgroup$
    – GJC20
    Oct 8, 2021 at 5:27
  • $\begingroup$ @GJC20 : I am glad this was of help. $\endgroup$ Oct 8, 2021 at 11:50

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