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.
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
The expectation follows from Fubini since $\mathbb{E}\left[\int_0^t B_s^2 \mathrm{d}s\right] = \int_0^t \mathbb{E}[B_s^2] \mathrm{d}s= \int_0^t s\mathrm{d}s = \frac{1}{2}t^2$.
The variance follows from Ito's Isometry and is answered here.
-
$\begingroup$ Thanks! But how to prove it is a random variable? $\endgroup$– jackmCommented Sep 29, 2019 at 18:27
-
$\begingroup$ $X_t=\int_0^t B_s\mathrm{d}s$ 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. $\endgroup$– KevinCommented Sep 29, 2019 at 18:44
-
$\begingroup$ Will $B_s^2$ cause any difference? $\endgroup$– jackmCommented Sep 29, 2019 at 18:49
-
$\begingroup$ 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 $\endgroup$– KevinCommented Sep 29, 2019 at 18:49