square integrable martingale Hilbert space
I read that $mathcal{M}^2 = {X:X text{ is a cadlag, square integrable martingale}}$ is isomorphic to $L^2(Omega, mathcal{F}_{infty}, mathbb{P})$ as hilbert spaces if we identify martingales that are modifications (since these are cadlag they are also indistinguishable).
The mapping is $phi(M_t) = M_{infty}$. I'm not sure why this is surjective. more precisely, I'm not sure why defining $X_t = mathbb{E}[X_{infty}|mathcal{F}_t]$ yields a cadlag or right continuous process, could someone point out the reason or pass me a reference?
martingales stochastic-analysis
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I read that $mathcal{M}^2 = {X:X text{ is a cadlag, square integrable martingale}}$ is isomorphic to $L^2(Omega, mathcal{F}_{infty}, mathbb{P})$ as hilbert spaces if we identify martingales that are modifications (since these are cadlag they are also indistinguishable).
The mapping is $phi(M_t) = M_{infty}$. I'm not sure why this is surjective. more precisely, I'm not sure why defining $X_t = mathbb{E}[X_{infty}|mathcal{F}_t]$ yields a cadlag or right continuous process, could someone point out the reason or pass me a reference?
martingales stochastic-analysis
Take a look at Section II.2 in Continuous martingales and Brownian motion by Revuz & Yor.
– saz
Nov 22 '18 at 20:29
add a comment |
I read that $mathcal{M}^2 = {X:X text{ is a cadlag, square integrable martingale}}$ is isomorphic to $L^2(Omega, mathcal{F}_{infty}, mathbb{P})$ as hilbert spaces if we identify martingales that are modifications (since these are cadlag they are also indistinguishable).
The mapping is $phi(M_t) = M_{infty}$. I'm not sure why this is surjective. more precisely, I'm not sure why defining $X_t = mathbb{E}[X_{infty}|mathcal{F}_t]$ yields a cadlag or right continuous process, could someone point out the reason or pass me a reference?
martingales stochastic-analysis
I read that $mathcal{M}^2 = {X:X text{ is a cadlag, square integrable martingale}}$ is isomorphic to $L^2(Omega, mathcal{F}_{infty}, mathbb{P})$ as hilbert spaces if we identify martingales that are modifications (since these are cadlag they are also indistinguishable).
The mapping is $phi(M_t) = M_{infty}$. I'm not sure why this is surjective. more precisely, I'm not sure why defining $X_t = mathbb{E}[X_{infty}|mathcal{F}_t]$ yields a cadlag or right continuous process, could someone point out the reason or pass me a reference?
martingales stochastic-analysis
martingales stochastic-analysis
edited Nov 22 '18 at 0:21
Bernard
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118k639112
asked Nov 21 '18 at 23:59
CeeersonCeeerson
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Take a look at Section II.2 in Continuous martingales and Brownian motion by Revuz & Yor.
– saz
Nov 22 '18 at 20:29
add a comment |
Take a look at Section II.2 in Continuous martingales and Brownian motion by Revuz & Yor.
– saz
Nov 22 '18 at 20:29
Take a look at Section II.2 in Continuous martingales and Brownian motion by Revuz & Yor.
– saz
Nov 22 '18 at 20:29
Take a look at Section II.2 in Continuous martingales and Brownian motion by Revuz & Yor.
– saz
Nov 22 '18 at 20:29
add a comment |
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Take a look at Section II.2 in Continuous martingales and Brownian motion by Revuz & Yor.
– saz
Nov 22 '18 at 20:29