While I was playing around with the Girsanov's Theorem I stumbled upon the following absurdity and I couldn't resolve it with the current knowledge of stochastic analysis that I have.
$B$ being standard Brownian motion (SBM) I define the stochastic process $X$ as $X_t = \mu t + B_t$ for some $\mu \geq 0$. Then for a fixed $m > 0$ I define the hitting time $\tau = \inf\{t\geq 0: X_t = m\}$.
If I apply the Girsanov's theorem with $Z_t = e^{-mu B_t -\frac{1}{2}\mu^2t}$, I have that $X$ is SBM under the measure $(Q_t)_{t\geq 0}$ defined as $Q_t(A) = E^P[Z_t\mathbf{1}_A]$.
I had derived the moment generating function of the single-barrier hitting time of SBM so I will just use that here. $$E^Q[e^{-\lambda\tau}] = e^{\sqrt{2\lambda}m}$$
But then $E^Q[e^{-\lambda\tau}] = E^P[e^{-\lambda\tau}Z_t]$. If I substitute the expression for $Z_t$ in this equality I get $$E^P[e^{-(\lambda\tau + \mu B_t)}] = e^{\sqrt{2\lambda}m}e^{\frac{1}{2}\mu^2t}\qquad \qquad \triangle$$ I know that $E^P[e^{-\mu B_t}] = e^{\frac{1}{2}\mu^2t}$. So then it looks like under the measure $P$ the joint moment generating function of $\tau$ and $B_t$ factors, which implies that they are independent. That cannot be right since
$$P\{\tau \leq t, B_t > m\} = P\{B_t > m\}$$
This would lead to the contradiction that $\tau \leq t$ almost surely. What is also nonsensical is that if I take $\mu = 0$ in $\triangle$, I get $E^P[e^{-\lambda\tau}] = e^{\sqrt{2\lambda}m}$. I just cannot convince myself that the moment generating function of $\tau$ could be the same under two different measures.
I am almost certain that there is something off in the way I am applying the Girsanov's theorem but I don't see where. I would appreciate it if someone could point out my error and possibly derive the true expression for $E^P[e^{-(\lambda\tau + \mu B_t)}]$.
You write $E^Q[e^{-\lambda \tau}] = e^{\sqrt{2\lambda} m}$. What measure $Q$ is this? Under $Q_t$ we know $X$ is a brownian motion on $[0,t]$, but your stopping time's moment generating function was for a brownian motion on $[0,\infty)$.
It is actually possible to get a measure $Q$ where $X$ is a brownian motion on $[0,\infty)$ assuming the underlying space has certain properties. In that case you would have $E^Q[e^{-\lambda \tau}] = e^{\sqrt{2\lambda} m}$, but you can only arrange $Q(A) = E[1_A Z_T]$ for all $A \in \mathscr{F}_T, T < \infty$. Your $\tau$ is $\mathscr{F}_\infty$ measurable, but not necessarily $\mathscr{F}_t$ measurable for any fixed $t$. Thus you cannot write $E^Q[e^{-\lambda \tau}] = E^P[e^{-\lambda \tau}Z_t]$ and have it be meaningful.