Let $Y(t) = t^2W_t - 2 \int_0^t sW_s \ ds$ where $W_t$ is brownian motion. I am trying to show it is a martingale by showing it is driftless. I set $Z(t,W_t) = t^2W_t$ and ito's gives $dZ = 2tW_t \ dt + t^2 d W_t$, integrating and noting $Z(0) = 0$ gives $Z(t) = 2\int_0^t sW_s \ dt + \int_0^t s^2 \ dW_s$ subbing htis back into $Y(t)$ I get $Y(t) = \int_0^t s^2 \ dW_s$. How do I show it has 0 drift from here?

  • $\begingroup$ What text are you learning from for this? Most will contain a statement like: If $f \in \mathcal{H}$, then $\int_0^t f(s) \, dW_s$ is a Gaussian process with mean $0$ and covariance function $c(t_1,t_2) = \int_0^{t_1 \wedge t_2} f(s) \,ds$. For instance, it is listed in proposition 15 here. $\endgroup$ – Marcus M Apr 29 '16 at 13:22
  • $\begingroup$ @MarcusM I don't have a particular text - it's an intro class, and all we know is a martingale is one which has drift zero $\endgroup$ – statshelp Apr 29 '16 at 13:53
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    $\begingroup$ When you say "a martingale is one which has drift zero," what is your "one" referring to? A diffusion? How have you defined drift? $\endgroup$ – Marcus M Apr 29 '16 at 14:10
  • $\begingroup$ @MarcusM my one is $Y(t)$, and I have drift defined as the coefficient infront of the $dt$ term, e.g. if $dX = Adt + BdW_t$ then $A$ is the drift, $B$ is the diffusion $\endgroup$ – statshelp Apr 29 '16 at 18:14
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    $\begingroup$ Your statement $Y_t = \int_0^t s^2 \,dW_s$ is the same as $dY = t^2 \, dW_t$. There is no $dt$ term here, so the drift is $0$. $\endgroup$ – Marcus M Apr 29 '16 at 18:49

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