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?

Any answer, comments and references are highly appreciated.

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?


1 Answer 1


$\newcommand\ep\varepsilon$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)}_s<g(0)\ \forall s\in[0,t]) \\ &=P(B) P(W^{(t)}_s<g(0)\ \forall s\in[0,t])>0, \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}).

  • $\begingroup$ 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! $\endgroup$
    – user420828
    Jan 27, 2022 at 6:16
  • $\begingroup$ @Philo18 : I am glad you liked this answer. $\endgroup$ Jan 28, 2022 at 1:17

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