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For each $T > 0$, let $B^T$ be a Brownian bridge on $[0, T]$, conditioned to start and end at $0$.

Question: Is it true that $\mathbb E[|\text{exp}\, (\sup_{0 \leq t \leq T} B^T_t) - 1|] \to 0$ as $T \to 0^+$?

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Without loss of generality, $B_t^T=B_t-\frac tT\,B_T$, where $B_\cdot$ is a standard Brownian motion. So, $$0\le\sup_{t\in[0,T]}B_t^T\le M_T+|B_T|,$$ where $M_T:=\sup_{t\in[0,T]}B_t$. So, in view of the Cauchy-Scwarz inequality, $$E|\exp\sup_{t\in[0,T]}B_t^T-1| =E\exp\sup_{t\in[0,T]}B_t^T-1 \\ \le\sqrt{E\exp(2M_T)} \sqrt{E\exp(2|B_T|)}-1.$$ By the reflection principle, $M_T$ equals $|B_T|$ in distribution. So, $$E|\exp\sup_{t\in[0,T]}B_t^T-1| \le E\exp(2|B_T|)-1=E\exp(2\sqrt T\,|B_1|)-1\to0$$ as $T\downarrow0$, by dominated convergence. $\quad\Box$

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  • $\begingroup$ Thank you very much! $\endgroup$
    – Nate River
    Jan 4, 2023 at 11:33

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