<p>The <em>augmentation of the filtration generated by $B$</em> is typically just called <em>the augmented filtration</em> and is defined using the below construction. It is more commonly referred to simply as <em>the standard Brownian filtration</em>. </p>

<ul>
<li>The collection $\mathcal{C}$ of all sets of probability $0$ in the sigma-field $\sigma\{B_{s} \: : \: s \leq t\}$</li>
<li>The collection of null-sets, $\mathcal{N}$, of all $A$ such that $A \subset B$ for $B \in \mathcal{C}$. It is assumed $\mathbb{P}(A)=0$ for all such $A \in \mathcal{N}$.</li>
</ul>

<p>Then the <em>augmented filtration</em> for Brownian motion to be the filtration given by $\{\mathcal{F}_{t}\}$, where for each $t$ we take $\mathcal{F}_{t}$ to be the smallest sigma-field containing $\sigma\{B_{s} \: : \: s \leq t\}$ and $\mathcal{N}$.</p>

<p>The point of this is to endow the filtration with some properties which are considered <em>helpful</em>.</p>

<p>The relationship to martingales is simply that $X_{t}:=\mathbb{E}(X\:|\:\mathcal{G}_{t})$ is a martingale.</p>

CC BY-SA 3.0
2015-01-01T02:20:42.300Z
Note that $X_t\equiv E[X|\mathcal{G}_t]$ being a martingale is equivalent to the law of iterated expectations: $E[X_{t+s}|\mathcal{G}_t]=E[E[X|\mathcal{G}_{t+s}]|\mathcal{G}_t]=E[X|\mathcal{G}_t]=X_t$
nominally rigid
CC BY-SA 3.0
2015-01-01T00:18:32.650Z
2015-01-01T00:18:32.650Z
Rusan Kax
<p><em>Note: This question is related to the following question about <a href="https://economics.stackexchange.com/questions/1616/complete-markets-in-continuous-time">complete markets in continuous time</a>. In the linked question, the answer mentions that complete markets in this setting is a result of the Martingale Representation theorem.</em></p>

<p>I'm trying to understand the statement of the theorem as given in <a href="https://en.wikipedia.org/wiki/Martingale_representation_theorem" rel="nofollow noreferrer">its Wikipedia article</a>:</p>

<blockquote>
  <p>Let $B_t$ be a Brownian motion on a standard filtered probability space $(\Omega, \mathcal F, \mathcal F_t, P)$, and let $\mathcal G_t$ be the augmentation of the filtration generated by $B$.
  If $X$ is a square integrable random variable measurable with respect to $\mathcal G_\infty$ then there exists a predictable process $C$ which is adapted with respect to $\mathcal G_t$, such that
  $$
X = E[X] + \int_0^\infty C_s dB_s.
$$
  Consequently,
  $$
E[X \mid G_t] = E[X] + \int_0^t C_s dB_s.
$$</p>
</blockquote>

<p>In this definition, where is the relationship to Martingales? I see that it is assumed that $X$ is square integrable with respect to $\mathcal G_\infty$ and I assume that it has something to do with the fact that $\mathcal G_t$ is an "augmentation of the filtration generated by $b$." Also, what is an "augmentation of a filtration?"</p>

<finance><martingales><probability>
Augmented Filtrations and Martingales in the Martingale Representation Theorem
jmbejara
# Augmented Filtrations and Martingales in the Martingale Representation Theorem

(Asked by: username_0 on 2014-12-31)

*Note: This question is related to the following question about [complete markets in continuous time](https://economics.stackexchange.com/questions/1616/complete-markets-in-continuous-time). In the linked question, the answer mentions that complete markets in this setting is a result of the Martingale Representation theorem.*

I'm trying to understand the statement of the theorem as given in [its Wikipedia article](https://en.wikipedia.org/wiki/Martingale_representation_theorem):

> Let $B\_t$ be a Brownian motion on a standard filtered probability space $(\Omega, \mathcal F, \mathcal F\_t, P)$, and let $\mathcal G\_t$ be the augmentation of the filtration generated by $B$.
> If $X$ is a square integrable random variable measurable with respect to $\mathcal G\_\infty$ then there exists a predictable process $C$ which is adapted with respect to $\mathcal G\_t$, such that
> $$
> X = E[X] + \int\_0^\infty C\_s dB\_s.
> $$
> Consequently,
> $$
> E[X \mid G\_t] = E[X] + \int\_0^t C\_s dB\_s.
> $$

In this definition, where is the relationship to Martingales? I see that it is assumed that $X$ is square integrable with respect to $\mathcal G\_\infty$ and I assume that it has something to do with the fact that $\mathcal G\_t$ is an "augmentation of the filtration generated by $b$." Also, what is an "augmentation of a filtration?"

**Tags:** finance, martingales, probability

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## 1 answers

### Answer by username_1 on 2015-01-01. Score: 2 | Accepted answer

The *augmentation of the filtration generated by $B$* is typically just called *the augmented filtration* and is defined using the below construction. It is more commonly referred to simply as *the standard Brownian filtration*.

* The collection $\mathcal{C}$ of all sets of probability $0$ in the sigma-field $\sigma\{B\_{s} \: : \: s \leq t\}$
* The collection of null-sets, $\mathcal{N}$, of all $A$ such that $A \subset B$ for $B \in \mathcal{C}$. It is assumed $\mathbb{P}(A)=0$ for all such $A \in \mathcal{N}$.

Then the *augmented filtration* for Brownian motion to be the filtration given by $\{\mathcal{F}\_{t}\}$, where for each $t$ we take $\mathcal{F}\_{t}$ to be the smallest sigma-field containing $\sigma\{B\_{s} \: : \: s \leq t\}$ and $\mathcal{N}$.

The point of this is to endow the filtration with some properties which are considered *helpful*.

The relationship to martingales is simply that $X\_{t}:=\mathbb{E}(X\:|\:\mathcal{G}\_{t})$ is a martingale.

> **Comments:**
> **username_2**: Note that $X_t\equiv E[X|\mathcal{G}_t]$ being a martingale is equivalent to the law of iterated expectations: $E[X_{t+s}|\mathcal{G}_t]=E[E[X|\mathcal{G}_{t+s}]|\mathcal{G}_t]=E[X|\mathcal{G}_t]=X_t$  

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