= Solution
Write $\tau=\tau_x$ and $T_t=t\wedge\tau$. Path continuity gives $|B_{T_t}|\leq x$. Stop the <martingale> $B_s^2-s$ at the bounded <stopping time> $T_t$ and apply the <optional stopping theorem>:
$$
\mathbb E T_t=\mathbb E B_{T_t}^2\leq x^2.
$$
By the <monotone convergence theorem>, $\mathbb E\tau\leq x^2$, so $\tau<\infty$ <almost surely>. Path continuity then gives $B_\tau\in\{-x,x\}$. The <dominated convergence theorem> for the bounded variables $B_{T_t}^2$ shows
$$
\boxed{\mathbb E\tau_x=x^2.}
$$
Next stop the quartic <Hermite polynomial martingale> from part (a), again only at $T_t$. Its <expectation> is zero, so
$$
3\mathbb E T_t^2
=6\mathbb E[T_tB_{T_t}^2]-\mathbb E B_{T_t}^4
\leq6x^2\mathbb E T_t\leq6x^4.
$$
<Monotone convergence> proves $\mathbb E\tau^2\leq2x^4$, establishing the needed second-moment integrability before the final passage to the limit. Since $T_tB_{T_t}^2\leq x^2\tau$ and $B_{T_t}^4\leq x^4$, <dominated convergence> gives
$$
3\mathbb E\tau^2=6x^2\mathbb E\tau-x^4=5x^4.
$$
Consequently the <Brownian symmetric interval-exit moments> are
$$
\boxed{\mathbb E\tau_x^2=\frac53x^4,\qquad
\operatorname{Var}(\tau_x)=\frac23x^4.}
$$
To obtain the <Laplace transform of symmetric Brownian interval-exit time>, put $a=\sqrt{2\lambda}$. The <Exponential martingale for Brownian motion> shows that
$$
H_t=e^{-\lambda t}\cosh(aB_t)
=\tfrac12\bigl(e^{aB_t-a^2t/2}+e^{-aB_t-a^2t/2}\bigr)
$$
is a <martingale> with $H_0=1$. Bounded-time stopping gives $\mathbb E H_{T_t}=1$. Its stopped values are bounded by $\cosh(ax)$, so <dominated convergence> applies as $t\to\infty$. Since $\cosh(aB_\tau)=\cosh(ax)$, it yields
$$
\boxed{\mathbb E[e^{-\lambda\tau_x}]
=\frac1{\cosh(x\sqrt{2\lambda})}\quad(\lambda>0).}
$$
Every use of stopping at $\tau$ has thus been justified through bounded stopping and an explicit integrability or domination argument.
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