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Answer by Martin Hairer for On martingale convergence
You can construct a continuous martingale $X$ such that $X(n) = n$ almost surely. Indeed, for $t < 1$, set $Y(t) = B(t/(1-t))$ for $B$ a Brownian motion and then $X(t) = Y(t \wedge \tau)$ with $\tau...
View ArticleAnswer by mike for On martingale convergence
If the answer to Nate Rivers' last question, and sorry that im not just linking to it, is yes, and people seem to think it is, then you can have convergence in probability to infinity, from which a...
View ArticleOn martingale convergence
Let $(X_t)_{t\ge0}$ be a martingale with continuous paths. It was previously shown here and here that then it is impossible that $X_t\to\infty$ almost surely as $t\to\infty$.Is it possible that there...
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