# Search for conditions of the positive probability that a stochastic process never hits zero

Consider a stochastic process $$X$$ defined by

$$X_t:=1+\int_0^t b(s,X_s) \, ds+ W_t,\quad \forall t\ge 0,$$

where $$(W_t)_{t\ge 0}$$ is a standard Brownian motion. Suppose that $$b:\mathbb R_+ \times \mathbb R \to \mathbb R$$ is Lipschitz and of linear growth so that $$X$$ is uniquely defined. Under what kind of conditions one has

$$\mathbb P\big(X_t>0 \text{ for all } t\ge 0\big)>0?$$

An obvious condition is $$\inf_{(t,x)}b(t,x)>0$$. My question is whether we have more general conditions for the above inequality, especially for the case where $$b$$ changes sign?

PS : I'm looking for sufficient conditions, and we may consider the simple case $$b\equiv b(t)$$. Denote by $$B(t):=\int_0^tb(s) \, ds$$ for $$t\ge 0$$. Then a necessary condition is $$\limsup_{t\to \infty} B(t)=\infty$$. Can we impose some suitable condition on the growth of $$B$$ such that the desired inequality holds?

$$\newcommand\ep\varepsilon$$The case of interest when $$b(t,x)=b(t)$$ depends only on $$t$$ is comparatively simple.

Indeed, let $$g(t):=\sqrt{(2t+1/2)\ln\ln(3+t)}$$ for real $$t\ge0$$. By the law of the iterated logarithm, $$\sup_{s\in[t,\infty)}\frac{W_s}{g(s)}\to1$$ as $$t\to\infty$$ almost surely and hence in probability.

So, for each real $$\ep>0$$ there is some real $$t=t_\ep>0$$ such that $$\begin{equation*} \tfrac\ep2\,g(t)>(1+\ep)g(0) \tag{1}\label{1} \end{equation*}$$ and $$\begin{equation*} P(B)>0, \tag{2}\label{2} \end{equation*}$$ where $$\begin{equation*} B:=\{W_s<(1+\ep/2)g(s)\ \forall s\in[t,\infty)\}. \end{equation*}$$ Let $$\begin{equation*} A:=\{W_s<(1+\ep)g(s)\ \forall s\in[0,t]\}. \end{equation*}$$ Note that the function $$g$$ is concave. So, for all $$s\in[0,t]$$ we have $$g(s)\ge g(0)+\frac st\,(g(t)-g(0))$$ and hence for all real $$u<(1+\ep/2)g(t)$$ \begin{equation*} \begin{aligned} &(1+\ep)g(s)-\tfrac st\,u \\ &\ge(1+\ep)g(0)+\tfrac st\,[(1+\ep)(g(t)-g(0))-u] \\ &\ge(1+\ep)g(0)+\tfrac st\,[(1+\ep)(g(t)-g(0))-(1+\ep/2)g(t)] \\ &=(1+\ep)g(0)+\tfrac st\,[\tfrac\ep2\,g(t)-(1+\ep)g(0)] \\ &\ge(1+\ep)g(0)>g(0), \end{aligned} \tag{3}\label{3} \end{equation*} in view of \eqref{1}.

Note also that the Brownian bridge $$W^{(t)}_\cdot$$ defined by the formula $$W^{(t)}_s:=W_s-\tfrac st\,W_t$$ for $$s\in[0,t]$$ is independent of $$W_t$$. Recalling also the symmetry of $$W_\cdot$$ and its Markov property, as well as \eqref{3}, we get \begin{equation*} \begin{aligned} &P(W_s>-(1+\ep)g(s)\ \forall s\in[0,\infty)) \\ &=P(W_s<(1+\ep)g(s)\ \forall s\in[0,\infty)) \\ &=P(A\cap B) \\ &=P(W_s<(1+\ep)g(s)\ \forall s\in[0,t],W_t<(1+\ep)g(t),B) \\ &=\int_{-\infty}^{(1+\ep)g(t)} P(W_t\in du,B) P(W_s-\tfrac st\,W_t<(1+\ep)g(s)-\tfrac st\,u\ \forall s\in[0,t]) \\ &=\int_{-\infty}^{(1+\ep)g(t)} P(W_t\in du,B) P(W^{(t)}_s<(1+\ep)g(s)-\tfrac st\,u\ \forall s\in[0,t]) \\ &\ge\int_{-\infty}^{(1+\ep)g(t)} P(W_t\in du,B) P(W^{(t)}_s0, \end{aligned} \tag{4}\label{4} \end{equation*} by \eqref{2} and because $$g(0)>0$$.

Note that $$X_s=1+B(s)+W_s$$. So,
\begin{equation*} \begin{aligned} &P(X_s>1+B(s)-(1+\ep)g(s)\ \forall s\in[0,\infty)) \\ &=P(W_s>-(1+\ep)g(s)\ \forall s\in[0,\infty))>0, \end{aligned} \end{equation*} by \eqref{4}. Thus, the condition that $$\begin{equation*} B(s)>(1+\ep)g(s)-1 \tag{5}\label{5} \end{equation*}$$ for some real $$\ep>0$$ and all $$s\in[0,\infty)$$ is sufficient for $$\begin{equation*} P(X_s>0\ \forall s\in[0,\infty))>0. \end{equation*}$$ Note finally that for any $$\ep\in(0,\frac1{g(0)}-1)=(0,3.61\ldots)$$ there is a positive continuous function $$b$$ such that the function $$B$$ given by the formula $$B(t)=\int_0^t b(s)\,ds$$ for all real $$t\ge0$$ satisfies condition \eqref{5} (however, of course, $$b$$ does not have to be everywhere positive or continuous in order for $$B$$ to satisfy condition \eqref{5}).

• Thanks a lot for your answer. I thought of the iterated logarithmic asymptotic of $W$ at infinity, and I have no idea how to proceed further. Very tricky reasoning and nice arguments!
– user420828
Commented Jan 27, 2022 at 6:16
• @Philo18 : I am glad you liked this answer. Commented Jan 28, 2022 at 1:17