Let $W=(\Omega,{\cal F},P)$ be the probability space of Brownian motion. Let $V=\Omega \rightarrow {\mathbb R}$ be the signature of random variables. Let $B_t(\omega)$ be a time-indexed family of random variables which are Brownian motions. $B$ has signature $B: {\mathbb R}^+ \rightarrow V$. Let $X:V$. Define by Lebesque integral the expectation ${\mathbb E}_{W} X=\int_{\omega \in \Omega} X(\omega) d P(\omega)$. So ${\mathbb E}_W: V\rightarrow {\mathbb R}$. Let $(X \star Y)(\omega)=X(\omega) Y(\omega)$. Let ${\cal G} \subseteq {\cal F}$. Define, by implicit construction, the conditional expectation $Z={\mathbb E}_W(X|{\cal G})$, as the random variable $Z:V$ such that for all $A \in {\cal G}$, ${\mathbb E}_W(Z \star {\mathbb 1}_A) = {\mathbb E}_W(X \star {\mathbb 1}_A)$. Let ${\cal F}_t \subset {\cal F}$ be the natural filtration of $B_t$. Can we show, by explicit construction, that ${\mathbb E}_W(B_t|{\cal F}_s)(\omega)=B_s(\omega), 0\leq s\leq t$?

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    $\begingroup$ Sorry but I fail to understand your question. You know that $E(W_t/mid\mathcal F_s)=W_s$, what more are you seeking? $\endgroup$ – Did Oct 26 '17 at 22:27
  • $\begingroup$ In the discrete case, $E[X|B]=\sum_{\omega\in B} X(\omega)P(\omega|B)$. So I am looking for the analogous continuous case where you replace the sum with an integral and $B$ with ${\cal F}$. $\endgroup$ – Lars Ericson Oct 27 '17 at 3:44
  • $\begingroup$ Then $$E(W_t\mid\mathcal F_s)=\int_0^sdW_u\ ?$$ $\endgroup$ – Did Oct 27 '17 at 5:23
  • $\begingroup$ Hi Did, that works. Doesn't give me a general expression though. Let me rephrase as "A Lebesque or Ito integral in the continuous case analogous to the sum in the discrete case of $E[X|B]$", where X is some unknown random expression, not just $W_t$. Do you get the gist? $\endgroup$ – Lars Ericson Oct 27 '17 at 13:39
  • $\begingroup$ No. My impression is that you have no question here... but please prove me wrong. $\endgroup$ – Did Oct 27 '17 at 15:41

(NOT AN ANSWER YET, JUST NOTES. This is the start of an answer to my question as posted, not an answer. Go ahead and downvote if you want. I'm not putting this in the statement of the question because it's going to be the answer, not the question.)


Let $\Omega$ be a set. A $\sigma$-algebra in $\Omega$ is a nonempty collection ${\cal G}$ of subsets of $\Omega$ such that

  • $\Omega \in {\cal G}$
  • If $A \in {\cal G}$ then $\Omega-A \in {\cal G}$
  • If $(\forall n \in {\mathbb N})\, A_n \in {\cal G}$, then $\cup_{i=1}^\infty A_i \in {\cal G}$

$\sigma$-algebra generated by a family of subsets of $\Omega$

Let $H$ be any family of subsets of $\Omega$. The unique smallest $\sigma$-algebra which contains every set in $H$, written $\sigma(H)$, is called the $\sigma$-algebra generated by $H$.

Measurable set

A pair $(S,\sigma(S))$ of a non-empty set and its $\sigma$-algebra is called a measurable space. a member $T \in \sigma(S)$ is called a measurable set.

A function $f:S\rightarrow T$, where $(S,\sigma(S))$ and $(T,\sigma(T))$ are measurable spaces is said to be $(\sigma(S),\sigma(T))$-measurable if for all $B \in \sigma(T)$ then $f^{-1}(B) \in \sigma(S)$. The set of real-valued measurable functions is called ${\cal L}^0$. Zitkovic


$\mu:{\cal F}\rightarrow {\mathbb R}$ is a signed real-valued measure in the measurable space $(\Omega,{\cal F})$ iff for any sequence of disjoint sets $\{A_n\}_{n=1}^\infty \in {\cal F}$, $\mu(\cup_{n=1}^\infty A_n) = \sum_{n=1}^\infty\mu(A_n)$.

