Lecture 3. Stochastic processes

Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space.

Definition. A ( $d$ -dimensional) stochastic process on $(\Omega, \mathcal{F}, \mathbb{P})$ , is a sequence $(X_t)_{t \ge 0}$ of $\mathbb{R}^d$ -valued random variables that are measurable with respect to $\mathcal{F}$ .

A stochastic process $(X_t)_{t \ge 0}$ can also be seen as an application

$X(\omega) \in \mathcal{A}( \mathbb{R}_{\ge 0}, \mathbb{R}^d), t \rightarrow X_t (\omega).$

The applications $t \rightarrow X_t (\omega)$ are called the paths of the process. The application

$X:(\Omega, \mathcal{F}) \rightarrow (\mathcal{A}( \mathbb{R}_{\ge 0}, \mathbb{R}^d), \mathcal{T}( \mathbb{R}_{\ge 0}, \mathbb{R}^d))$

is easily seen to be measurable, where $\mathcal{A}( \mathbb{R}_{\ge 0}, \mathbb{R}^d)$ denotes the set of functions $\mathbb{R}_{\ge 0} \to \mathbb{R}^d$ endowed with the $\sigma$ -field generated by the cylinders (see Lecture 2). The probability measure defined by

$\mu (A)=\mathbb{P} (X^{-1}(A)), A \in \mathcal{T}(\mathbb{R}_{\ge 0}, \mathbb{R}^d)$

is then called the law (or distribution) of $(X_t)_{t \ge 0}$ .

For $t \geq 0$ , we denote by $\pi_t$ the application that transforms $f \in \mathcal{A}( \mathbb{R}_{\ge 0}, \mathbb{R}^d)$ into $f(t)$ . The stochastic process $(\pi_t)_{t \in \ \mathbb{R}_{\ge 0}}$ which is defined on the probability space $(\mathcal{A}( \mathbb{R}_{\ge 0},\mathbb{R}), \mathcal{T}( \mathbb{R}_{\ge 0}, \mathbb{R}^d),\mu)$ is called the canonical process associated to $X$ . It is a process with distribution $\mu$.

Definition. A process $(X_t)_{t \ge 0}$ is said to be measurable if the application

$(t,\omega) \rightarrow X_t (\omega)$

is measurable with respect to the $\sigma$ -algebra $\mathcal{B}(\mathbb{R}_{\ge 0} ) \otimes \mathcal{F}$ that is, if

$\forall A \in \mathcal{B}(\mathbb{R}^d), \{ (t,\omega), X_t (\omega) \in A \} \in \mathcal{B}(\mathbb{R}_{\ge 0} ) \otimes \mathcal{F}.$

$\mathcal{B}(\mathbb{R}_{\ge 0} )$ denotes here the Borel $\sigma$ -field on $\mathbb{R}_{\ge 0}$ .

The paths of a measurable process are, of course, measurable functions $\mathbb{R}_{\ge 0} \rightarrow \mathbb{R}^d$ .

Definition. If a process $X$ takes its values in $\mathcal{C}(\mathbb{R}_{\ge 0} , \mathbb{R}^d)$ , that is if the paths of $X$ are continuous functions, then we say that $X$ is a continuous process.

If $(X_t)_{t \ge 0}$ is a continuous process then the application

$X:(\Omega, \mathcal{F}) \rightarrow (\mathcal{C}(\mathbb{R}_{\ge 0} , \mathbb{R}^d), \mathcal{B}(\mathbb{R}_{\ge 0} , \mathbb{R}))$

is measurable and the distribution of $X$ is a probability measure on $(\mathcal{C}(\mathbb{R}_{\ge 0} , \mathbb{R}^d), \mathcal{B}(\mathbb{R}_{\ge 0} , \mathbb{R}^d))$ . Moreover, a continuous process is measurable:

Proposition
A continuous stochastic process is measurable.

Proof.

Let $(X_t)_{t \ge 0}$ be a continuous process. Let us first prove that if $A$ is a Borel set in $\mathbb{R}$, then

$\{ (t,\omega) \in [0,1]\times \Omega, X_t (\omega) \in A \} \in \mathcal{B}(\mathbb{R}_{\ge 0}) \otimes \mathcal{F}.$

For $n \in \mathbb{N}$ , let

$X_t^n=X_{\frac{[2^n t]}{2^n}}, t \in [0,1],$

where $[ x]$ denotes the integer part of $x$ . Since the paths of
$X^n$ are piecewise constant, we have

$\{ (t,\omega) \in [0,1]\times \Omega, X^n_t (\omega) \in A \} \in \mathcal{B}(\mathbb{R}_{\ge 0} ) \otimes \mathcal{F}.$

Moreover, $\forall t \in [0,1], \omega \in \Omega$ , we have

$\lim_{n \rightarrow +\infty} X^n_t (\omega)=X_t (\omega),$

which implies

$\{ (t,\omega) \in [0,1]\times \Omega, X_t (\omega) \in A \} \in \mathcal{B}(\mathbb{R}_{\ge 0}) \otimes \mathcal{F}.$

In the same way we obtain that $\forall k \in \mathbb{N}$ ,

$\{ (t,\omega) \in [k,k+1]\times \Omega, X_t (\omega) \in A \} \in \mathcal{B}(\mathbb{R}_{\ge 0} ) \otimes \mathcal{F}.$

Observing

$\{ (t,\omega) \in \mathbb{R}_{\ge 0} \times \Omega, X_t (\omega) \in A \}=\cup_{k \in \mathbb{N}} \{ (t,\omega) \in [k,k+1] \times \Omega, X_t (\omega) \in A \},$
yields the sought of conclusion. $\square$

7 Responses to Lecture 3. Stochastic processes

Viktor says:

October 31, 2013 at 6:57 pm

Hello, I’m sorry if I’m wrong, but I think, that there may be a typo-mistake in that last observing.
But I seem not to understand one thing in the text: In the second definition, 4th row it is said, that for whichever A from T(R^d) [which i understand like a function from algebra T] all (t,\omega) such, that X_t (\omega) is an element of A %stands something%. But how can X_t (\omega) [I understand a function of $t$ for some \omega] be an element of another previously chosen function A?
Thanks a lot for explanation!

