How to evaluate $ \int s dB_s $?

Question

Is there a way to evaluate the stochastic integral (we integrate f(x)=x against brownian motion) With evaluate I mean to write it in an explicit form, as we can write: $$ \int_0^t B_s dB_s = B_t^2/2-t/2$$

Can we also find an explicit form of this:

$$ \int _0 ^t s dB_s $$

It seems to me that Ito's formula is of no use because the integrand is not a local martingale.

If there is no easy form for the integral, can we still say something about the quantity

$$ E[ (\int_0 ^t s d B_s)^2 ]$$

Based on Did's comment I would say that: \begin{align} E[( \int_0^t s dB_s)^2] &= E[(\int_0^\infty s \mathbb 1\{s<t\} dB_s)^2\ ] \\ &= E[(\int_0^\infty s \mathbb 1\{s<t\} dB^t_s)^2\ ] \\&= E[ \int_0^\infty (s^2 \mathbb 1\{s<t\} d[B^t]_s) ] \\ & = E [ \int _0 ^t s^2 ds] \\ &=t^3/2\end{align} Here the third equality is he definition of Ito Isometry as I learnt it, that for suitable processes H, and when the inegrator is an L2 bounded martingale $M_t$ , which $B^t_s:= B_t^s$ is,

$$E[(\int_0^\infty H_s dM_s)^2]=E(\int_0^\infty H_s^2 d[M]_s). $$

For the second equality I want to change the integrator to an L2 bounded martingale, by stopping the BM after time t. Intuitively this doesn't change the integral as the integrad is simply zero after time t, but I'm unsure how to prove this.

I hope someone can comment on my solution.

Answer 1

Hint There is no expression for the integral in terms of $B_t$ or whatever, but its distribution is known exactly. Since the increments $sdB_s$ you're adding up are normal with mean zero and variance $s^2ds,$ $\int_0^t sdB_s$ is normal with variance....

Answer 2

Take $s_i^{(n)} = \frac{i}{n}\cdot t$ and then form the approximation $$ I = \int_0^t s dB_s \approx \sum_{i=0}^{n-1}s_i^{(n)} \left( B(s_{i+1}^{(n)}) - B(s_i^{(n)}\right) $$ now using the identity $$ c(b-a) = (db - ca) - b(d-c) $$ with a judicious choice of $d$ we have $$ I \approx \sum_{i=0}^{n-1} \left[ s_{i+1}^{(n)}B(s_{i+1}^{(n)})-s_{i}^{(n)}B(s_i^{(n)}) \right] - \sum_{i=0}^{n} B(s_{i+1}^{(n)})\left( s_{i+1}^{(n)} - s_i^{(n)} \right) $$ Will leave it to you to finish tidying those sums and then take the appropriate limits and make everything rigorous.