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Jan 31, 2021 at 8:52 vote accept BCLC
Jun 17, 2020 at 8:33 history edited CommunityBot
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Apr 13, 2017 at 12:46 history edited CommunityBot
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Jul 16, 2016 at 8:12 history edited BCLC CC BY-SA 3.0
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Jul 11, 2016 at 13:46 comment added Quantuple Sure. Don't worry for me. But maybe your question will find an answer on math or mathoverflow SE? Also it would help if you narrow down the questions you want an answer to, or at least better highlight them.
Jul 11, 2016 at 13:39 comment added BCLC @Quantuple I might set a bounty but at least you got your +10
Jul 10, 2016 at 16:48 vote accept BCLC
Jul 11, 2016 at 13:38
Jul 10, 2016 at 16:33 comment added Quantuple done, but you also raise interesting questions that I do not answer to :)
Jul 10, 2016 at 16:27 answer added Quantuple timeline score: 7
Jul 10, 2016 at 15:05 comment added Quantuple IMHO the problem isn't stated correctly indeed, in the sense that the Radon-Nikodym derivative provided as the "solution" is not the unique way to define a measure $\mathbb{Q}$ equivalent to $\mathbb{P}$ and under which $X_t$ is a martingale (just take $d\mathbb{Q}/d\mathbb{P} =\mathcal{E}(-\int_0^t \cos(s) dW_s + a)$, for any $a \in \mathbb{R}$. I think the exercice should have been written the other way around as you mention: show that with the given Radon-Nikodym derivative, the measure $\mathbb{Q}$ is equivalent to the original measure and such that $X_t$ is a $\mathbb{Q}$-martingale.
Jul 10, 2016 at 14:17 history tweeted twitter.com/StackQuant/status/752144770491973637
Jul 10, 2016 at 4:58 history asked BCLC CC BY-SA 3.0