A function $\mu$ is a measure iff $\mu(A)\geq 0$ for all $A \in{\cal F}$, and then we say $(\Omega,{\cal F},\mu)$ is a measure space. The measure $\mu(A)$ defines an integral of a nonnegative measurable function $f(\omega)$ by

$$\int_{\omega\in\Omega} f(\omega) d\mu(\omega) = \lim_{h\rightarrow 0} \lim_{N \rightarrow \infty} \sum_{n=0}^N n h \mu(\{\omega \in \Omega: nh \leq f(\omega) \leq (n+1) h\})$$

Probability measure

A positive measure $P:{\cal F}\rightarrow[0,1]$ such that $P(\Omega)=1$ is called a probability measure.

Lebesgue integral, ${\cal L}^1$

Let $f\in{\cal L}_+^{Simp,0}$. The Lebesque integral $$\int f d\mu = \sum_{i=1}^n \alpha_k \mu(A_k) \in [0,\infty)$$ where $$f=\sum_{k=1}^n \alpha_k {\mathbb 1}_{A_k}$$ is a simple-function representation of $f$.

Let $f\in{\cal L}_+^0([0,\infty])$. The Lebesque integral

$$\int f d\mu = \sup \{ \int gd\mu: g \in {\cal L}_+^{Simp,0}, g(x) \leq f(x), \forall x \in S\} \in [0,\infty]$$

Let $f^+=\max{(f,0)},f^-=\max{(-f,0)}$. A function $f\in{\cal L}^0$ is said to be integrable if $\int f^+ d\mu < \infty$ and $\int f^- d\mu < \inf$. The collection of all integrable functions in ${\cal L}^0$ is written ${\cal L}^1$. Zitkovic

Brownian trajectories and events

The space $\Omega$ of elementary events for Brownian motion is the set of all continuous real functions $\Omega = \{\omega: {\mathbb R}_+ \rightarrow {\mathbb R}\}$. We refer to the $\omega(t)$ as Brownian trajectories.

A cylinder set of Brownian trajectories for a finite set of times $0\leq t_1<t_2<\cdots<t_n$ and real intervals $I_k=(a_k,b_k),1\leq k \leq n$ is constructed as

$$C(t_1,\ldots,t_n;I_1,\ldots,I_n)=\{\omega\in \Omega: \forall\,1\leq k \leq n: \omega(t_k) \in I_k\}$$

The space of Brownian events ${\cal F} \subseteq 2^\Omega$ consists all sets of Brownian trajectories obtained from cylinders by operations of countable unions, intersections and complement. ${\cal F}$ is a $\sigma$-algebra of $\Omega$:

$${\cal F}=\sigma(\{C(t_1,\ldots,t_n;I_1,\ldots,I_n): n \in {\mathbb Z}^+: 0\leq t_1<\cdots<t_n, I_k=(a_k,b_k),1\leq k \leq n\})$$

By definition, the elements of ${\cal F}$ are measurable sets.

Weiner probability measure

Let the Weiner probability measure $P$, where $I=(a,b)$ and $0=t_0<t_1<t_2<\cdots<t_n$ and $I_k,1\leq k\leq n$ are real intervals, be:

$$P(C(t;I))=\frac{1}{\sqrt{2\pi t}}\int_a^b e^{\frac{-x^2}{2 t}} dx$$

$$P(C(t_1,t_2,\cdots,t_n;I_1,I_2,\ldots,I_n))=\int_{I_1} \int_{I_2}\cdots\int_{I_n} \prod_{k=1}^n \frac{dx_k}{\sqrt{2\pi(t_k-t_{k-1})}} \exp\big(-\frac{(x_k-x_{k-1})^2}{2(t_k-t_{k-1})}\big)$$

Brownian motion probability space

The probability space of Brownian motion, $W=(\Omega,{\cal F}\subseteq 2^\Omega,P:{\cal F}\rightarrow [0,1])$.

Brownian motion

Let $B_t(\omega)$ be a time-indexed family of random variables which are Brownian motions. $B$ has signature $B: {\mathbb R}^+ \rightarrow V$. Let $X:V$.

Then mathematical Brownian motion $B_t(\omega) = \omega(t)$ is an infinite family of random variables $B:{\mathbb R}_+ \rightarrow V$ for which, in the Brownian motion probability space $W$,

  1. $B_0(\omega)=0$ with probability 1

  2. $B_t(\omega)$ is almost surely a continuous function of $t$

  3. $\forall t,s \geq 0$, the increment $\Delta B_s(\omega)=B_{t+s}(\omega)-B_t(\omega)$ is independent of ${\cal F}_t$ is a zero mean Gaussian random variable with variance ${\mathbb E}|\Delta B_s|^2 = s$.

