Consider the Itō integral $X_t := \int_0^t s \,dB_s$.
Here is my attempt. Let $f(t,x) = tx$. By Itō's formula, $$ d f(t, B_t) = B_t dt + t dB_t $$ so $$ t B_t = \int_0^t B_s\, ds + \int_0^t s \,dB_s. $$ But how is $\int_0^t B_s\,ds$ calculated?
Are $\int_0^t f(s) \,dB_s$ and $\int_0^t f(s, B_s)\,ds$ not solvable for any $f$, and generally left as they are when computing Itō integrals and solving Itō SDEs?
Thanks and regards!
I don't think there is an explicit formula for $\int_0^t B_s ds$ but as it is a limit of normal random variables it is still a normal random variable and thus you just need to calculate its variance and mean. It is not difficult to see that $E\left( \int_0^t B_s ds \right)$ is zero. The variance is a little more tricky we have
\begin{align}E\left( \left( \int_0^t B_s ds \right)^2 \right) &= E\left( \int_0^t \int_0^t B_u B_s \, du \, ds \right) \\ \, &= \int_0^t \int_0^t E(B_u B_s) \, du \, ds \\ \, &= \int_0^t \int_0^t \min(u, s) \, du \, ds = \frac{t^3}{3} \end{align} The Fubini steps can be all well justified.