Let $(B_t)_{t\geq 0}$ be a brownian motion, $P$ the measure of the probability space which satisfys the usual conditions and $\mathbb E$ the expected value. I like to show $$P\left(\inf_{t\geq0}\int_0^t e^{-s}\mathrm d B_s\geq -1\right)>0$$ or $$P\left(\inf_{t\geq0}\int_0^t e^{-s}\mathrm d B_s>-1\right)>0,$$ if one of them is easier to show than the other. Let $$Z_t:=\int_0^t e^{-s}\mathrm d B_s.$$ I already used the Ito formula to show $Z_t=B_te^{-t}+\int_0^t B_se^{-s}\mathrm d s$ and the Ito isometry to show that $Z_t$ is a Martingale with $\sup_t \mathbb E Z^2_t<\infty$. Since $\mathbb E|B_t|=\sqrt{\frac{2t}{\pi}}$ we can further deduce by some calculation that $Z_t$ converges a.s. Further I know that $Z_t$ is $\mathcal N(0,\int_0^t e^{-2s}\mathrm ds)$ distributed. But since I'm taking the infimum I don't know how to go on. Is there any possibility to caculate the distribution of $\inf_{t\geq0} Z_t$?
The attempt taking Markov's inequality for showing $$P(\sup_{t\geq0}(-Z_t)\geq 1)\leq P(|\sup_{t\geq0}(-Z_t)|\geq 1)\leq \mathbb E|\sup_{t\geq0}(-Z_t)|\leq \mathbb E\sup_{t\geq0}|Z_t|\overset{!}{<}1 $$ failed since the last inequality is false, as I saw after some numerical simulations.
But I'm not sure whether any of this above helps me to show the equation I wanted. I am grateful for any help. It needn't to be the full solution, any advice which inequality I can used to show problems like this would be great help.
One way you could show this is with the optional stopping theorem. Since you showed $Z$ is a true martingale and $\sup_t \mathbb{E}[Z^2] < \infty$, $\{Z_T: T \text{ is a stopping time} \}$ is uniformly integrable. Set $\tau := \inf\{t \ge 0: Z_t \le -1\}$ and suppose working towards a contradiction that $\mathbb{P}(\tau < \infty) = 1$. Then $\mathbb{E}[Z_\tau] = \mathbb{E}[-1] = -1 \ne 0 = Z_0$, a contradiction! Hence $\mathbb{P}(\tau = \infty) > 0$, and so $\mathbb{P}(\inf_t Z_t > -1) = \mathbb{P}(\tau = \infty) > 0$.