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Jun 15, 2020 at 7:27 history edited CommunityBot
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S Feb 19, 2014 at 18:46 history bounty ended CommunityBot
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Feb 17, 2014 at 18:41 comment added John Jiang I suggest asking Prof. Darrell Duffie about this. He is very good at mathematical rigor in finance.
Feb 14, 2014 at 14:35 history edited hulik CC BY-SA 3.0
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S Feb 11, 2014 at 17:15 history bounty started hulik
S Feb 11, 2014 at 17:15 history notice added hulik Draw attention
Feb 10, 2014 at 20:36 history edited hulik CC BY-SA 3.0
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Feb 10, 2014 at 12:36 comment added hulik @NateEldredge Thanks for you comment. I edited my question. I was also trying to prove uniform integrability. All we know is $\lim_{A\to\infty}\sup_{n}E_{Q_n}E[S_k\mathbf1_{Sk>A}]=0$ for all $k=0,…,N−1$. This is stated in the same proof before equation (3.17). But this estimate is used to prove tightness of the $\{Q_n\}$
S Feb 10, 2014 at 12:10 history suggested Davide Giraudo CC BY-SA 3.0
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Feb 10, 2014 at 12:09 review Suggested edits
S Feb 10, 2014 at 12:10
Feb 10, 2014 at 8:56 history edited hulik CC BY-SA 3.0
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Feb 10, 2014 at 8:34 history edited hulik CC BY-SA 3.0
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S Feb 10, 2014 at 3:14 history suggested gaoxinge CC BY-SA 3.0
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Feb 10, 2014 at 3:01 review Suggested edits
S Feb 10, 2014 at 3:14
Feb 9, 2014 at 23:35 comment added Nate Eldredge Something doesn't seem to be right with your definition of $\mathcal{W}_n$. Anyway, it seems to me that we are supposed to show the uniform integrability (under $P$) of $\{g(S_N(X_n))\}$, which since it is just a statement about their distributions, would follow from the uniform integrability under $Q_n$ of $g(S_N)$. It's sufficient to find a uniform bound on their $L^r$ norms for some $r > 1$, but I don't quite see how to get that from the given conditions. If we had something like $|g(x)| \le C(1+x^{p-\epsilon})$ I think it would work.
Feb 9, 2014 at 16:22 review First posts
Feb 9, 2014 at 16:33
Feb 9, 2014 at 16:05 history asked hulik CC BY-SA 3.0