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 Jan8 revised Strictly local martingales: what is the intuition behind them? added 3 characters in body Jan8 answered Strictly local martingales: what is the intuition behind them? Dec11 awarded Yearling Oct9 comment Show that $E[B_t|\mathscr{F}_s] = B_s$ Yep, totally correct. Oct9 comment Show that $E[B_t|\mathscr{F}_s] = B_s$ Technical point: the pde condition on f only guarantees that it will be a local martingale. Integrability has to be checked separately. Sep30 awarded Explainer Sep24 awarded Autobiographer May22 comment Existence of a hedging portfolio and martingale property ok, so the supermartingale is the value process corresponding to an american option. if i understand your question, then at a theoretical level, the answer is yes: doob-meyer decomposition to split the supermartingale into a martingale and decreasing process, and then martingale representation to get a hedging strategy for the martingale part. May22 comment Existence of a hedging portfolio and martingale property what do you mean by a super martingale price process? May18 comment Difference betweem martingale property and adapted filteration Also, this is something I think about from time to time. You can definite martingality intrinsically, by taking the filtration to be the natural one generated by $X$. Does this mean that martingality is an intrinsic property? May18 comment Difference betweem martingale property and adapted filteration Continuity and measurability are defined slightly differently than what you said. For a continuous function, inverse image of an open set is open, and similarly with measurable. If you unpack the $\epsilon-\delta$ definition of continuity, you'll see that this is what it's logically equivalent to. May18 answered Difference betweem martingale property and adapted filteration May3 awarded Custodian May3 reviewed No Action Needed What is the hedging underlying of MBS Apr22 comment Distribution of Brownian Bridge The starting point doesn't matter. You're conditioning on the Brownian Motion being $a$ at time $T_1$, so there's no variance there. Just do the calculation on $[0, T_2 - T_1]$. Apr9 comment unique equivalent martingale measure in incomplete markets yeah. you apply $f(x)$ pointwise. Apr8 revised unique equivalent martingale measure in incomplete markets added 9 characters in body Apr8 comment unique equivalent martingale measure in incomplete markets yeah, thanks. this was a while ago, i have to try and understand what i wrote. Apr8 comment How can the Wiener process be nowhere differentiable but still continuous? Continuity is a weaker property than differentiability. Apr8 comment How do you calibrate a poisson arrival rate process? Don't you know $\delta$? You have to estimate $A$, where $A \leq 1$?