Timeline for Discrete approximation of one step martingale
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 17, 2021 at 7:03 | vote | accept | Nate River | ||
| May 17, 2021 at 7:02 | answer | added | Anthony Quas | timeline score: 1 | |
| May 17, 2021 at 6:55 | comment | added | Nate River | Ah, right we have a.s. convergence thanks to Doob’s theorem. Thanks! Would you like me to post this as an answer? So that it doesn’t stay “open”. | |
| May 17, 2021 at 6:48 | comment | added | Anthony Quas | I think that follows from the pointwise convergence of $Y^n_i$ to $X_i$. We have $\|Y^n_i\|_2^2+\|Y^n_i-X_i\|_2^2=\|X_i\|_2^2$; and since $Y^n_i\to X_i$, we have by Fatou $\liminf \|Y^n_i\|_2^2\ge \|X_i\|_2^2$. | |
| May 17, 2021 at 6:46 | comment | added | Nate River | In the case where the $X_i$ are absolutely continuous I think a “by hand” argument suffices, but I’m not sure how to proceed if they’re more singular. | |
| May 17, 2021 at 6:39 | comment | added | Nate River | Oh, and the existence of such a sequence would be because $\mathcal F(X_i)$ is countably generated. How does one get $L^2$ convergence from convergence along the filtration though? | |
| May 17, 2021 at 6:31 | comment | added | Anthony Quas | Why can't you just take a sequence of finite $\sigma$-algebras $\mathcal F^n_i$ increasing to $\mathcal F(X_i)$ (the $\sigma$-algebra generated by $X_i$) and define $Y^n_1=\mathbb E(X_1|\mathcal F^n_1\vee \mathcal F^n_0)$ and $Y^n_0=\mathbb E(X_1|\mathcal F^n_0)$? | |
| May 17, 2021 at 4:38 | history | edited | Nate River | CC BY-SA 4.0 |
added 96 characters in body
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| May 17, 2021 at 4:32 | history | asked | Nate River | CC BY-SA 4.0 |