the probability for a stochastic process $\mathbb{P}(\text{sup}_{0\leq t\leq2}X_t\geq1)$, where $X_t = \int _0^t\frac{dW_s}{\sqrt{1+s}}$

Question

$X_t = \int _0^t\frac{dW_s}{\sqrt{1+s}}$, the probability $\mathbb{P}(\text{sup}_{0\leq t\leq2}X_t\geq1)$

Answer 1

Since $ X_t=\int_0^t\frac{dW_s}{\sqrt{1+s}} $, then $ \{X_t,t\ge 0\} $ is a continuous Gaussian process with independent increments, and $ \mathsf{E}[X_t]=0$, $ \mathsf{E}[X_t^2]=\int_{0}^{t}\frac{ds}{1+s}=\log(1+t) $. Meanwhile, $ \{W_{\log(1+t)},t\ge 0\} $ is also a continuous Gaussian process with independent increments, and $ \mathsf{E}[W_{\log(1+t)}]=0$, $ \mathsf{E}[W_{\log(1+t)}^2]=\log(1+t)$ too. Hence $ \{X_t,t\ge 0\} $ and $ \{W_{\log(1+t)},t\ge 0\} $ are same distributed. Furthermore, \begin{align*} &\mathsf{P}\left(\sup_{0\le t \le 2}X_t \ge 2\right) =\mathsf{P}\left(\sup_{0\le t \le 2}W_{\log(1+t)} \ge 2\right)\\ &\quad =\mathsf{P}\left(\sup_{0\le s \le \log3}W_{s} \ge 2\right) =2\mathsf{P}(W(\log3)\ge 2) \qquad \text{by reflection principle}\\ &\quad = \sqrt{\frac{2}{\pi\log3}}\int_{2}^{\infty}e^{-u^2/(2\log3)}\,\mathrm{d}u. \end{align*}