We have $T<\infty$ a.s. for T stopping time. Let $X_t$ be a continuous Martingale.
Is this enough to conclude $E[X_T]=E[X_0]$?
2026-03-30 06:44:12.1774853052
optional stopping theorem conditions satisfied?
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No: let $X_t$ be a standard Brownian motion (starting at zero), and define $$ T=\inf\{t\geq 0:X_t=1\}$$ Then $T<\infty$ almost surely, hence $X_T=1$ almost surely, so $$ \mathbb{E}[X_T]=1\neq 0=\mathbb{E}[X_0] $$