Brownian Motion Hitting Time of functions $y=t$ and $y=t^2$

Question

Let $W(t)$ be a standard Brownian motion. Let $y(t)=t$. Define the following stopping time:

$T_h = \inf \{t>0: W(t)=y(t) \}$

What would be the probability:

$$\mathbb{P}(T_h\leq k)$$ for some constant $k>0$?

Now defining $y(t)=t^2$, what would be the probability?

Is there any literature on these types of problems? (I of course know how to compute the probability for the function $y(t)=constant$, but I haven't found any material for non-constant functions).

Answer 1

Hint

On can prove the following proposition

If $(W_s)_s$ is a Brownian motion on a probability space $(\Omega ,\mathcal F, \mathbb P)$, then $$ \mathbb P\left\{W_t\in \mathrm d x,\sup_{0\leq s\leq t}W_s\in \mathrm d y\right\}=\frac{2(2y-x)}{\sqrt{2\pi t^3}}\exp\left\{-\frac{(2y-x)^2}{2t}\right\}\boldsymbol 1_{(-\infty ,y]}(x)\,\mathrm d x\,\mathrm d y. \label{1}\tag{1}$$

Recall Girsanov theorem

Let $(W_t)$ a Brownian motion and let $\theta \in L^2([0,T])$. Under the measure $$\mathbb Q(\mathrm d \omega )=\exp\left\{\int_0^T \theta _s\,\mathrm d W_s-\frac{1}{2}\int_0^T \theta _s^2\,\mathrm d s\right\}\mathbb P(\mathrm d \omega ),$$ the process $$\tilde W_t:=W_t-\int_0^t\theta _s\,\mathrm d s,\quad t\in [0,T],$$ is a Brownian motion.

By Girsanov theorem, under the measure $$\mathbb Q(\mathrm d \omega )=\exp\left\{W_t-\frac{1}{2}t\right\}\mathbb P(\mathrm d \omega )$$ the process $$\tilde W_t:=W_t-t,$$ is a Brownian motion. Then

\begin{align*} \mathbb P\{T_h>t\}&=\mathbb P\left\{\sup_{0\leq s\leq t}(W_s-s)<0\right\}\\ &=\mathbb E_{\mathbb Q}\left[\boldsymbol 1_{\left\{\sup_{0\leq s\leq t}\tilde W_s>0\right\}}\exp\left\{-\tilde W_t-\frac{3}{2}t\right\}\right]. \end{align*} where $\mathbb E_{\mathbb Q}$ denote the expectation under the measure $\mathbb Q$. Using \eqref{1} allow you to conclude.

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For $y(t)=t^2$, same idea using $$\mathbb Q(\mathrm d \omega )=\exp\left\{2\int_0^ts\,\mathrm d W_s-\frac{2}{3}t^3\right\}\mathbb P(\mathrm d \omega ),$$ and the process $$\tilde W_t=W_t-t^2.$$ But at the end, I'm not so sure that you can find a nice closed form as the previous case.