I would like to compute the expectation of the stopped local time $L_t^x$ with underlying Brownian Motion $B_t$ with $B_0=x$. Set
$T_a=\inf\{ s \geq 0 \mid B_s=a\}$
with $y<x<a$.
The target of interest is
$\mathbb{E}_x[L_{T_a}^y]$. I would like apply optional stopping theorem. We have by Tanaka's Formula
$$|B_t-y|-|B_0-y|=\int_0^t \operatorname{sgn}(B_s-y)\,dB_s+L_t^y.$$
But for this I have to prove $\int_0^{T_a} \operatorname{sgn}(B_t-y)\,dB_t$ is bounded.
Nevertheless assuming this. We obtain
$|a-y|-|x-y|=\mathbb{E}_x[L_{T_a}^y]$.
We have:
$|a-y|-|x-y|=a-y-|x-y|=a-x$.
So I'm not very sure about this computation. Thanks for every comment
There is some other Question which arises. If we have the case $x<y<a$ we would obtain
$\mathbb{E}_x[L_{T_a}^y]=a+x-2y$. For example for $x=0$ and $y=a3/4$ we get
$\mathbb{E}_x[L_{T_a}^y]=-a/2$
so there is a contradiction. But where is the mistake?
It seems the assumption fails.
So how one can handle
$\mathbb{E}_x[L_y^{T_a}]$?.
It is quite strange here I used the same
expectation of stopped local time
and the result was true.
I know the Formula expectation of stopped local time
holds true nevertheless my mistake (see site 171, Diffusion process and their sample path).
Now let $a <x<y<b$ then we have
$\mathbb{E}_x[L_y^{T_a \wedge T_b}]=2 \dfrac{(x-a)(b-y)}{(b-a)} \leq \mathbb{E}_x[L_y^{T_a }] $
because the local time is increasing.
Now, we have
$\mathbb{E}_x[L_y^{T_a \wedge T_b}]
=2 \dfrac{(x-a)(b-y)}{(b-a)}= 2 \dfrac{b(x-a)(1-y/b)}{b(1-a/b)}= 2 \dfrac{(x-a)(1-y/b)}{(1-a/b)} \rightarrow 2 {(x-a)}$
for $b \rightarrow \infty$.
Now, the continuity of the local time, dominated convergence one obtains
$\mathbb{E}_x[L_y^{T_a}]=2(x-a)$.
We also have
$\mathbb{E}_x[L_y^{T_a \wedge T_b}]=2 \dfrac{a(x/a-1)(b-y)}{a(b/a-1)} =2 \dfrac{(x/a-1)(b-y)}{(b/a-1)} \rightarrow 2(b-y)$ for $a \rightarrow -\infty$.
This leads to
$\mathbb{E}_x[L_y^{ T_b}]=2(b-y)$
By the way.
I hope this computation is on the right way.
I need this to compute the walk Dimension for gap diffusions.
Best Answer
Because the integrand is bounded, the stochastic intergal is uniformly integrable. So $\int_0^{T_a}\operatorname{sgn}(B_s-y)\,dB_s$ has mean zero. The rest of your computation is fine.