$E[(\int_0^t 1_{y_k} (X_s) X_s ds)^2]=0$, where $X_s$ is a geometric Brownian motion.

Question

I want to show that $E[(\int_0^t 1_{y_k} (X_s) X_s ds)^2]=0$, where $1_{y_k}$ is an indicator function on a single point and $X_s$ is a geometric Brownian motion. I want to use Fubini, in the case of $dB_s$ I can use hte Ito isometry to solve this but what theorem can I use in this case to handle the square? I would greatly appreciate any help.

Answer 1

Since $X_s$ has a continuous distribution for any $s > 0$, $1_{y}(X_s) = 0$ a.s. By Fubini and a bound on $E|X_s|$, $\int_0^t 1_{y}(X_s) X_s\; ds = 0$ a.s.