What's [itex]d\int_0^t s dB(s)[/itex], where B(t) is Wiener process.

[operationsres]

Is it true that d\int_0^t s dB(s) = tB(t)? If not, why not? If not, what's the correct answer?

I've googled extensively for how to find the derivative of an integral, but all sources talk about preceding the integral with \frac{d}{dx}, but I'm preceding it with only d.

 

[jackmell]

Is it true that d\int_0^t s dB(s) = tB(t)? If not, why not? If not, what's the correct answer?

I've googled extensively for how to find the derivative of an integral, but all sources talk about preceding the integral with \frac{d}{dx}, but I'm preceding it with only d.

Well, d means the derivative and it can be with respect to any variable so if you have several, really need to specify which. However, say we just want to invert the integral by writing:

d\int f(x)dx=f(x)

by the Fundamental Theorem of Calculus right. But if we have an expression containing two variables like:

\int_0^t f(x)dx

then need to specify which variable the derivative is taken with respect to like:

\frac{d}{dt}\int_0^t f(x)dx

So that we have:

\frac{d}{dt} \int_0^t f(x)dx=f(t)

\frac{d}{dt} \int_0^t sf(s)ds=tf(t)

\frac{d}{dt}\int_0^t d(f(s))=\frac{d}{dt}\left(f(s)\biggr|_0^t\right)= \frac{d}{dt}\left(f(t)-f(0)\right)=f'(t)

and:

\frac{d}{dt} \int_0^t sd\left(f(s)\right)=tf'(t)

Now try all of those with a real function like f(x)=2+2z^2-3z and see if they work.