Formulations of a stochastic process

Consider a state space $\stateSpace$ as some nonempty subset of a finite-dimensional vectorspace $\vectorSpace$ with inner-product $\langle\cdot,\cdot\rangle$ and subsequent norm $|\cdot|$. This state space may be equipped with a relative algebra $\stateAlgebra \subseteq \vectorAlgebra$, where $\vectorAlgebra$ is the Borel algebra induced by the topology from $\langle\cdot,\cdot\rangle$. $$ \begin{gathered} \vectorAlgebra = \sigma\Big( \{ v' \in \vectorSpace : |v'-v| < \delta \} \;\big|\; \delta > 0, v \in \vectorSpace \Big) \\ \stateAlgebra = \big\{ A \cap \stateSpace | A \in \vectorAlgebra \big\} \end{gathered} $$ In the weakest form, a stochastic process taking values in $\stateSpace$ at times in the interval $[0,\TT]$ can be seen as a collection $\X = (\X_t)_{t\in[0,\TT]}$ of random variables $\X_t: \Omega \rightarrow \stateSpace$. Unpacking this statement, we are assuming $(\Omega, \Sigma, \Prb)$ is a probability space with $\Sigma$ rich enough to ensure variables $\X_{t_1}, \ldots, \X_{t_n}$ are jointly measurable. $$ \Big\{ \X_{t_1} \in A_1, \ldots, \X_{t_n} \in A_n \Big\} \in \Sigma, \quad t_1, \ldots, t_n \in [0, \TT], ~ A_1, \ldots, A_n \in \stateAlgebra $$ Explicitly, this is the same as declaring a product algebra $$ \stateAlgebra^n = \sigma\Big( A_1 \times \cdots \times A_n | A_1, \ldots, A_n \in \stateAlgebra \Big) $$ and insisting the product map $\X_{\underline t}=(\X_{t_1},\ldots, \X_{t_n}): \Omega \rightarrow \stateSpace^n$ is $\Sigma/\stateAlgebra^n$-measurable for each $\underline t = (t_1, \ldots, t_n) \in [0,\TT]^n$. We provide a proof for this, with the help of a lemma.

Lemma 0.  Suppose $(\bbA, \scrA)$ and $(\bbB, \scrB)$ are measurable spaces with $\scrB = \sigma\calC$ for some collection of sets $\calC$. If a function $f: \bbA \rightarrow \bbB$ is such that $f^{-1}C \in \scrA$ for all $C \in \calC$, then $f$ is $\scrA/\scrB$ measurable.

Show proof.

Proof.  Let $f_\#\scrA \subseteq 2^\bbB$ denote the pushforward algebra. $$ f_\#\scrA = \big\{ B \subseteq \bbB : f^{-1}B \in \scrA \big\} $$ This is a $\sigma$-algebra which contains $\calC$ by hypothesis, so it must too contain $\sigma\calC = \scrB$. Since $\scrB \subseteq f_\#\scrA$, $f$ is $\scrA/\scrB$ measurable. Hide proof.

Result 0.  The joint measurability property of $\X_{t_1}, \ldots, \X_{t_n}$ for some $\underline t = (t_1, \ldots, t_n)$ is the same as $\Sigma/\stateAlgebra^n$-measurability of $\X_{\underline t}$.

Show proof.

Proof.  We first recognize that the following sets are equal. $$ \Big\{ \X_{t_1} \in A_1, \ldots, \X_{t_n} \in A_n \Big\} = \X_{\underline t}^{-1}(A_1 \times \cdots \times A_n) $$ If $\X_{\underline t}$ is $\Sigma/\stateAlgebra^n$-measurable, then the right-hand side of the equality above is an element of $\Sigma$, thus showing the joint measurability property. Meanwhile, if the joint measurability property holds, the equality above shows $\X_{\underline t}^{-1}B \in \Sigma$ for boxes $B$ as above. Since $\stateAlgebra^n$ is generated by such boxes, Lemma 0 proves $\X_{\underline t}$ is $\Sigma/\stateAlgebra^n$-measurable. Hide proof.

Of course, we may equivalently view a stochastic process $\X$ as a random object which realizes paths, i.e. $\omega \mapsto (t \mapsto \X_t(\omega))$. In this case, the process $\X$ is seen as a map $\X: \Omega \rightarrow \stateSpace^{[0,\TT]}$ into the space $\stateSpace^{[0,\TT]}$ of paths $\pathVar: [0,\TT] \rightarrow \stateSpace$. The joint measurability property is then equivalent to declaring projection maps $\pi_{\underline t}: \stateSpace^{[0,\TT]} \rightarrow \stateSpace^n$, $$ \pi_{\underline t}(\pathVar) = \big(\pathVar(t_1),\ldots,\pathVar(t_n)\big), \quad \underline t = (t_1, \ldots, t_n) \in [0,\TT]^n $$ and insisting $\X$ is $\Sigma/\stateAlgebra^{[0,\TT]}$-measurable, where $\stateAlgebra^{[0,\TT]}$ is the weak algebra induced by our projection maps. $$ \stateAlgebra^{[0,\TT]} = \sigma\Big( \pi_{\underline t} \;\big|\; \underline t = (t_1, \ldots, t_n) \in [0,\TT]^n, n \in \bbN \Big) $$ Again, the proof of this is quite simple, with the help of a lemma.

