On a definition of a probability measure on the space of all continuous functions.

Question

I am following Continuous Time Markov Processes by Thomas M. Liggett. and I reached chapter 1.7 where he introduces formally the Markov property. To do so he first wants to define a probability measure on the space of continuous functions:

Let $\Omega$ be the set $C[0, \infty)$ be the set of all continuous functions $\omega $ on $[0, \infty)$, the $\sigma$-algebra $\mathcal{F}$ is taken to be the smallest one for which the projection $\omega \rightarrow \omega(t)$ is measurable.

We now have a family $\{ P^x \}$ of probability measures on $(\Omega, \mathcal{F})$ indexed by $x \in \mathbb{R}$. The probability measure $P^x$ is the distribution of $x + B(\cdot)$ where $B$ is a standard Brownian motion.

This last sentence I don't understand, if I have a set $A \in \mathcal{F}$ what would be $P^x(A)$ ?

Also afterwards it is stated that $$E^x[ f(B(t)) ] = E[f(x+B(t))]$$ where $E^x$ is the expectation wrt the probability measure $P^x$. So first we consider $f(B(t))$ as a function from $\omega \in \Omega \rightarrow \mathbb{R}$ and this becomes a regular expectation on the Brownian motion? How is this true?

Answer 1

On $\Omega = C[0,\infty)$ we consider the smallest $\sigma$-algebra $\mathcal{F}$ which makes all projections $\omega \mapsto \pi_t(\omega):=\omega(t)$ measurable. Consequently, the cylinder sets, i.e. the sets of the form

$$\{\omega \in \Omega; \omega(t_1) \in A_1,\ldots,\omega(t_n) \in A_n\} = \bigcap_{j=1}^n \{\pi_{t_j} \in A_j\} \tag{1}$$

where $t_j \geq 0$ and $A_j$ are Borel sets, are a $\cap$-stable generator of $\mathcal{F}$. By the uniqueness of measure theorem, this means that every probability measure $\mathbb{P}$ on $(\Omega,\mathcal{F})$ is uniquely determined by its values on the cylinder sets,

$$\mathbb{P} \left( \bigcap_{j=1}^n \{\pi_{t_j} \in A_j\} \right).$$

Roughly speaking, this means that we "know" the measure $\mathbb{P}$ if we know its values on the cylinder sets.

Aim 1: Find a probability measure $\mathbb{P}$ on $(\Omega,\mathcal{F})$ such that the canonical process $\omega \mapsto B_t(\omega) := \omega(t)$ is a standard Brownian motion on $(\Omega,\mathcal{F},\mathbb{P})$.

Let $(W_t)_{t \geq 0}$ be a Brownian motion on some probability space $(\Omega',\mathcal{F}',\mathbb{Q})$ and assume that $(W_t)_{t \geq 0}$ has exclusively continuous sample paths. Now consider the mapping

$$\Omega' \ni \omega' \mapsto W_{\bullet}(\omega') \in C[0,\infty), \tag{2}$$

which maps each $\omega' \in \Omega'$ to its sample path $W_{\bullet}(\omega')$. Using that the cylinder sets $(1)$ are a generator of $\mathcal{F}$, it is not difficult to check that this mapping is measurable (if we equip $\Omega'$ with $\mathcal{F}'$ and $\Omega$ with $\mathcal{F}$). Since the mapping is measurable, it induces a measure on $(\Omega,\mathcal{F})$,

$$\mathbb{P}(F) := \mathbb{Q}(\{\omega' \in \Omega'; W_{\bullet}(\omega') \in F\}), \qquad F \in \mathcal{F}; \tag{3}$$

this is the push-forward of $\mathbb{Q}$ under the mapping $(2)$. Equivalently, $\mathbb{P}$ is the distribution of $W_{\bullet}$ (with respect to $\mathbb{Q}$). Now let $F$ be a cylinder set of the form $(1)$ and set $B_t(\omega):=\omega(t)$. Then

\begin{align*} &\mathbb{P}(\{\omega \in \Omega; B_{t_1}(\omega) \in A_1,\ldots,B_{t_n} \in A_n\}) \\ &= \mathbb{P}(\{\omega \in \Omega; \pi_{t_1}(\omega) \in A_1,\ldots, \pi_{t_n}(\omega) \in A_n\}) \\ &= \mathbb{P}(F) \\ &\stackrel{(3)}{=} \mathbb{Q}(\{\omega' \in \Omega'; W_{t_1}(\omega') \in A_1,\ldots,W_{t_n}(\omega') \in A_n\}). \end{align*} This shows that $(B_t)_{t \geq 0}$ and $(W_t)_{t \geq 0}$ have the same finite-dimensional distributions, and, as a consequence, $(B_t)_{t \geq 0}$ is a Brownian motion on $(\Omega,\mathcal{F},\mathbb{P})$.

Aim 2: Introduce the probability measures $\mathbb{P}^x$ on $(\Omega,\mathcal{F})$ .

