# How to apply Ito's Formula to show that this is a martingale?

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?

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}

• Thanks for your wonderful answer. I have to say, I grossly overlooked the fact that $(1+\lambda|x|)e^{-\lambda|x|}$ is differentiable simply because it did not look so due to the modulus appearing. Apr 23 at 18:01
• @Dovahkiin you're welcome. Indeed that was the only trick needed in my opinion, which is likely why the author did not mention more details about that passage. Apr 23 at 18:05
• @ Snoop: great answer! I also have a question on Stochastic Processes - can you please take a look at this if you have time? math.stackexchange.com/questions/4903926/… thank you so much! Apr 24 at 0:59

Since you are following Le Gall, you can just use the Multidimensional Ito's Formula that appears on page 113 under the heading "Ito's Formula".

You have that for $$X_{t}^{1}$$ and $$X_{t}^{2}$$ being Semi-Martingales \begin{align}F(X_{t}^{1},X_{t}^{2})-F(X_{0}^{1},X_{0}^{2})=\sum_{k=1}^{2}\int_{0}^{t}\dfrac{\partial F}{\partial x^{k}}(X_{s}^{1},X_{s}^{2})\,dX_{s}^{k}+\\\frac{1}{2}\sum_{i,j=1}^{2}\int_{0}^{t}\dfrac{\partial^{2}F}{\partial x^{i}\partial x^{j}}(X_{s}^{1},X_{s}^{2})\,d\langle X^{i},X^{j}\rangle_{s}\end{align}

as per the notations from that book.

Also, since it wasn't directly apparent to you(nor me to be honest) that $$(1+\lambda|x|)e^{-\lambda|x|}$$ was differentiable,

Now, you set $$F(x^{1},x^{2})=f(x^{1})g(x^{2})$$ where $$f(x)=(1+\lambda x)e^{-\lambda x}$$ and $$g(x)=e^{\lambda x}$$ And also set

$$X_{t}^{1}=|B_{t}|$$ and $$X_{t}^{2}=\int_{0}^{t}\text{sgn}(B_{s})\,dB_{s}$$ .

Note that $$X_{t}^{1}$$ is a Semi-Martingale by Tanaka's Formula as you have stated and $$X_{t}^{2}$$ is a Martingale (in fact it's a Standard Brownian Motion).

Also, the Martingale part of both $$X_{t}^{1}$$ and $$X_{t}^{2}$$ are equal by Tanaka's Formula and since both are the same Brownian Motion, you have $$\langle X^{1},X^{2}\rangle_{t}=t$$

Now, you only need to worry about the terms where integrals are not with respect to $$dB_{t}$$

Note that in the first term, you get $$\int_{0}^{t}\lambda^{2} |B_{s}|e^{-\lambda|B_{s}|}e^{\lambda\int_{0}^{t}\text{sgn}(B_{s})\,dB_{s}} d|B_{s}|$$

But, as by Tanaka's Formula you have $$d|B_{t}|=\text{sgn}(B_{t})\,dB_{t}+dL_{t}^{0}$$ and you have that $$dL_{t}^{0}$$ is supported on $$\{t:B_{t}=0\}$$

Thus, the problematic integral wrt $$dL_{s}^{0}$$ vanishes due to the presence of the factor $$|B_{s}|$$ in the integrand.

Now, it is a matter of direct computation, that the integral wrt $$ds$$ vanishes.

This is due to the fact that a elementary computation yields that \begin{align}&\sum_{i,j=1}^{2}\dfrac{\partial^{2}F}{\partial x^{i}x^{j}}\bigg(|B_{t}|\,,\int_{0}^{t}\text{sgn}(B_{s})\,dB_{s}\bigg)\\ \\ &=\lambda^{2}|B_{s}|e^{\lambda\int_{0}^{t}\text{sgn}(B_{s})\,dB_{s}}e^{-\lambda|B_{s}|}-\frac{\lambda^{2}}{2}e^{-\lambda|B_{s}|}\cdot e^{\lambda\int_{0}^{t}\,\text{sgn}(B_{s})\,dB_{s}}\bigg(\lambda|B_{s}|-1\bigg)\\\\ &+\frac{\lambda^{2}}{2}e^{\lambda\int_{0}^{t}\text{sgn}(B_{s})\,dB_{s}}\bigg(1+\lambda|B_{s}|\bigg)=0\end{align}

This shows that the $$F(X_{t}^{1},X_{t}^{2})-F(X_{0}^{1},X_{0}^{2})=\int_{0}^{t}\bigg(\cdots\bigg)\,dB_{s}$$ which shows Martingale Property.

• Thanks for your answer. It really helped that you used the notation from the book. Apr 23 at 20:09