Fabrice Baudoin says:

November 3, 2013 at 7:45 pm

That was a typo, I just corrected. Thanks for pointing this out.

zihui says:

February 26, 2014 at 9:23 pm

I am trying to understand the relationships between $X$ and $\pi$. At first this is really confusing since many concepts with different names seemed like the same thing. What are we trying to achieve here?

Suppose we are very clear about the meanings of the following terms: $(\Omega, \mathfrak{F}), (\mathbb{R}, \mathfrak{B}_\mathbb{R}), ([0,\infty), \mathfrak{B}_{[0,\infty)}), (A, \mathfrak{T})$. We clearly understand which $\sigma$-algebra is associated with which space, and later I will often omit writing the $\sigma$-algebras to make notations more concise.

We can view the stochastic process $X$ from different angles, namely

\begin{enumerate}
\item{ $X_t: \Omega \rightarrow \mathbb{R}$, (fix any $t$, $X_t$ is a random variable). }
\item{ $X_\omega: [0,\infty) \rightarrow \mathbb{R}$, (fix any $\omega$, $X_\omega$ is a path in $\mathbb{R}$). }
\item{ $X: [0,\infty) \times \Omega \rightarrow \mathbb{R}$, ($X$ is a process, with two arguments $t$ and $\omega$). }
\item{ $X: \Omega \rightarrow A$, ($X$ is a random variable taking values in the path space $A$). }
\item{ $X: [0,\infty) \rightarrow M$, ($X$ is a path in $M$, where $M$ is the space of all $\mathfrak{F}$-measurable random variables from $\Omega$ to $\mathbb{R}$). }
\end{enumerate}

Based on view 4, $X$ induces a measure (denote it by $\mu$) on $A$, it is called the \textit{law} (or \textit{distribution}) of $X$. (Now we have obtained $\mu$, this is a very important object)

Next we want to introduce $\pi (t,\omega) = \omega_t$, where $\omega \in A$. Similar to the different views we held for $X$, we can also view $\pi$ in the following different ways

\begin{enumerate}[i.]
\item{ $\pi_t: A \rightarrow \mathbb{R}$ }
\item{ $\pi_\omega: [0,\infty) \rightarrow \mathbb{R}$ }
\item{ $\pi: [0,\infty) \times \Omega \rightarrow \mathbb{R}$ }
\item{ $\pi: A_1 \rightarrow A_2$ }, ($\pi$ is the identity map)
\item{ $\pi: [0,\infty) \rightarrow M’$, where $M’$ is the space of $\mathfrak{T}$-measurable random variables from $A$ to $\mathbb{R}$. }
\end{enumerate}

(Notice that up until now we have not mentioned $X$ in the discussion of $\pi$.)

Consider view iv, if we endow the measure $\mu$ on $A_1$ (this is where $X$ kicks in!), then $\pi$ induces law $\mu$ on $A_2$. Since $\pi: A_1 \rightarrow A_2$ and $X: \Omega \rightarrow A$ induce exactly the same laws on $A$, we say that $\pi$ is the \textit{canonical process associated to $X$}.

After all above endeavors it seems view 4 and iv are the key to understanding the relationships between $X$ and $\pi$, therefore we could only focus on these two views and ignore the rest. Am I making the whole thing too complex?

zihui says:

February 26, 2014 at 10:21 pm

Thank you Professor Baudoin, I think I finally understand. When you say “if X is a continuous process then the application …. is measurable” you mean X is measurable viewed as a map from $\Omega$ to $C$; and when you say “a continuous process is measurable” you mean X is measurable viewed as a map from $[0, \infty) \times \Omega$ to $\mathbb{R}$, with the properly associated $\sigma$-algebras of course.

zihui says:

February 27, 2014 at 4:39 pm

When you say “if X is a continuous process then the application X (from \Omega to C) is measurable”, it seems to me that the condition “continuous” is almost irrelevant here, am I wrong?

If X is just any process (not necessarily continuous), and we view it as a function from \Omega to A, with A endowed with the \sigma-algebra T (generated by cylinder sets), then X is automatically measurable, isn’t it? (because for each fixed t, X_t is F-measurable, and cylinder sets only care about finitely many t’s)

If X is a continuous process, i.e. instead of mapping from \Omega to A it is now from \Omega to C, with C endowed with the \sigma-algebra B (again generated by cylinder sets), then X is still measurable, nothing has changed compared to the previous step. Measurable or not it all depends on what \sigma-algebra to choose in the image space (C or A).

When we claim “a continuous process X is measurable”, we mean X is measurable viewed as a function from [0,\infty) x \Omega to R. This is a completely different claim from above therefore needs proof.

luc says:

November 6, 2014 at 2:12 pm

Bonjour,

in the last line, you missed ” + ” , —-> \[ \mathbb{R}_+\]

- Fabrice Baudoin says:
  
  November 6, 2014 at 10:47 pm
  
  Thanks. Corrected.