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If $f(t)$ is a deterministic function of $t$ and $B_{n}$ is a brownian motion and:

$Z =\int^t_0 f(s)dB(s)$

How does one take the partial derivatives wrt to $t$ and $B_n$ on an integral like this?

I know $dZ = f(t)dB(t)$

Is this just?...

$\frac{\partial z}{\partial t} = f(t)$

and

$\frac{\partial z}{\partial B} = f(t)dB(t)$

Looking to apply the Ito formula on a bigger problem but stuck on this. Thanks.

1 Answers 1