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?


2 Answers 2


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: $$\begin{aligned}d(f(B_t)e^{\lambda Y_t})&=f(B_t)d(e^{\lambda Y_t})+e^{\lambda Y_t}df(B_t)+d\langle f(B),e^{\lambda Y}\rangle_t\\ &=\lambda (1+\lambda |B_t|)e^{-\lambda |B_t|}e^{\lambda Y_t}\textrm{sgn}(B_t)dB_t+\frac{\lambda^2}{2}(1+\lambda |B_t|)e^{-\lambda |B_t|}e^{\lambda Y_t}dt\\ &-\lambda^2B_te^{-\lambda |B_t|}e^{\lambda Y_t}dB_t+\frac{\lambda^2}{2}(\lambda |B_t|-1)e^{-\lambda |B_t|}e^{\lambda Y_t}dt\\ &-\lambda^3|B_t|e^{-\lambda |B_t|}e^{\lambda Y_t}dt\\ &=(...)dB_t \end{aligned}$$

  • $\begingroup$ 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. $\endgroup$
    – Dovahkiin
    Apr 23 at 18:01
  • 1
    $\begingroup$ @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. $\endgroup$
    – Snoop
    Apr 23 at 18:05
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    $\begingroup$ @ 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! $\endgroup$
    – stats_noob
    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.

  • 1
    $\begingroup$ Thanks for your answer. It really helped that you used the notation from the book. $\endgroup$
    – Dovahkiin
    Apr 23 at 20:09

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