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$X_t = e^{-t}B(e^{2t})$ is an Ornstein-Uhlenbek process, and it is above a fixed level when the Brownian motion is above a square root boundary. The probability you want is the probability that it i …

consider $\sum X_n e^{int}$ where the $X_n$ have the property that $X_n$ are independent and eventually 0. Then every sample path is a trig polynomial and infinitely differentiable, however by arrang …

this problem, and it's analogue for Brownian motion have been solved as solved as they can get, which may not be as solved as you want, by the same technique, which is an eigenfunction expansion of t …

For α≥1 you can use scaling plus a 0-1 to show that it is. The 0-1 law says that the probability of immediately going negative is 0 or 1, similarly for positive. Let a be small. $ P \lbrace \inf_{0 …

No, 4 values of $k_i$ will determine that the distribution is tightly concentrated around $\pm 1$. Set $x_a = E[(B_{\sigma}- a)^+] $. Then $x_{-1 - \epsilon} = 1 + \epsilon $ implies $B_{\sigma} > …

if x = y the value is 0 but if $x \ne y$ it is 2. By a large deviation estimate the sum involving the x terms will concentrate on an interval $k \in ((x-\epsilon) n, (x + \epsilon n))$ and similarly …