Timeline for Do convex and decreasing functions preserve the semimartingale property?
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
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| Oct 24, 2010 at 17:47 | comment | added | The Bridge | Ok, thanks for your comment, I got your point and I am sorry not to be able to help further. Regards | |
| Oct 24, 2010 at 16:46 | comment | added | George Lowther | ...or something like that. Maybe it is looking at the quadratic variation of f(X), which is of the form $\int L^x_t (f^{\prime}(x))^2\,dx$. | |
| Oct 24, 2010 at 16:39 | comment | added | George Lowther | One further point: I don't have Protter's book to hand right now but, if you look at the proof of the result you just quoted, I expect that it is writing the drift term of f(X) as $\int L^x_t f^{\prime\prime}(x)\,dx$ where $L^x_t$ is the local time at x. This explodes if $L^0>0$ in your example but, if f was either convex or concave, $f^{\prime\prime}$ would be locally integrable and you get that f(X) is indeed a semimartingale. | |
| Oct 24, 2010 at 16:29 | comment | added | George Lowther | No. f is concave individually on the intervals $[0,\infty)$ and $(-\infty,0]$. It is neither convex nor concave on any open interval about zero. And that is the point of the result you refer to. In any case, a local martingale restricted to being either nonnegative everywhere or nonpositive everywhere must remain at 0 once it hits zero. It can only escape 0 if it can only take both positive and negative values, so the behaviour of f in a neighbourhood of 0 is what is important here. | |
| Oct 24, 2010 at 16:19 | comment | added | The Bridge | Sorry to insit but $f$ is concave though (right ?),so taking $-f$ should do the trick. I must miss something about the convex function definition here. Regards | |
| Oct 24, 2010 at 15:59 | comment | added | George Lowther | Actually, I think what I just called the Ito-Tanaka formula is the same thing as Tanaka-Meyer's theorem that you just mentioned. It is also called the Ito-Tanaka-Meyer formula. | |
| Oct 24, 2010 at 15:56 | comment | added | George Lowther | The function $f(x)=\vert x\vert^\alpha$ is not convex for $\alpha < 1$. Convex functions do preserve the semimartingale property (see the Ito-Tanaka formula for example). It is only when you have convex in space and decreasing in time that it gets difficult. | |
| Oct 24, 2010 at 15:48 | history | answered | The Bridge | CC BY-SA 2.5 |