# integration of squared brownian motion w.r.t time

How to prove $$\int_0^1 B_s^2ds$$ is a random variable and compute its first two moments? From excercise 1.15 on the book martingales and brownian motion.

The expectation follows from Fubini since $$\mathbb{E}\left[\int_0^t B_s^2 \mathrm{d}s\right] = \int_0^t \mathbb{E}[B_s^2] \mathrm{d}s= \int_0^t s\mathrm{d}s = \frac{1}{2}t^2$$.
• $X_t=\int_0^t B_s\mathrm{d}s$ defines a stochastic process $(X_t)$. It is just the limit of the corresponding Riemann sums. A Riemann sum is just a linear combination of Brownian motions (this is measurable) and so is the limit. For every fixed time poin $t$, a stochastic process $(X_t)$ is just a random variable. Sep 29 '19 at 18:44
• Will $B_s^2$ cause any difference? Sep 29 '19 at 18:49
• No, sorry, it was just a typo, I actually wanted to write $X_t=\int_0^t B_s^2\mathrm{d}s$. My fault, sorry Sep 29 '19 at 18:49