Random variable

A random variable $X(\omega) \in (\Omega, {\cal F}, P)$ is a real function $X:\Omega \rightarrow {\mathbb R}$ such that ${\omega \in \Omega : X(\omega) \leq x} \in {\cal F}, \forall\,x \in {\mathbb R}$. Write $V=\Omega \rightarrow {\mathbb R}$ for the signature of random variables and write $X:V$.


A topology on a set $S$ is a family $\tau$ of subsets of $S$ which contains $\emptyset$ and $S$ and is closed under finite intersections and countable or uncountable unions. The elements of $\tau$ are called open sets. The structure $(S,\tau)$ is called a topological space.

The topology of real numbers

Let ${\cal O}$ be the set of all open sets $(a,b)$ where $-\infty < a < b < \infty$. Then $({\mathbb R},{\cal O})$ is a topology of real numbers.

Borel $\sigma$-algebra

If $(S,\tau)$ is a topological space, then $\sigma(\tau)$, generated by all open sets, is called the Borel $\sigma$-algebra on $S$.

Borel algebra on the reals

$${\cal B}({\mathbb R}) = \sigma({\cal O})$$

$\sigma$-algebra generated by a random variable

Let $X$ be a random variable. The $\sigma$-algebra generated by $X$ is

$$\sigma(X) = \{X^{-1}(A) : A \in {\cal B}({\mathbb R})$$


A real-valued integral in $W=(\Omega,{\cal F},P)$ of a random variable $X(\omega)$ with respect to probability measure $P$ is called an expectation and written

$${\mathbb E}_W X=\int_{\omega \in \Omega} X(\omega) d P(\omega)$$

The signature is ${\mathbb E}_W: V \rightarrow{\mathbb R}$.

Indicator function

For any set $A\in\Omega$, the indicator of $A$ is a random variable defined as

$${\mathbb 1}_A(\omega) = \cases{ 1 & if $\omega \in A$\\ 0 & otherwise }$$

The signature is ${\mathbb 1}: {\cal F} \rightarrow \Omega \rightarrow \{0,1\}$.

Multiplication of two random variables

Let $(X \star Y)(\omega)=X(\omega) Y(\omega)$.

Event which is an atom

Let $\mu$ be a measure. A set $A \subseteq \Omega$ is called an atom if $\mu(A) > 0$ and for any measurable subset $B \subset A$ with $\mu(B) < \mu(A)$, $\mu(B) = 0$.

Atoms of a random variable

Let $Y$ be a random variable. The atoms of $Y$ are

$${\cal A}(Y) =\{\{x : Y(x) = a\}: a \in Y^{-1}(\Omega)\}$$

TBD: Show that each element of ${\cal A}(Y)$ is an event which is an atom.

Conditional expectation

Conditional expectation: Implicit construction

Assume we are working in $W=(\Omega,{\cal F},P)$. Let

  • ${\cal G}$ be a $\sigma$-algebra contained in ${\cal F}$

  • $X$ be a random variable

  • $Z$ be ${\cal G}$-measurable

Define the conditional expectation of $X$ with respect to ${\cal G}$, written $Z={\mathbb E}_W(X|{\cal G})$, to be a random variable $Z\in {\cal L}^1$ such that Zitkovic

  • $Z$ is ${\cal G}$-measurable

  • For all $A \in {\cal G}$, ${\mathbb E}_W(Z \star{\mathbb 1}_A) = {\mathbb E}_W(X \star {\mathbb 1}_A)$.

This definition of ${\mathbb E}_W(X|{\cal G})$ is implicit. The signature is ${\mathbb E}_W(X|{\cal G}):V$.

