= Solution
Use the positive <exponential Brownian martingale>
$$
M_t=\exp(2\lambda W_t-2\lambda^2t).
$$
The <strong law for Brownian motion>, $W_t/t\to0$, makes its exponent tend to $-\infty$ and hence $M_t\to0$. If $S=\sup_{t\geq0}(W_t-\lambda t)$, then $\sup M=e^{2\lambda S}$. Part (b) gives, for $x>0$,
$$
\mathbb P(S>x)=e^{-2\lambda x}.
$$
Thus the maximum is exponentially distributed with rate $2\lambda$:
$$
\boxed{f_S(x)=2\lambda e^{-2\lambda x}\mathbf1_{\{x>0\}}.}
$$
There is no atom at zero, by letting $x\downarrow0$ in the tail. This is the <infinite-horizon crossing probability for Brownian motion with negative drift>.
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