I am considering the following question and want to convince myself that the stochastic process $X$ has continuous sample paths. I hope someone could give me some hints or references, many thanks!

Suppose that $\{B_t\}_{t\ge 0}$ is a standard Brownian motion and a stochastic process $\{X_t\}_{t\ge 0}$ is defined as $$dX_t=\mathbf 1_{\{X_t\le a\}}dt+dB_t, X_0=x_0 \,\,a.s.$$ By intuition, I think $\{X_t\}_{t\ge 0}$ has continuous sample paths, and it seems that the key is to prove that for each $T\ge 0$ and for almost every $\omega\in \Omega$, the function $$F(T)=\int_0^T \mathbf 1_{\{X_t(\omega)\le a\}}\,dt$$ is continuous


1 Answer 1


Let $F(T)$ be defined in the way you've defined it. Consider $\epsilon > 0$, we have that $$\begin{align} F(T+\epsilon) &= \int_0^{T+\epsilon}\boldsymbol{1}_{\{X_t(\omega)\le a\}}dt\\ &= \int_0^{T}\boldsymbol{1}_{\{X_t(\omega)\le a\}}dt + \int_{T}^{T+\epsilon}\boldsymbol{1}_{\{X_t(\omega)\le a\}}dt \\ &\le F(T) + \int_{T}^{T+\epsilon}1\cdot dt = F(T) + \epsilon. \end{align}$$

Similarly, we can show that $F(T+\epsilon) \in [F(T),F(T)+\epsilon]$ and $F(T-\epsilon) \in [F(T)-\epsilon,\epsilon].$

Hence, we have shown that $|T'-T|\le \epsilon \Rightarrow |F(T')-F(T)|\le \epsilon$. Hence, $F(T)$ is continuous pointwise in $\omega$ for each $T\geq 0$. And since $B_t(\omega)$ is a.s. continuous, we have that $X_t$ is the sum of a pointwise and a.s. continuous function, hence a.s. continuous.


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