Conditional expectation: Explicit construction in discrete case

Let's look at an example from Zitkovic in the finite domain for insight that may lead us to an explicit construction for ${\mathbb E}_W(X|{\cal G})$:

Let $W=(\Omega,{\cal F},{\mathbb P})$ where $\Omega=\{a,b,c,d,e\}$, $F=2^\Omega$ and ${\cal P}$ is uniform. Let $X=\{a\mapsto 1,b\mapsto 3,c\mapsto 3,d\mapsto 5,e\mapsto 5,f\mapsto 7\}$, $Y=\{a\mapsto 2,b\mapsto 2,c\mapsto 1,d\mapsto 1,e\mapsto 7,f\mapsto 7\}$. Let ${\cal G}=\sigma(Y)$. Then $Z={\mathbb E}_W(X|\sigma(Y))=\{a\mapsto 2,b\mapsto 2,c\mapsto 4,d\mapsto 4,e\mapsto 6,f\mapsto 6\}$ is explicitly constructed as follows:

  1. Construct the set of atoms ${\cal A}(Y) =\{\{x : Y(x) = a\}: a \in Y^{-1}(\Omega)\}$.

  2. For each $A \in {\cal A}(Y)$, for each $a \in A$, set $Z(a) = \frac{{\mathbb E}_W(X\star {\mathbb 1}_A)}{{\mathbb P}(A)}$

In the example above, the atoms of ${\cal G}=\sigma(Y)$ are ${\cal A}(Y)=\{\{a,b\}, \{c,d\},\{e,f\}\}$ and ${\mathbb P}(A)=\frac{1}{3}$ for $A \in {\cal A}(Y)$. Also ${\mathbb E}_W(X\star {\mathbb 1}_{\{a,b\}})=\frac{2}{3}$, ${\mathbb E}_W(X\star {\mathbb 1}_{\{c,d\}})=\frac{4}{3}$ and ${\mathbb E}_W(X\star {\mathbb 1}_{\{e,f\}})=\frac{6}{3}$. So for $a \in \{a,b\}$, $Z(a)=\frac{\frac{2}{3}}{\frac{1}{3}}=2$, and similarly for $\{c,d\}$ and $\{e,f\}$, so finally $Z=\{a\mapsto 2,b\mapsto 2,c\mapsto 4,d\mapsto5,e\mapsto 6,f\mapsto 6\}$.

Conditional expectation: Explicit construction in continuous case

Let $W=(\Omega,{\cal F},{\mathbb P})$ be the Brownian probability space. Let ${\cal G} \subseteq {\cal F}$ be a sub-$\sigma$-algebra of ${\cal F}$. Write $A_{\cal G}(\omega) \in {\cal G}$, for all $\omega \in \Omega$, for the atom containing $\omega$ in ${\cal G}$. Then

$${\mathbb E}_W(X|{\cal G})(\omega)=\frac{{\mathbb E}_W(X\star {\mathbb 1}_{A_{\cal G}(\omega)})}{{\mathbb P}(A_{\cal G}(\omega))}$$

Natural filtration

Define the family of $\sigma$-algebras ${\cal F}_t$ for $t \geq 0$ by cylinder sets confined to times $0\leq s < t$ for some fixed $t$. So ${\cal F}_s \subset {\cal F}_t \subset {\cal F}$ if $0 \leq s < t < \infty$. We call ${\cal F}_t$ the Brownian filtration generated by the Brownian events up to time $t$. Constructively, we can say that

$${\cal F}_t=\sigma(\{C(t_1,\ldots,t_n;I_1,\ldots,I_n): n \in {\mathbb Z}^+: 0\leq t_1<t_2<\cdots<t_n <t, I_k=(a_k,b_k),1\leq k \leq n\})$$

Original problem

Show, using the explicit construction of continuous conditional expectation, that for all $0\leq s \leq t$, ${\mathbb E}_W(B_t|{\cal F}_s)(\omega)=B_s(\omega), 0\leq s\leq t$. So, show that

$${\mathbb E}_W(B_t\star {\mathbb 1}_{A_{{\cal F}_s}(\omega)}) = B_s {\mathbb P}(A_{{\cal F}_s}(\omega))$$

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    $\begingroup$ Everytime you edit a post, it gets bumped up on the front page and crowds out other answers. If you want to compose a lengthy post requiring many edits, you might want to use the sandbox on meta. $\endgroup$ – Michael Greinecker Jan 8 '18 at 8:43
  • $\begingroup$ Sorry I didn't know about that. I'm not looking to crowd people out. I am working on a question which may be trivial to the well-trained but is still foggy in my mind, and not of too much interest to others. It's taking me a while to tease out all of the definitions to make the question well-posed and well-founded. $\endgroup$ – Lars Ericson Jan 8 '18 at 15:22
  • $\begingroup$ No problem, I just wanted to make you aware of the possibility to use the sandbox. $\endgroup$ – Michael Greinecker Jan 8 '18 at 16:46

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