Let $\beta$ be a Brownian motion.
How would you solve $ \int^t_0 s \ d\beta(s)$ ?
Motivation: I wanted to deduce that $\int \beta(s) \ ds \sim N(0, t^3/3)$. Therefore rewrote as $\int^t_0 \beta(s) \ ds= t\beta(t)- \int^t_0 s \ d\beta(s) $
I've tried to rewrite as:
- $\lim_{\Delta t \to 0} \sum_k t_k (\beta(t_{k+1})-\beta(t_k))$ but then I got stuck. If I already had a candidate, then I could check the mean-squared convergence. However, I'm not sure how the candidate will be. I've tried $t$ and $\beta(t)$, but none seemed to be the limit.
- $dx=t d\beta$, but this only seems to postpone the problem, since the solution to this SDE requires me to solve a very similar integral...
Proposition If $f$ is a deterministic square-integrable function, then $\int_0^t f d\beta$ is a Gaussian process.
Proof. By considering characteristic functions of finite-dimensional distribution and the definition of the Itô integral, one observes that the $L^2$-limit of Gaussian processes (which arise in the stochastic integral of step functions) is again a Gaussian process.
With that in mind, you are now very close. You might use this to observe that $M_t =\int_0^t s d\beta (s)$ is a mean-zero Gaussian process with covariance $$\mathbb{E}[M_t M_s] = \mathbb{E}\left[ \int_0^t r d\beta (r) \int_0^s r d\beta (r) \right] = \int_0^{\min (t,s)} r^2 dr = \frac{\min (t,s)^3}{3}$$ where the second equality is the covariance formula that follows from Itô's isometry. From this you get that $M_t \sim N(0, t^3 / 3)$.
To "solve" the original integral you intended, you are also very close:
$$\int_0^t \beta(s) ds = t \beta (t) - \int_0^t s d \beta (s) = \int_0^t (t-s) d\beta (s)$$ Which is a mean-zero process with variance $\int_0^t (t-s)^2 ds = \frac{t^3}{3}$,