I have a short question regarding this problem:
$B$=$B_t$ is standard Brownian motion
Process $q$ = $q_t$ defined by $q_t$ = $B_t^2$-t
Show that $E$[$q_t$]=$0$ and $Var$[$q_t$]=$2t^2$
What am I looking for here? I tried some different approaches but I'm not really sure how to solve this. I've been trying since 3-4 days which is rather frustrating considerung that the solution will probably have like 2 lines :D
Well anyways, I'm not necessarily looking for the solution straight away but if anyone could guide me into the right directiong that would be immensly helpful! Thanks
What do you know about a Brownian motion? $B_t = B_t - B_0$ is an increment so its $\mathcal{N}(0,t)$ distributed
So $E[B_t] = 0$ and $Var(B_t) = E[B_t^2] - E[B_t] = E[B_t^2] = t$
So you have: $E[q_t] = E[B_t^2] - t = t-t=0$
Because you know the distribution of $B_t$ you know it's density $f_{B_t}$ and can calculate $E[g(B_t)]$ for any measurable function by $$\int_{-\infty}^\infty g(x) f_{B_t}(x) dx$$
Reacognising $Var(q_t) = Var(B_t^2 - t) = Var(B_t^2) = E[B_t^4] - E[B_t^2]^2 = E[B_t^4] - t^2$ gives you the variance by calculation $E[B_t^4]$ what you can do as described above setting $g(x) = x^4$ (or checking Wikipedia for the 4th moment of a normal rv)