# Partial Derivative of an Integral

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

$$Z =\displaystyle\int^t_0 f(s)d\left(B(s)\right)$$

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?...

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

and

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

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

• You're right I think. I was trying to work out a characteristic function for this and if you set dt to zero it computes. Thanks. FYI ${\partial z \over \partial B} = f(t)$ I believe... – Dirk Calloway Nov 27 '12 at 12:32
• Do you know if/why my answer is incorrect (if yours is correct)? I know that it's false to say that "you can't write partial derivatives w.r.t. BM". For example apply Ito's lemma to the function $f(t,x)$, where $x$ is $B(t)$ (in fact this is one of the most standard routines in stochastic finance). Applying Ito's lemma you will end up having to evaluate $\frac{\partial f}{\partial x}(t,x)$. – Jase Dec 4 '12 at 7:27