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 Yearling
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May
27
answered Multidimensional Ito's Lemma for Vector-Valued functions
Apr
27
comment Creating correlated Brownian motions from independent ones
Just compute the quadratic variation of each process.
Apr
27
comment Why the expected return rate of a stock has nothing to do with its option price?
Because the pricing is by no arbitrage.
Jan
8
revised Strictly local martingales: what is the intuition behind them?
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Jan
8
answered Strictly local martingales: what is the intuition behind them?
Dec
11
awarded  Yearling
Oct
9
comment Show that $E[B_t|\mathscr{F}_s] = B_s$
Yep, totally correct.
Oct
9
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.
Sep
30
awarded  Explainer
Sep
24
awarded  Autobiographer
May
22
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.
May
22
comment Existence of a hedging portfolio and martingale property
what do you mean by a super martingale price process?
May
18
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?
May
18
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.
May
18
answered Difference betweem martingale property and adapted filteration
May
3
awarded  Custodian
May
3
reviewed No Action Needed What is the hedging underlying of MBS
Apr
22
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 $[0, T_2 - T_1]$.
Apr
9
comment unique equivalent martingale measure in incomplete markets
yeah. you apply $f(x)$ pointwise.
Apr
8
revised unique equivalent martingale measure in incomplete markets
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