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First the exit time has sufficiently many finite moments. Over any interval of length $a^2$, a Brownian motion increment has probability $q=\mathbb{P}(|B_{a^2}|>2a)>0$ of absolute size larger than $2a$. If the process is still inside $(-a,a)$, such an increment forces it to exit before the next interval endpoint. Independence of increments at deterministic times therefore givesThus $T<\infty$ almost surely and all its positive moments are finite. This justifies extracting moments from its transform rather than presupposing them.

For $\lambda >0$, put $\theta =\sqrt{2\lambda }$. The processis a martingale, being the average of two exponential martingales for Brownian motion. Apply the optional stopping theorem at the bounded time $T\wedge t$. Since $|B_{T\wedge t}|\le a$, the stopped values are bounded by $\cosh (\theta a)$. Continuity gives $B_T\in \{-a,a\}$. Letting $t\to \infty$ by the dominated convergence theorem yieldsHence the Laplace transform of symmetric Brownian interval-exit time isThe value at zero follows from almost-sure finiteness. Expanding the reciprocal givesCompare with part (a), now applicable because of the geometric tail bound. The Brownian symmetric interval-exit moments are

Write $\tau ={\tau }_x$ and $T_t=t\wedge \tau$. Path continuity gives $|B_{T_t}|\le x$. Stop the martingale $B_s^2-s$ at the bounded stopping time $T_t$ and apply the optional stopping theorem:By the monotone convergence theorem, $\mathbb{E}\tau \le x^2$, 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

Next stop the quartic Hermite polynomial martingale from part (a), again only at $T_t$. Its expectation is zero, soMonotone convergence proves $\mathbb{E}{\tau }^2\le 2x^4$, establishing the needed second-moment integrability before the final passage to the limit. Since $T_t{B}_{T_t}^2\le x^2\tau$ and $B_{T_t}^4\le x^4$, dominated convergence givesConsequently the Brownian symmetric interval-exit moments are

To obtain the Laplace transform of symmetric Brownian interval-exit time, put $a=\sqrt{2\lambda }$. The Exponential martingale for Brownian motion shows thatis 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 $\cosh (a{B}_{\tau })=\cosh (ax)$, it yieldsEvery use of stopping at $\tau$ has thus been justified through bounded stopping and an explicit integrability or domination argument.