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Let $X_t$ be a brownian motion

define: $Y(t) = t^2X_t - 2 \int_0^t sX_s \ ds$ Is $Y$ a martingale? I am trying to use Ito's lemma, and show that the drift is 0, however I am having troubles differentiating the integral.

I can do it if I use fundamental thoerem of calculus but I am told there is an easier way.

Any help please

cupes0
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1 Answers1

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I will assume that $X$ is a standard brownian motion (without drift and starting from $0$.) In differential notation we have $dY_t = 2t X_t dt + t^2 d X_t -2 t X_t dt = t^2 d X_t$ so that $d\langle Y,Y\rangle_t = t^2 d\langle X,X\rangle_t = t^2 dt$ (Lévy) as $X$ is a brownian motion. So $Y$ cannot be a brownian motion, as were it be we would have (Lévy) $d\langle Y,Y\rangle_t = dt$. But $Y$ is nevertheless a martingale, as it has no drift (the "something $\times dt$ part) as we just saw.

Olórin
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  • sorry but how are you integrating it? afaik you cannt integrate that integrand $sX_s$ – cupes0 Dec 10 '15 at 22:52
  • I am not integrating it, I am especially switching to the differential notation to avoid the $\int$, and calculation $d$ of $t^2 X_t$ using Itô's lemma applied to $f(t,X_t)$ with $f(x,y)=x^2 y$. I am also using Paul Lévy's characterization of the brownian motion. – Olórin Dec 10 '15 at 22:54
  • I am unfamiliar with this differential notation - could you explain how you got the first line and what $<Y,Y>_t$ means (I have not studied Levy's characterisation. – cupes0 Dec 10 '15 at 22:58
  • Look here: http://math.stackexchange.com/questions/523724/why-do-people-write-stochastic-differential-equations-in-differential-form For the brackets, they denote the quadratic variation of a process. For Lévy's characterization, you'll find references with google. ;-) – Olórin Dec 10 '15 at 23:01
  • OK Thanks, I have read over that, but I still don't see how you got $dY_t = 2tX_t dt + t^2 d X_t - 2t X_t dt$ I get that you are using itos for $t^2X_t$ but how did you get the other term, and how can you combine itos and that term? – cupes0 Dec 10 '15 at 23:10
  • $d(\int_0^t Z_s ds) = Z_s$... Think of the classic differential/integral case (with functions) and of the definition of the stochastic integral... – Olórin Dec 10 '15 at 23:11
  • if I let $Y_s = \int_0^tsX_s \ ds$ then I get $dY_s = Y(s+ds) - Y(s)$ from my definition. I understand using the fundamental theorem of calculus I can get this for regular functions.. but I am not sure for this – cupes0 Dec 10 '15 at 23:17