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For an equation given below: $$ \int_0^t\phi(s)dW(s) $$ where $W(s)$ is a Brownian Motion. Can I calculate the derivate w.r.t $t$ as follows: $$ \frac{d}{dt} \left( \int_0^t\phi(s)dW(s)\right) = (\phi(t)-\phi(0))(W(t)-W(0)) $$ ? Or is this wrong?

coffee-raid
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1 Answers1

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It is wrong. The derivative does not exist. If $\phi$ is a locally square-integrable deterministic function, the process $\int_0^\cdot \phi(s) dW_s$ has the same distribution as the process $W_{\int_0^\cdot \phi^2(s) ds}$. It is not differentiable in general since $W$ is not dirrerentiable.

Christophe Leuridan
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  • Thank you. So if I want to calculate $dX/dt$ where $X=B(t).\int_0^t\phi(s)dW(s)$ how should I go about it? – coffee-raid Nov 18 '22 at 11:52
  • You cannot calculate it, it does not exist. – Christophe Leuridan Nov 18 '22 at 11:59
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    The derivative of $X$ does not exist but there is an integration-by-parts formula that gives $dX_t=A_t,dB_t+B_t,dA_t+d\langle A,B\rangle_t$. Now plug in the $dW$-integral for $A$ and calculate the covariation between $A$ and $B$. – Kurt G. Nov 18 '22 at 12:08