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$d(g(t)B(t)e^{B(t)})$ how can I calculate this using Ito's formula. I keep getting wrong answers all the time. I am using the Ito's chain rule for $e^{B(t)}$. Obtaining $e^{B(t)}dB(t)+1/2e^{B(t)}dt$.

Chaos
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GAUSS
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1 Answers1

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Let $f(t,B(t)):=g(t)e^{B(t)}$ and then

$$df=g'(t)e^{B(t)}dt+g(t)e^{B(t)}dB(t)+\frac 1 2 e^{B(t)}dt$$

Now you must use the following

$$d( B(t)\cdot f(t,B(t))=B(t)df+fdB(t)+(df)(dB(t)).$$

Chaos
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  • Can ()()^() be a martingale wrt the filtration generated by the brownian motion for some choice of non-random g(t) that is not zero – GAUSS Sep 30 '21 at 12:41
  • @GAUSS I don't think so, an Itô process is a martingale if and only if the drift term is a.s. equal $0$, in this case I don't think you could achieve that. – Chaos Sep 30 '21 at 13:23
  • Agree. We can make it zero by letting g(t)=0 being zeros but then the diffusion part also vanishes, so I guess we cannot, as you also intended. – GAUSS Sep 30 '21 at 13:39