In the book Brownian Motion, Martingales and Stochastic Calculus by J.F. Le Gall, in order to give an alternatice derivation of the distribution of $L_{U_{a}}^{0}(B)$ where $L^{0}_{t}(B)$ is the Local-Time at $0$ of a Standard Brownian Motion and $U_{a}=\inf\{t:|B_{t}|\geq a\}$, he states as a remark that
"use Itô’s formula to verify that, for every $\lambda>0$, $$(1+\lambda |B_{t}|)e^{-\lambda L_{0}^{t}(B)}$$ is a continuous Martingale (local)".
My question: What is the function that we are supposed to apply Ito's Formula to?
I can write $L_{0}^{t}(B)=|B_{t}|-\int_{0}^{t}\text{sgn}(B_{s})\,dB_s$ by using Tanaka's Formula. In that case, I will get
$$(1+\lambda|B_{t}|)\exp\bigg(-\lambda |B_{t}|+\lambda\int_{0}^{t}\text{sgn}(B_{s})\,dB_{s}\bigg)$$
But the problem is I cannot express it in the form of $f(X_{t})$ where $f$ is some function and $dX_{t}=\mu_{t}\,dt+\sigma_{t}\,dB_{t}$ inorder to apply Ito's Formula.
Can someone help me out with this?
Best Answer
Note $f(x)=(1+\lambda |x|)e^{-\lambda |x|}$ is twice continuously differentiable. Indeed: $f'(x)=-\lambda^2xe^{-\lambda |x|}$ and $f''(x)=\lambda^2e^{-\lambda |x|}(\lambda |x|-1)$. Furthermore, if $Y_t:=\int_0^t\textrm{sgn}(B_s)dB_s$ then by Ito $d(e^{\lambda Y_t})=\lambda e^{\lambda Y_t}\textrm{sgn}(B_t)dB_t+\lambda^2e^{\lambda Y_t}dt/2$ and so $$d\langle f(B),e^{\lambda Y}\rangle_t=-\lambda^3B_t\textrm{sgn}(B_t)e^{-\lambda |B_t|}e^{\lambda Y_t}dt=-\lambda^3|B_t|e^{-\lambda |B_t|}e^{\lambda Y_t}dt$$ and we conclude: $$\begin{aligned}d(f(B_t)e^{\lambda Y_t})&=f(B_t)d(e^{\lambda Y_t})+e^{\lambda Y_t}df(B_t)+d\langle f(B),e^{\lambda Y}\rangle_t\\ &=\lambda (1+\lambda |B_t|)e^{-\lambda |B_t|}e^{\lambda Y_t}\textrm{sgn}(B_t)dB_t+\frac{\lambda^2}{2}(1+\lambda |B_t|)e^{-\lambda |B_t|}e^{\lambda Y_t}dt\\ &-\lambda^2B_te^{-\lambda |B_t|}e^{\lambda Y_t}dB_t+\frac{\lambda^2}{2}(\lambda |B_t|-1)e^{-\lambda |B_t|}e^{\lambda Y_t}dt\\ &-\lambda^3|B_t|e^{-\lambda |B_t|}e^{\lambda Y_t}dt\\ &=(...)dB_t \end{aligned}$$