# Is this stochastic Picard iterator well-defined?

Preliminaries

• Let $$x_0 \in \mathbb{R}^d$$.
• Let $$T \in (0, \infty)$$.
• Let $$\sigma: \mathbb{R}^d \rightarrow \mathbb{R}^{d \times d}$$ and $$\mu: \mathbb{R}^d \rightarrow \mathbb{R}^{d}$$ be affine linear transformations (which implies, that they are smooth and globally Lipschitz).

• Let $$(\Omega, \mathcal{G}, (\mathcal{G}_t)_{t \in [0,T]}, \mathbb{P})$$ be a complete probability space with a complete, right-continuous filtration $$(\mathcal{G}_t)_{t \in [0,T]}$$.

• Let $$B : [0,T] \times \Omega \rightarrow \mathbb{R}^d$$, $$(t,\omega) \mapsto B_t(\omega)$$, be a standard $$d$$-dimensional $$(\mathcal{G}_t)_{t \in [0,T]}$$-adapted Brownian motion on $$\mathbb{R}^d$$, such that $$B_0 = 0$$ and such that, for every pair $$(t,s) \in \mathbb{R}^2$$ with $$0 \leq t < s$$, the random variable $$B_s-B_t$$ is independent of $$\mathcal{G}_t$$.

Question

Consider the set $$\begin{equation} \begin{gathered} \mathcal{S}_{T} := \{ S: [0,T] \times \Omega \rightarrow \mathbb{R}^d \mid \\ S \ \text{is} \ (\mathcal{G}_t)_{t \in [0,T]} \text{-adapted and has \mathbb{P}-a.s. continuous sample paths} \} . \end{gathered} \end{equation}$$ I would like to define a type of Picard iterator function via $$\begin{equation} \begin{gathered} \mathcal{I}_{\text{Pic}}^{(x_0, \sigma, \mu)} : \mathcal{S}_{T} \rightarrow \mathcal{S}_{T}, \\ \mathcal{I}_{\text{Pic}}^{(x_0, \sigma, \mu)}(S)_t = x_0 + \int_{0}^{t} \mu (S_s) ds + \int_{0}^{t} \sigma(S_s) dB_s. \end{gathered} \end{equation}$$ But is $$\mathcal{I}_{\text{Pic}}^{(x_0, \sigma, \mu)}$$ truly well-defined for all $$t$$ and $$\omega$$? Are both integrals in the expression always well defined for all $$t$$ and $$\omega$$? And, most importantly, does $$\mathcal{I}_{\text{Pic}}^{(x_0, \sigma, \mu)}$$ really map $$\mathcal{S}_{T}$$ into itself?

(I am still learning the basics of stochatic calculus.)

• The exact details might depend on how you defined the integral against Brownian motion. What is your definition? – Rhys Steele Apr 22 '19 at 22:01
• I am using the definition described in Le Gall's "Brownian Motion, Martingales, and Stochastic Calculus" ; the integral against Brownian motion here is supposed to be an Ito (rather than a Stratonovich ) integral. – Joker123 Apr 22 '19 at 22:18
• Good to know. Unfortunately, the reason I asked wasn't to determine whether you meant Ito or Stratonovich, but rather to check how the Ito integral is being constructed. I'll take a look at what Le Gall does when I get a chance. – Rhys Steele Apr 22 '19 at 22:20
• There are various ways. They all lead to essentially the same object but in principle differ depending on when you identify equivalence classes of functions and on whether your construction requires you to take modifications to get a continuous process. These things are mainly pedagogical, I just wanted to avoid confusing you by mentioning technicalities from a different construction. Once all is done, each construction will yield something that take equivalence classes of progressive integrands to a suitable local martingale and so you can just think of the construction of Le Gall. – Rhys Steele Apr 22 '19 at 22:57
• You might be interested in my answer here which discusses two different constructions given in books by Oksendal and by Karatzas and Shreve. The one given by Karatzas and Shreve is iirc at least very close to the one given by Le Gall. – Rhys Steele Apr 22 '19 at 22:59

In this answer, following your comment, I assume the Ito integral has been constructed as in Le Gall's book. Since we work on a bounded time interval, Brownian motion is a continuous $$L^2$$-bounded martingale there and so we only need that part of the construction.

