2
$\begingroup$

Given two SDEs $X^1$, $X^2$ :

$$X^i_t=1+t+\int_0^t\sigma_i(s)dW_s,\quad \forall t\ge 0,$$

where $\sigma_i:\mathbb R_+\to [1/2,1]$ are non-decreasing s.t. $\sigma_1(t)\le \sigma_2(t)$ for all $t\ge 0$. Can we prove $\mathbb P[\inf_{0\le s\le t}X^1_s>0]\ge \mathbb P[\inf_{0\le s\le t}X^2_s>0]$ for all $t\ge 0$?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

Yes. This follows by time change: The process $(X^i_t)$ equals the process $(1+t+W_{\tau_i(t)})$ in distribution, where $$\tau_i(t):=\int_0^t\sigma_i(s)^2\,ds,$$ so that $\tau_2\ge\tau_1$ and hence for the corresponding inverse functions we have $\tau_2^{-1}\le\tau_1^{-1}$.

It follows that $$\begin{aligned} &\inf_{s\in[0,t]}(1+s+W_{\tau_2(s)}) \\ &=\inf_{u\in[0,\tau_2(t)]}(1+\tau_2^{-1}(u)+W_u) \\ &\le\inf_{u\in[0,\tau_1(t)]}(1+\tau_1^{-1}(u)+W_u) \\ &=\inf_{s\in[0,t]}(1+s+W_{\tau_1(s)}). \end{aligned}$$

So, for all real $x$, $$ \begin{aligned} &P(\inf\limits_{s\in[0,t]}X^1_s>x) \\ &=P(\inf_{s\in[0,t]}(1+s+W_{\tau_1(s)})>x) \\ &\ge P(\inf_{s\in[0,t]}(1+s+W_{\tau_2(s)})>x) \\ &=P(\inf\limits_{s\in[0,t]}X^2_s>x). \end{aligned}$$ Thus, the desired result follows.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Very nice observation! Thank you so much for the answer. Do you think this monotonicity can be helpful to derive the uniqueness of the solution at mathoverflow.net/questions/410069/…? $\endgroup$
    – GJC20
    Commented Dec 13, 2021 at 20:20
  • $\begingroup$ @GJC20 : It may. Unfortunately, I am not an expert in SDE's, especially for non-Markov processes, and I don't have a good idea for that question. But I will have it in mind. $\endgroup$
    – Iosif Pinelis
    Commented Dec 13, 2021 at 21:33
  • $\begingroup$ Thanks a lot for your consideration. Indeed, by defining the operator $\Gamma$ with $\Gamma[m](t):=\mathbb P[\inf_{0\le s\le t}Y^m_s>0]$, where $Y^{\alpha}_t:=X_0+t+\int_0^t dW_s/(1+m(s))$, one may construct an iterated sequence. More precisely, taking $m_0\equiv 1$ (resp. $m_0\equiv 1$) and setting $m_{n}:=\Gamma[m_{n-1}]$ for $n\ge 1$, we obtain a decreasing (resp. increasing) sequence lying between $0$ and $1$, so the limit must exist. In particular, we show the existence of the maximal solution $\overline m$ and the minimal solution $\underline m$. $\endgroup$
    – GJC20
    Commented Dec 14, 2021 at 6:14
  • $\begingroup$ If $\overline m=\underline m$, the uniqueness follows $\endgroup$
    – GJC20
    Commented Dec 14, 2021 at 6:14
  • $\begingroup$ @GJC20 : Nice argument for the existence (a couple of typos there, though). I have been thinking about the linked uniqueness problem, having somewhat similar ideas. Will check if they hold and, if so, will write down an answer. $\endgroup$
    – Iosif Pinelis
    Commented Dec 14, 2021 at 15:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .