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In some financial application, we are given that

$$X(t)=B(t)\times[{X(0)+\int_{0}^{t}\lambda(s)dW(s)}]$$

and need to calculate the $dX(t)$ using Ito's product rule. Assuming $X(0)$ is given, then $d[X(0)+\int_{0}^{t}\lambda(s)dW(s)]$ is simply $\lambda(t)dW(t)$ and if we know that $dB(t)=\alpha(t)B(t)dt$ then we have

$$dX(t)=\alpha(t)B(t)dt\times[{X(0)+\int_{0}^{t}\lambda(s)dW(s)}] + B(t)\times \alpha (t)B(t)dt$$

(no third term as $dtdW(t)$ cancel out.)

Is there a way to simplify the first term $\alpha(t)B(t)dt\times[{X(0)+\int_{0}^{t}\lambda(s)dW(s)}]$, i.e. how do you deal with $(...)dt\times \int_{0}^{t}...dW(s)$?

  • $$\alpha(t)X(t)dt\ ?$$ – Did Mar 12 '17 at 13:48
  • Not sure i understand - are you saying that $a(t)B(t)dt \times [X(0) + \int_{0}^{t}\lambda(s)dW(s)$ is $a(t)X(t)dt$? If so, could you please explain? –  Mar 12 '17 at 16:25
  • Sorry but you simply might want to reread your own question. – Did Mar 12 '17 at 17:13

0 Answers0