# Brownian Motion and stochastic integration on the complete real line

I'm struggling to understand stochastic integration over intervals containing zero, i.e. integrals of the form $\int_{a}^{b} X_s \, d B_s$ where $-\infty \leq a < b \leq \infty$, $(X_t)_{t \in \mathbb{R}}$ is some adapted process and $(B_t)_{t \in \mathbb{R}}$ is a Brownian motion. Integrals of this type appear over and over again in several articles (of established authors) and the usual martingale techniques (Burkholder-Davis-Gundy, Ito-isometry, etc.) are happily applied as if $a>0$, but always without any justification, so I guess that there's some "standard way" these type of integrals are understood. I've searched for literature on this subject but couldn't find anything, any reference on this is highly appreciated!

As far as I understand, one can define filtrations, (adapted) stochastic processes and martingales in the usual way as all these definitions don't depend on the index set. A Browian motion on $\mathbb{R}$ is then simply an almost surely continuous stochastic process with independent increments such that $B_0=0$ and $B_t-B_s \sim \mathcal{N}(0,t-s)$ for $s \leq t$. Apparently, one way to realize it is by constructing a two-sided Brownian motion $B$ by taking two independent Brownian motions $(B_{1,t})_{t \geq 0}$, $(B_2,t)_{t \geq 0}$ and letting them run in opposite directions: $B_t := B_{1,t}$ if $t \geq 0$ and $B_t := B_{2,-t}$ if $t < 0$.

Here is my first problem: Is there a filtration (on $\mathbb{R}$) in which the two-sided Brownian Motion is a Brownian motion on $\mathbb{R}$? The two filtrations of the two-sided Brownian motion increase in opposite directions...

Now, assuming that the filtration problem can be solved somehow, a naive way of defining a stochastic integral with respect to a two sided Brownian motion $B$ on an interval $(a,b)$ containing zero would be $$\int_{a}^{b} X_s \, d B_s := \int_{a}^0 X_s \, d B_s + \int_0^a X_s d B_s,$$ where $$\int_a^0 X_s \, d B_s := - \int_0^{-a} X_{-s} \, d B_2(s),$$ but with this definition I see no way of associating a martingale on $\mathbb{R}$ with this integral. Is there another definition?

To give you an example where such a stochastic integral is used, see https://sites.google.com/site/giovannipeccati/Home/Publications-by-G-Peccati/PEC1.pdf On page 7 (Thm. 3.1) a Brownian motion on the real line is introduced and on page 10 in the proof a stochastic integral over an interval containing zero appears.

• I guess the stochastic integral on the negative part of the real line becomes a backward martingale and this is why the martingale techniques still apply. Do you have a reference to an article where the stochastic integral on the real line is used? – saz Aug 12 '14 at 16:27
• @saz This would partially make sense. So given a non-decreasing filtration $\mathcal{E}_x$ and a nice process $Y_x$ on $\mathbb{R}$ you would define something like $\int_{-a}^a X_u \, dB_u$ by the sum $\int_{-a}^0 X_u dB_u + \int_0^{a} X_u d B_u$, where the second integral is a backward martingale. This still doesn't explain how to define a Brownian motion on $\mathbb{R}$ and also wouldn't allow you to apply for example Burkholder-Davis-Gundy (because it is not linear) - at least not without additional justification. I've added a reference at the end of my question. – r_faszanatas Aug 14 '14 at 9:59
• As far as I can see, only stochastic integrals of the form $\int_{-a}^{b} \ldots dB_s$ where $a>\infty$ are used there. From my point of view, it is much more easier to define these integrals than the stochastic integral on the whole real line. (But I agree with you; it is an interesting question whether it is possible to define a stochastic integral on the real line with the usual properties.) – saz Aug 14 '14 at 15:10

## 1 Answer

There are two notions that are getting mixed up here.

When people speak of a Brownian motion on the real line (or more generally of a martingale on the real line) they usually refer to a martingale that is indexed by $\mathbb{R}$ with $\lim_{t \to - \infty} X_{t} = 0$. In that case the fact that you index your process with $[0,\infty)$ or $\{-\infty\} \cup \mathbb{R}$ does not change anything and all the formula you know for martingales or Brownian motion stay valid in this setting. This is what Peccati is referring to in his paper you mentioned.

On the other hand, a two-sided Brownian motion cannot be a martingale nor even a local martingale in any filtration and you therefore cannot perform stochastic integration with respect to it. There was some research in that direction in this paper if you are interested.