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For standard Brownian motion started at zero, the strict first-passage times form a subordinator in the level parameter $a$. The Strong Markov property gives independent increments and stationary increments; the strict inverse of the continuous running maximum has càdlàg paths. The Brownian first-passage Laplace transform gives Laplace exponent $\Phi (\lambda )=\sqrt{2\lambda }$, so the process is strictly stable of index $1/2$. Choosing the non-strict hitting times preserves each fixed-level law but generally loses right continuity at random levels, as in the fixed-level versus simultaneous Brownian passage-time equality.

For $a>0$, the Brownian reflection principle gives, for $t>0$,where $\Phi$ is the standard normal distribution function. Hence $H_a<\infty$ almost surely; for $a=0$, $H_0=0$.

For $u\ge 0$, use the Exponential martingale for Brownian motion $Z_t=\exp (u{B}_t-u^{2}t/2)$. At $H_a\wedge t$, it is bounded by $e^{ua}$ because $B_s<a$ before the first hit and $B_{H_a}=a$ by continuity. The bounded optional stopping theorem gives $\mathbb{E}{Z}_{H_a\wedge t}=1$. Its limit is $e^{ua-u^2{H}_a/2}$, and the dominated convergence theorem yields $\mathbb{E}{e}^{ua-u^2{H}_a/2}=1$. Equivalently, the Brownian first-passage Laplace transform isTaking $\lambda =u^2/2$ proves the displayed exponential identity. The cases $u=0$ and $a=0$ are included. The restriction $u\ge 0$ is what gives the upper bound on the stopped exponential.

Specify the sign convention for the Lévy characteristic exponent byConditioning on $T_a$ and using the Brownian first-passage Laplace transform givesThereforeThe absolute value is essential for negative $u$. The process is the standard symmetric Cauchy process; for $a>0$, $X_a$ has the Cauchy distribution with location zero and scale $a$. If the exponent convention instead uses $\mathbb{E}{e}^{iu{X}_a}=e^{a\psi (u)}$, the answer is $\psi (u)=-|u|$.