Lemma 1.  Suppose $(\bbA, \scrA)$ and $(\bbB, \scrB)$ are measurable spaces with $\scrB = \sigma\calM$ for some collection of maps $\calM$ of the form $g: \bbB \rightarrow \bbB_g$, where each $\bbB_g$ has some associated $\sigma$-algebra $\scrB_g$. If a function $f: \bbA \rightarrow \bbB$ is such that $g \circ f$ is $\scrA/\scrB_g$-measurable for all $g \in \calM$, then $f$ is $\scrA/\scrB$-measurable.

Show proof.

Proof.  Note that the definition of $\sigma\calM$ is as follows. $$ \sigma\calM = \sigma\big( g^{-1}B : B \in \scrB_g, g \in \calM \big) $$ Because $g \circ f$ is measurable for each $g \in \calM$, we have $$ f^{-1}(g^{-1}B) = (g\circ f)^{-1}B \in \scrA $$ for each $B \in \scrB_g$. Thus, Lemma 0 tells us that $f$ is $\scrA/\scrB$-measurable. Hide proof.

Result 1.  The joint measurability property of $\X_{t_1}, \ldots, \X_{t_n}$ for all $\underline t = (t_1, \ldots, t_n)$ is the same as $\Sigma/\stateAlgebra^{[0,\TT]}$-measurability of $\X$

Show proof.

Proof.  We have already shown the joint measurability property is equivalent to each finite-dimensional distribution $\X_{\underline t}$ being $\Sigma/\stateAlgebra^n$-measurable. This means that each $\pi_{\underline t} \circ \X = \X_{\underline t}$ is $\Sigma/\stateAlgebra^n$-measurable, so Lemma 1 tells us that $\X$ is $\Sigma/\stateAlgebra^{[0,\TT]}$-measurable. Hide proof.

Distribution of a stochastic process

The perspective of $\X$ as a $\Sigma/\stateAlgebra^{[0,\TT]}$-measurable map allows us to explicitly formalize the distribution $\X_\#\Prb$ of the process $\X$, a measure on $\stateAlgebra^{[0,\TT]}$. $$ \X_\#\Prb(A) \defeq \Prb(\X^{-1}(A)) = \Prb(\X \in A), \quad A \in \stateAlgebra^{[0,\TT]} $$ Because $\stateAlgebra^{[0,\TT]}$ is generated by the projection maps $\pi_{\underline t}$, a measure $\mu$ on $\stateAlgebra^{[0,\TT]}$ is determined by how its pushforwards $(\pi_{\underline t})_\#\mu$ evaluate boxes. $$ (\pi_{\underline t})_\#\mu(A_1 \times \cdots \times A_n) = \mu\big( \pi_{\underline t}^{-1}(A_1 \times \cdots \times A_n) \big) $$

Result 2.  Provided two measures $\mu, \nu$ on $\stateAlgebra^{[0,\TT]}$, if $(\pi_{\underline t})_\#\mu = (\pi_{\underline t})_\#\nu$ for all vectors of times $\underline t$, then $\mu = \nu$.

Show proof.

Proof.  Suppose $\mu, \nu$ are measures on $\stateAlgebra^{[0,\TT]}$ whose finite-dimensional distributions agree. $$ (\pi_{\underline t})_\#\mu A = (\pi_{\underline t})_\#\nu A, \quad \underline t = (t_1, \ldots, t_n) \in [0, \TT]^n, A \in \stateAlgebra^n $$ Denote $\calD \subseteq \stateAlgebra^{[0,\TT]}$ the collection of sets $B$ in which $\mu B = \nu B$. Denote $\calP \subseteq \stateAlgebra^{[0,\TT]}$ the set of finite intersections of preimages of projection maps. $$ \calP = \Big\{ \bigcap_{k=1}^m \pi_{\underline{t}_k}^{-1}A_k : m \in \bbN, \underline{t}_k = (t_{k1}, \ldots, t_{kn_k}) \in [0,\TT]^{n_k}, A_k \in \stateAlgebra^{n_k} \Big\} $$

