Timeline for A Stochastic Taylor Expansion/Asymptotics
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Jun 6, 2016 at 20:33 | history | edited | Joris Bierkens | CC BY-SA 3.0 |
corrected typos based on OP's comments
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| Jun 6, 2016 at 18:44 | comment | added | Hans | OK, Joris. You probably missed it, but as I have mentioned in my earlier comment, you left out a $y_t$ in the $dB_t$ term of $z_t$. The $b$ there should be $\mu$. We also need $f$ and the second derivatives of $f$ and $g$ to be bound (continuous). I was awaiting your response to my comment before I accept your answer. I have now. Thank you. | |
| Jun 6, 2016 at 18:29 | vote | accept | Hans | ||
| Jul 27, 2018 at 21:38 | |||||
| Jun 6, 2016 at 9:45 | comment | added | Joris Bierkens | Hans, I of course agree with Yemon Choi. If you find any true errors or confusions, please point those out in a comment, and I am happy to change my answer accordingly. Also, if my approach (essentially) answers your problem, why don't you accept it as 'answer'? | |
| Jun 4, 2016 at 8:58 | comment | added | Hans | Joris Bierkens: As I have expressed in my first comment, I like your answer very much. I took the liberty and edited your answer --- correcting a typo --- in a way I thought would be clearer at least to me. But as Yemon Choi rightly reminded me that I had overstepped my bound. He has reverted it back to your original version. Would you be so kind as to take a look at my edit and see if you like it? If you do, you may elect to use my last edit. If not, I will just write another answer which will clearly state that it is merely my rephrasing of your answer. | |
| Jun 4, 2016 at 8:43 | comment | added | Hans | @YemonChoi: You are right. I thought my edit adhered closely along his line of reasoning but was clearer at least to me. I have said as much in my first comment. But I will ask for his opinion and permission explicitly first. Thank you for reminding me. | |
| Jun 4, 2016 at 4:10 | comment | added | Yemon Choi | @Hans, I do not think it is very polite to make such heavy edits to a user's answer, when he has not approved this. Many of your changes seemed subjective and unnecessary. If you think there are errors in the answer then it would be more polite to leave a comment notifying Joris. | |
| Jun 4, 2016 at 4:09 | history | rollback | Yemon Choi |
Rollback to Revision 1
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| S Jun 4, 2016 at 4:06 | history | suggested | Hans | CC BY-SA 3.0 |
Eliminate the logarithmic transform. Explained why the $dr_tdy_t$ term is not included.
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| Jun 4, 2016 at 2:25 | review | Suggested edits | |||
| S Jun 4, 2016 at 4:06 | |||||
| S Jun 3, 2016 at 23:59 | history | suggested | Hans | CC BY-SA 3.0 |
Tidy up the derivation. Deleted the part about interchanging differentiation and expectation as it is not needed.
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| Jun 3, 2016 at 23:16 | review | Suggested edits | |||
| S Jun 3, 2016 at 23:59 | |||||
| Jun 3, 2016 at 22:59 | comment | added | Hans | Yours is a clever way to circumvent the difficulty of taking second derivatives of the function of a stochastic variable. Or rather, it is a frontal attack for the second derivative. The expectation works to smooth out the non-differentiable part. Thank you! I have corrected my typo. You left out a $y_u$ in the $dBu$ term of the integral expansion of $z_t$. I have taken the liberty to make the derivation integral expansion of $y_t$ more explicit at least for me. I hope you would like it. | |
| S Jun 3, 2016 at 22:52 | history | suggested | Hans | CC BY-SA 3.0 |
Made the integral expansion of $y_t$ more explicit.
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| Jun 3, 2016 at 22:29 | review | Suggested edits | |||
| S Jun 3, 2016 at 22:52 | |||||
| Jun 3, 2016 at 13:10 | history | answered | Joris Bierkens | CC BY-SA 3.0 |