Suppose we have a process of the form $X_t=\int_0^t g(x,t)dB_t$, where $g$ is a continously differentiable function (on both parameters). Let $f(t,x)$ and $F(t,x)$ denote the density and the cumulative density functions of the process at time $t$, respectively. Then it is true that:
$$\mathbb{E}[L_t^x]=\int_0^tg(x,s)^2f(x,s)ds$$
I have managed to prove it when $g \equiv 1$:
\begin{align*} \mathbb{E}[L_t^x]&= \mathbb{E}\bigg[\lim\limits_{\epsilon \to 0} \frac{1}{2\epsilon} \int_0^t \chi_{\{x-\epsilon \leq X_s \leq x + \epsilon\}}d[X]s\bigg] \\ &=\lim\limits_{\epsilon \to 0} \frac{1}{2\epsilon} \int_0^t \mathbb{E}[\chi_{\{x-\epsilon \leq X_s \leq x + \epsilon\}}]ds \\ &=\lim\limits_{\epsilon \to 0} \frac{1}{2\epsilon} \int_0^t [F(s,x+\epsilon)-F(s,x-\epsilon)] ds \\ &=\int_0^t \lim\limits_{\epsilon \to 0} \frac{1}{2\epsilon}[F(s,x+\epsilon)-F(s,x-\epsilon)] ds \\ &=\int_0^t f(s,x) ds \\ \end{align*}
Can someone help me do it for more general $g$'s? Thank you.