Consider the shift $\tau_x(\omega) := x+\omega$ for $\omega \in \Omega$. The mapping $\tau_x: (\Omega,\mathcal{F}) \to (\Omega,\mathcal{F})$ is measurable, and, as above, we now introduce a new measure as a push-forward. More precisely, we set

$$\mathbb{P}^x(F) := \mathbb{P}(\tau_{x}^{-1}(F)), \qquad F \in \mathcal{F}, \tag{4}$$

where $\mathbb{P}$ is the probability measure from the first part of this answer. Note that $$\mathbb{P}^0=\mathbb{P}. \tag{5}$$ By the very definition of $\tau_{-x}$, we have $$\tau_{x}^{-1}(F) = \{\omega \in \Omega; \omega+x \in F\} = \{\omega \in \Omega; \omega \in F-x\},$$

and so $(4)$ can be equivalently written as

$$\mathbb{P}^x(F) = \mathbb{P}(F-x) \stackrel{(5)}{=} \mathbb{P}^0(F-x). \tag{6}$$

Take a cylinder set $F$ of the form $(1)$ and consider the canonical process $B_t(\omega)=\omega(t)$. Then

\begin{align*} &\mathbb{P}^x \left( \{\omega \in \Omega; B_{t_1}(\omega) \in A_1, \ldots,B_{t_n}(\omega) \in A_n\} \right)\\ \quad &= \mathbb{P}^x \left( \{\omega \in \Omega; \pi_{t_1}(\omega) \in A_1,\ldots,\pi_{t_n}(\omega) \in A_n\} \right) \\\quad &= \mathbb{P}^x(F) \\ \quad&\stackrel{(6)}{=} \mathbb{P}(F-x) \\ \quad &= \mathbb{P} \left( \{\omega \in \Omega; x+B_{t_1}(\omega) \in A_1, \ldots,x+B_{t_n}(\omega) \in A_n\} \right). \tag{7} \end{align*}

For $n=1$ this shows $$\mathbb{P}^x(B_t \in A) = \mathbb{P}(x+B_t \in A)$$ for any Borel set $A$, and so $$\mathbb{E}^x(f(B_t)) = \mathbb{E}(f(x+B_t)). \tag{8}$$ (Note that there are no conditional expectations around here; these are all "classical" expectations with respect to a probability measure.) Moreover, the calculation in $(7)$ shows that $(B_t)_{t \geq 0}$ is a Brownian motion on $(\Omega,\mathcal{F},\mathbb{P}^x)$ which starts ($\mathbb{P}^x$-)almost surely at $B_0=x$. Since $(B_t)_{t \geq 0}$ is the canonical process, we find from $(6)$ that

$$\mathbb{P}^x(F) = \mathbb{P}(F-x) = \mathbb{P}(B_{\bullet} \in F-x) = \mathbb{P}(x+B_{\bullet} \in F),$$

i.e. $\mathbb{P}^x$ is nothing but the distribution of $x+B_{\bullet}$ (w.r.t. $\mathbb{P}$).

Answer 2

Q: "...if I have a set $A \in \mathcal{F}$ what would be $P^x (A)$?"

A: $A$ is a subset of $C[0, \infty)$ (in particular, $A$ lies in $\mathcal{F}$), it's a set of continuous functions. So $P^x (A)$ is the probability that a realized sample path of Brownian motion starting at $x$ would lie in $A$.

Q: "...$E^x[ f(B(t)) ] = E[f(x+B(t))]$..."

A: $t \mapsto B(t)$ should be viewed as an element of $C[0, \infty)$, i.e. an $\omega \in \Omega$. So $x + B(t)$ shifts the function by $x$.

Take, for example, $f$ to be the indicator function $1_{[a,b]}$. Then the LHS is the probability that a sample path from Brownian motion starting at $x$ would lie in $[a,b]$ at time $t$. The RHS is the probability that a sample path from standard Brownian motion starting at $0$ would lie in $[a,b]$ at time $t$, if it were shifted by $x$.

More generally, with bounded measurable $f$, the equality says that the law of $P^x$ at time $t$ is the same as the law of $P^0$ at time $t$ shifted by $x$. (They're just normal distributions with the same variance $t$, and means $x$ and $0$ respectively.)

Answer 3

I try to be more concise than my predecessors:

The probability measure $P^x$ is the distribution of $x + B(\cdot)$ where $B$ is a standard Brownian motion. This last sentence I don't understand, if I have a set $A \in \mathcal{F}$ what would be $P^x(A)$ ?

We would have $P^x(A) = P\big((x+B(\cdot)) \in F\big)$.

This is the general definition of "* the distribution of a random variable*". The distribution of some random variable $X$ in a measurable space $E$ is the measure $\mu$ on $E$ defined by $\mu(A) = P(X\in A)$.

Also afterwards it is stated that $$E^x[ f(B(t)) ] = E[f(x+B(t))]$$

There is an convenient abuse of notation here: on the LHS $B$ denotes the identity function on $\Omega$, i.e. $B(\omega) = \omega$. Then $$ E^x[f(B(t))] = \int f(B(\omega)(t)) P^x(d\omega) = \int f(\omega(t)) P^x(d\omega) = E[f(x+B(t))].$$ The last equality is the classical fact that if $\mu$ is the distribution of $X$ and $\phi$ measurable, then $\int \phi(u) \mu(du) = E[\phi(X)]$, applied here to $X = x+ B(\cdot)$ and $\phi : \Omega \to \Bbb R, \omega \mapsto f(\omega(t))$.