The main technicality is that the Ito integral there is really defined for equivalence classes of integrands. That is, Le Gall defines the Ito integral for integrands in the space $$L^2(B):=L^2(\Omega \times [0,T], \mathcal{P}, d\mathbb{P} \otimes ds)$$ where $$\mathcal{P}$$ is the progressive $$\sigma$$-algebra. This space consists of equivalence classes of $$d\mathbb{P} \otimes ds$$-a.e. equal processes (I can tell you haven't done this since your processes are only almost surely continuous). So in order to take an Ito integral of $$S \in \mathcal{S}_T$$ we first should identify $$S$$ with its $$d\mathbb{P} \otimes ds$$ equivalence class. Let's agree that we implicitly do this when we write the Ito integral.

Another technicality is that the integral $$\int_0^t \mu(S_s)(\omega) ds$$ need not be defined for all $$\omega$$. For example if $$d = 1, \mu = \operatorname{Id}$$ this is essentially asking to define the Riemann integral for an arbitrarily irregular path, which we cannot do. So we need to agree what to do on the measure $$0$$ set of discontinuous paths. One way around this issue is to agree to integrate a representative of $$S$$ for which every sample path is continuous. Of course, there are many such representatives. However, the integrals we obtain this way can only differ on a measure $$0$$ set and so are versions of each other.

Having done this, since Le Gall's construction of the Ito integral against a continuous $$L^2$$-bounded martingale is a continuous $$L^2$$-bounded martingale and the Riemann integral is continuous, the process $$\mathcal{I}_{\text{Pic}}^{(x_0, \sigma, \mu)}(S)$$ is then defined as an equivalence class of continuous processes (all of which are versions of each other) and in particular are defined for all $$t$$.

Finally, by approximating each integral by Riemann sum type approximations (which are adapted) we see that each member of this equivalence class is adapted.

The short version of this is that the question is a bit fiddly because often in this subject we write down a process when we really mean a whole equivalence class of processes. What the above says is that if you replace $$\mathcal{S}_T$$ by $$\mathcal{S}_T / \sim$$ where $$S \sim \hat{S}$$ if and only if $$\hat{S}$$ is a version of $$S$$, then $$\mathcal{I}_{\text{Pic}}^{(x_0, \sigma, \mu)}(S): \mathcal{S}_T \to \mathcal{S}_T.$$ If you don't do this, there's a bit of a type mismatch and you need to pick a member of an equivalence class in a somewhat arbitrary way. In truth, the usual way to deal with these issues is to just agree we've always identified processes in the right manner from the start.

• Usually, a continuous version of the Ito integral is chosen, so it is not really an equivalence class. – zhoraster Apr 23 '19 at 8:39
• @zhoraster Maybe I was insufficiently clear. My equivalence class consists of all different continuous versions of the same process, rather than processes that are only equal $ds \otimes d \mathbb{P}$ everywhere. I don't want to pick one distinguished continuous version of the process since then I'd have no reason to hope that the way I picked these versions makes the Ito integral a linear map. – Rhys Steele Apr 23 '19 at 15:58
• Makes sense. There is an extensive literature devoted to linear selectors for the conditional expectation, but I do not remember having seen the same on stochastic integrals. Also I wonder why one wouldn't be satisfied with almost sure linearity. – zhoraster Apr 23 '19 at 18:21
• @zhoraster there might be something I miss but all that's obvious to me when you choose representatives is that for each linear combination you can choose a null set such that off that null set you get linearity for that specific linear combination. It seems like this is probably in practice enough (I cant think of a case where I'd want more than this) but seems also to me to be weaker than almost sure linearity. Do you know if it actually ensures almost sure linearity? – Rhys Steele Apr 23 '19 at 18:56
• "you get linearity for that specific linear combination". Right. Then you get almost sure linearity for countably many linear combinations. And this is usually enough for most problems. But not for all. E.g. in solving optimal control problems, these subtleties may show up. – zhoraster Apr 24 '19 at 14:28