We first show that $\calP \subseteq \calD$. To see this, suppose we had such an intersection $B$; then, denote $\underline t$ as the concatenated vector of times. $$ \underline t = (t_{11}, \ldots, t_{1n_1}, \ldots, t_{m1}, \ldots, t_{mn_m}) \in [0,\TT]^{n_1 + \ldots + n_m} $$ Note that we may rewrite our intersection as follows. $$ \begin{aligned} B&=\bigcap_{k=1}^m \pi_{\underline{t}_k}^{-1}A_k \\ &= \bigcap_{k=1}^m \pi_{\underline t}^{-1}\big( \stateSpace \times \cdots \times \stateSpace \times A_k \times \stateSpace \times \cdots \times \stateSpace\big)\\ &=\pi_{\underline t}^{-1} \prod_{k=1}^m A_k \end{aligned} $$ Since the product set in the final expression above is in $\stateAlgebra^{n_1 + \cdots + n_m}$, we can say $$ \mu B = (\pi_{\underline t})_\#\mu \prod_{k=1}^m A_k = (\pi_{\underline t})_\#\nu \prod_{k=1}^m A_k = \nu B. $$

Now, we simply recognize that $\calP$ is a $\pi$-system[undefined] by construction, and $\calD$ is a $\lambda$-system[undefined]. By the $\pi$-$\lambda$ theorem[undefined], we may now say $\sigma\calP \subseteq \calD$. We now have $\stateAlgebra^{[0,\TT]} = \sigma\calP \subseteq \calD$, so $\mu$ and $\nu$ are identical measures.

Hide proof.

In the context of $\X_\#\Prb$, this means that a the distribution of $\X$ is determined by its finite-dimensional distributions. $$ (\pi_{\underline t})_\#\X_\#\Prb(A_1 \times \cdots \times A_n) = \Prb\big( \X_{t_1} \in A_1, \ldots, \X_{t_n} \in A_n \big) $$ In effect, this result acts as a dimensionality reduction: knowing the distribution of $\X$ only amounts to knowing that of each $\X_{\underline t}$. Note that the same is not true, knowing only the distribution of each marginal $\X_t$, $t \in [0,\TT]$.

Example.  There exists a probability space $(\Omega, \Sigma, \Prb)$ equipped with stochastic processes $X$ and $Y$ such that there exists an event set $A \in \stateAlgebra^{[0,\TT]}$ such that $$ \Prb(X \in A) \neq \Prb(Y \in A), $$ despite the following equality for all $t \in [0,\TT]$ and $B \in \stateAlgebra$. $$ \Prb(X_t \in B) = \Prb(Y_t \in B) $$

Show proof.

Proof.  For those already somewhat comfortable with stochastic processes, the answer is easy. Suppose $X$ is a Brownian motion and $Z$ is a standard normal. Define $Y_t = \sqrt{t} Z$ and see that $$ \begin{aligned} &\Prb\big(X_{69} \in [0,\infty), X_{420} \in (-\infty, 0)\big) \\ &\quad= \int_0^\infty \int_{-\infty}^0 (702\pi)^{-1/2} \exp\Big(-\frac{(x_2-x_1)^2}{702}\Big) (138\pi)^{-1/2} \exp\Big( -\frac{x_1^2}{138} \Big)\rmd x_2 \rmd x_1 \\ &\quad> 0 \\ &\quad= \Prb\big(Y_{69} \in [0,\infty), Y_{420} \in (-\infty, 0) \big) \end{aligned}$$ Note that the last probability is $0$ simply due to the fact that if $Y_{69} \geq 0$, then $Z \geq 0$, so $Y_{420} \geq 0$ as well. Note that the inequality tells us that $$ \Prb(X \in A) > \Prb(Y \in A) $$ for $A = \pi_{69}^{-1}[0,\infty) \cap \pi_{420}^{-1}(-\infty,0)$, despite the fact that $$ \Prb(X_t \in B) = \int_B (2\pi t)^{-1/2} \exp\Big(-\frac{x^2}{2t}\Big) {\rm d}x = \Prb(Y_t \in B)$$ for all $t \in [0,\TT]$ and $B \in \scrR$ (where $\scrR$ is the Borel algebra on the real number line).

