Formalization of a Brownian motion and of stochastic integrals in Lean

2 Stochastic processes

Let \(T\) be an index set and \(\Omega \) a measurable space, with measure \(\mathbb {P}\). A stochastic process is a function \(X : T \to \Omega \to E\), where \(E\) is another measurable space, such that for all \(t \in T\), \(X_t : \Omega \to E\) is \(\mathbb {P}\)-a.e. measurable.

Definition 2.1 Law of a stochastic process
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The law of a stochastic process \(X\) is the measure on the measurable space \(E^T\) obtained by pushing forward the measure \(\mathbb {P}\) by the map \(\omega \mapsto X(\cdot , \omega )\).

Lean remark: we don’t use a Lean definition for the law, but write the map in full.

Definition 2.2 Modification
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We say that a stochastic process \(Y\) is a modification of another stochastic process \(X\) if for all \(t \in T\), \(Y_t =_{\mathbb {P}\text{-a.e.}} X_t\).

Lean remark: we don’t use a Lean definition for being a modification, but write explicitly the condition \(\forall t \in T,\ Y_t =_{\mathbb {P}\text{-a.e.}} X_t\) .

Definition 2.3 Indistinguishable
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We say that a stochastic processes \(Y\) is a indistinguishable from \(X\) if \(\mathbb {P}\)-a.e., for all \(t \in T\), \(X_t = Y_t\).

A summary of the next few lemmas is this:

  • indistinguishable \(\implies \) modification \(\implies \) same law,

  • modification and continuous with \(T\) separable \(\implies \) indistinguishable.

Lemma 2.4
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If \(Y\) is indistinguishable from \(X\), then \(Y\) is a modification of \(X\).

Proof ▶

Obvious.

Lemma 2.5
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Let \(X, Y : T \to \Omega \to E\) be two stochastic processes that are modifications of each other. Then for all \(t_1, \ldots , t_n \in T\), the random vector \((X_{t_1}, \ldots , X_{t_n})\) has the same distribution as the random vector \((Y_{t_1}, \ldots , Y_{t_n})\). That is, \(X\) and \(Y\) have same finite-dimensional distributions.

Proof ▶

By the modification property, almost surely \(X_{t_i} = Y_{t_i}\) for all \(i \in [n]\). Thus the function \(\omega \mapsto (X_{t_1}(\omega ), \ldots , X_{t_n}(\omega ))\) is equal to \(\omega \mapsto (Y_{t_1}(\omega ), \ldots , Y_{t_n}(\omega ))\) almost surely, hence the maps of \(\mathbb {P}\) by these two functions are equal.

Let \(X, Y : T \to \Omega \to E\) be two stochastic processes. Then \(X\) and \(Y\) have same finite-dimensional distributions if and only if they have the same law.

Proof ▶

TODO: consider the \(\pi \)-system of cylinder sets.

Lemma 2.7
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Let \(T\) and \(E\) be topological spaces and suppose that \(T\) is separable Hausdorff. Let \(X, Y : T \to \Omega \to E\) be two stochastic processes that are modifications of each other and are almost surely continuous. Then \(X\) and \(Y\) are indistinguishable.

Proof ▶

Since \(T\) is separable, it has a countable dense subset \(D\). Since \(D\) is countable,

\begin{align*} (\forall t \in D, \mathbb {P}\text{-a.e.}, X_t = Y_t) \iff (\mathbb {P}\text{-a.e.}, \forall t \in D, X_t = Y_t) \end{align*}

Hence by the modification property we have that almost surely, for all \(t \in D\), \(X_t = Y_t\). Then almost surely \(X\) and \(Y\) are continuous functions which are equal on a dense subset of \(T\): those two functions are equal everywhere.