Let $B_t$ be a standard BM w.r.t. $(\Omega, \{\mathcal F_t\}, \Bbb P)$, $\mu$ a constant, $X_t:=\mu t+ B_t$, and $M_t:=\max_{0\le\tau\le t} X_\tau$. Find $\Bbb P(M_t\le a)$ for $a\ge 0$.
I have used the Girsanov's theorem to find that
$$\Bbb P(M_t \le a)=2\left(\Phi\left(\frac{a - \mu t}{\sqrt{t}}\right) - \Phi\left(-\frac{\mu t}{\sqrt{t}}\right)\right)$$
where $\Phi$ is the CDF of standard normal. But I'm not exactly sure whether my approach is sound, especially about the application of the Girsanov. Would you please check if there's anything wrong? Thanks.
Proof:
By Girsanov's theorem, if we let $$E_t:=\exp(-\mu B_t - \frac12\mu^2t)$$ Then under the equivalent measure $\Bbb Q$ defined by the Radon-Nikodym derivative $d\Bbb Q = E_td\Bbb P$, we have $X_t=\mu t+ B_t$ is a standard Brownian motion. We thus denote $B_t^{\Bbb Q} := X_t$. And we know that under $\Bbb Q$, the distribution of $M_t$ would be particularly simple, i.e. $$\Bbb Q(M_t\le a) = \Bbb Q(|B_t^{\Bbb Q}|\le a)=\Phi(a/\sqrt{t})-\Phi(-a/\sqrt{t})$$ Thus it is desirable to compute everything under $\Bbb Q$. However, we are asked to compute the probability under $\Bbb P$. So we will need the inverse Radon Nikodym derivative $d\Bbb P/d\Bbb Q$.
Now comes what I'm not very sure of: noting that $B_t=B_t^{\Bbb Q} - \mu t$, we have $$\frac{d\Bbb P}{d\Bbb Q} = \exp(\mu B_t^{\Bbb Q} - \frac12\mu^2t).$$ Can we invert RN derivative like this? If not, then how?
Assuming the above inversion is correct. We then have $$ \begin{align} \Bbb P(M_t\le a) &= \int_{\{M_t\le a\}}d\Bbb P = \int_{\{M_t\le a\}}\frac{d\Bbb P}{d\Bbb Q} d\Bbb Q = \int_{M_t\le a}\exp(\mu B_t^{\Bbb Q} - \frac12\mu^2t)d\Bbb Q\\ &=\int_{\max_{0\le\tau\le t}B_\tau^{\Bbb Q}\le a}\exp(\mu B_t^{\Bbb Q} - \frac12\mu^2t)d\Bbb Q\\ &=\int_0^a \exp(\mu x - \frac12\mu^2t)f(x) dx \end{align} $$ where $f(x)$ is the PDF of the scaled half normal $\sqrt{t}|Z|$ where $Z\sim N(0,1)$. Explicitly, we have that $f(x) = 2h(x)$ where $h(x):=(2\pi t)^{-1/2}\exp(-x^2/2t)$ is the PDF of $\sqrt{t}|Z|$. Thus $$ \begin{align} \Bbb P(M_t\le a) &= 2\int_0^a \exp(\mu x - \frac12\mu^2t)h(x) dx\\ & =2\Bbb P(\sqrt{t}Z+\mu t\in [0,a])\\ & = 2\left(\Phi\left(\frac{a - \mu t}{\sqrt{t}}\right) - \Phi\left(-\frac{\mu t}{\sqrt{t}}\right)\right) \end{align} $$