We translate this argument to measures, for the mathematicians that do not know what a Brownian motion is. As we will describe in the next section, there exist measures $\mu$ and $\nu$ on $\scrR^{[0,\TT]}$ (where $\scrR$ is the Borel algebra on $\bbR$) that evaluate on the specific family of subsets as follows. $$\begin{aligned} \mu\big( \pi_{\underline t}^{-1}A\big) &= \int_A \big(2\pi(t_n-t_{n-1})\big)^{-1/2} \exp\Big( -\frac{(x_n-x_{n-1})^2}{2(t_n-t_{n-1})} \Big) \\ &\hspace{3em} \times\cdots\times \big(2\pi(t_2-t_1)\big)^{-1/2} \exp\Big( -\frac{(x_2-x_1)^2}{2(t_2-t_1)} \Big)\\ &\hspace{3em} \times \big(2\pi t_1\big)^{-1/2} \exp\Big(-\frac{x_1^2}{2t_1}\Big) \rmd x_n \cdots \rmd x_1 \\[1em] \nu\big(\pi_{\underline t}^{-1}A\big) &= \int_\bbR 1_A\big(\sqrt{t_1}x, \ldots, \sqrt{t_n} x \big) (2\pi)^{-1/2} \exp\Big(-\frac{x^2}{2}\Big)\rmd x \\ \end{aligned}$$ Now, we observe that $\mu \neq \nu$. $$\begin{aligned} &\mu\big( \pi_{(69,420)}^{-1}\big([0,\infty) \times (-\infty, 0)\big) \big) \\ &\quad= \int_0^\infty\int_{-\infty}^0 (702\pi)^{-1/2} \exp\Big(-\frac{(x_2-x_1)^2}{702}\Big) (138\pi)^{-1/2} \exp\Big( -\frac{x_1^2}{138} \Big) \rmd x_2 \rmd x_1 \\ &\quad>0 \\ &\quad=\int_\bbR 1_{[0,\infty) \times (-\infty,0)}(\sqrt{69} x, \sqrt{420}x) \big(2\pi\big)^{-1/2} \exp\Big(-\frac{x^2}{2}\Big) \rmd x \\ &\quad= \nu\big( \pi_{(69,420)}^{-1}\big([0,\infty) \times (-\infty, 0) \big)\big) \end{aligned}$$ Similar to before, the final integral is $0$ because no $x$ satisfies $\sqrt{69}x \geq 0$ and $\sqrt{420}x < 0$. However, for each $t \in [0,\TT]$ and $B \in \scrR$, $$ \begin{aligned} \mu \pi_t^{-1}B &= \int_B (2\pi t)^{-1/2} \exp(-x^2/(2t)) \rmd x \\ &= \int_\bbR 1_B(\sqrt{t}x) (2\pi)^{-1/2} \exp(-x^2/2) \rmd x \\ &= \nu \pi_t^{-1} B \end{aligned}$$ To bring this back to the language of stochastic processes, we simply construct our probability space from $\mu$ and $\nu$, as follows. $$\begin{gathered} \Omega = \bbR^{[0,\TT]} \times \bbR^{[0,\TT]},\\ \Sigma = \scrR^{[0,\TT]} \otimes \scrR^{[0,\TT]}, \\ \Prb = \mu \otimes \nu, \\ X(\xi, \upsilon) = \xi, \\ Y(\xi, \upsilon) = \upsilon \end{gathered} $$

Hide proof.

Construction of a stochastic process

When tasked with constructing a stochastic process $\X$, we must provide a measure $\mu$ on $\stateAlgebra^{[0,\TT]}$ to serve as the distribution. Of course, then $$(\Omega, \Sigma, \Prb, \X) = (\stateSpace^{[0,\TT]}, \stateAlgebra^{[0,\TT]}, \mu, \operatorname{identity})$$ serves as a suitable space. At first thought, proposing a family measures $(\mu_{\underline t})$ which are to be associated to the finite-dimensional distributions $\mu_{\underline t} = (\pi_{\underline t})_\#\mu$ seems sufficient in declaring $\mu$, but it is quickly apparent that these finite-dimensional distributions must be declared in a consistent fashion. Namely, if our family of finite-dimensional distribution proposals $(\mu_{\underline t})$ satisfies the following projection and symmetry properties, $$ \begin{aligned} \mu_{(t_1, \ldots, t_n, t_{n+1})}(A_1 \times \cdots \times A_n \times \stateSpace) &= \mu_{(t_1, \ldots, t_n)}(A_1 \times \cdots \times A_n) \\ \mu_{(t_{\gamma 1}, \ldots, t_{\gamma n})}(A_1 \times \cdots \times A_n) &= \mu_{t_1, \ldots, t_n}(A_{\gamma^{-1}1} \times \ldots \times A_{\gamma^{-1}n}), &\quad \gamma \text{ a permutation} \end{aligned} $$ we may say that there exists some measure $\mu$ on $\stateAlgebra^{[0,\TT]}$ with finite-dimensional distributions $(\pi_{\underline t})_\#\mu = \mu_{\underline t}$. This is a result known as the Kolmogorov extension theorem[undefined].

In our earlier example, we defined measures $\mu$ and $\nu$ on $\scrR^{[0,\TT]}$ by declaring how they acted on preimages $\pi_{\underline t}^{-1} A$. Note that $\mu\pi_{\underline t}^{-1}A = (\pi_{\underline t})_\#\mu A$, so we were effectively defining the finite-dimensional distributions. It is easy to verify that each of our $\mu, \nu$ satisfied the consistency conditions of the Kolmogorov extension theorem, hence why they were well-defined.

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