I have the probability $P(e^{σB_t+ αt}>Kf_t)$, where σ, α, K constants, f is a function of t and $B_t$ brownian motion. This probability must be independent of t. So why is $f_t$ chosen such as $f_t = 1/K*e^{σc\sqrt{t} + αt}$? Why is $B_t$ always greater than $c\sqrt{t}$? Sorry if it's obvious, but I might have missed this.
2026-04-13 07:21:46.1776064906
Inequality brownian motion
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Since $B_t$ is distributed like $\sqrt{t}B_1$, the event under consideration has the probability of the event $$[\exp(\sigma\sqrt{t}B_1+\alpha t)\gt Kf_t]=[B_1\gt g_t],\qquad \text{where}\ \exp(\sigma\sqrt{t}g_t)= Kf_t\mathrm e^{-\alpha t}. $$ To impose that $g_t$ does not depend on $t$ is to ask that $f_t$ has the form in the question.