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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.

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I think Z(t) can not be differentiated with respect to time. All you can do is use Ito's lemma as you have already correctly done. Why I say the partial derivative does not exist is Brownian motion is not smooth and the paths of Brownian motion are almost surely non differentiable hence Z too is non differentiable.

http://www2.math.uu.se/~takis/L/BMseminar/BMnotes03_continuity.pdf

If you post the complete problem I might be able to help more.

regards.

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  • $\begingroup$ 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... $\endgroup$ – Dirk Calloway Nov 27 '12 at 12:32
  • $\begingroup$ Hi, You can't write partial derivatives with respect to BM. Only differentials are defined in BM, meaning it makes sense to talk about dZ=f(t)dB but does not make sense to talk about dZ/dB (or partials). Because derivative is a limit. It's for the same reason that BM is not differentiable as if you could write dZ/dB then we know dB/dZ=1/(dz/dB) which is not defined. It is important to differentiate between derivative and differential in stochastic calculus. $\endgroup$ – MathewG Nov 27 '12 at 17:04
  • $\begingroup$ 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 $\frac{\partial f}{\partial x}(t,x)$. $\endgroup$ – Jase Dec 4 '12 at 7:27

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