The exponential process $Z_t=\exp(-B_t-t/2)$ is a martingale. Under the <Girsanov theorem> change of measure $d\mathbb Q|_{\mathcal F_t}=Z_t,d\mathbb P|_{\mathcal F_t}$, the process $W_t=B_t+t$ is Brownian motion. Its first hitting time $T_a$ of $a>0$ is finite $\mathbb Q$-almost surely. On $\{T_a\leq t\}$, the <optional sampling theorem> gives
$$
\mathbb Q(T_a\leq t)
=\mathbb E_{\mathbb P}[Z_{T_a}\mathbf1_{\{T_a\leq t\}}]
=e^{-a}\mathbb E_{\mathbb P}
\left[e^{T_a/2}\mathbf1_{\{T_a\leq t\}}\right],
$$
because $B_{T_a}=a-T_a$. Letting $t\to\infty$ and applying the <monotone convergence theorem> yields
$$
\mathbb E e^{T_a/2}=e^a.
$$
This is the <Critical exponential moment of a drifted Brownian hitting time>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 202 / 3 / b / Solution

{c}

For $a>0$ and $T_a=\inf\{t\geq0:B_t+t=a\}$,
$$
\mathbb E e^{T_a/2}=e^a.
$$
Changing measure with the density $e^{-B_t-t/2}$ turns $B_t+t$ into Brownian motion. At $T_a$ the density equals $e^{-a+T_a/2}$, and Brownian motion hits $a$ almost surely.


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 Incoming links: Critical exponential moment of a drifted Brownian hitting time