$X_n,n\geq1$ is a martingale w.r.t to $\mathcal{G}=\mathcal{\{G}_n\}_{n=1}^{\infty}$.

Let $\mathcal{F}=\mathcal{\{F}_n\}_{n=1}^{\infty}$. Where $F_n=\sigma(X_1,X_2, \ldots,X_n)$.

  1. Show that $\mathcal{F}_n \subseteq \mathcal{G}_n$ and that $X_n$ is a martingale w.r.t $\mathcal{F}$.

  2. Is the same true for any filtration $\mathcal{H}=\mathcal{\{H}_n\}_{n=1}^{\infty}$ such that $\mathcal{H}_n \subseteq \mathcal{G}_n$ that is, is $X_n$ a martingale w.r.t to $\mathcal{H}$?

So in the first question I assume that $X_n$ is a martingale w.r.t. $\mathcal{F}$ because of the way $\mathcal{F}_n$ is constructed which leads to $X_n$ being $\mathcal{F}_n$-measurable and as $\mathcal{F}_n$ is part of the collection of sigma algebras in $\mathcal{F}$ thus $X_n$ is $\mathcal{F}$-measurable and through that a martingale w.r.t $\mathcal{F}$?

Also if $X_n$ is a martingale w.r.t to $\mathcal{G}$ must mean that $X_n$ is $\mathcal{G}$-measurable which means that there exists a sigma algebra that is part of the collection $\mathcal{G}$ that is constructed the same way $\mathcal{F}_n$? Thus leading to?

$$\mathcal{F}_n \subseteq \mathcal{G}_n$$

In the second question the set $\mathcal{H}_n$ is equal to or a subset of $\mathcal{G}_n$ which kinda puts me in a situation where I can't really say much.


I'm kinda new to all of this so I'm trying.

  • 1
    $\begingroup$ $F_n$ is the smallest sigma algebra by definition w.r.t $X_n$ is measurable. So the second is only true if $F_n \subseteq H_n \subseteq G_n$. $\endgroup$ – Jimmy R. Feb 15 '16 at 19:17
  • $\begingroup$ So my thought on the first one was kinda correct? I understand that $\mathcal{F}_n$ is the smallest algebra which then leads to it being a subset of $\mathcal{G}_n$? The answer to the second question then is that the condition you pointed out has to be fullfilled. $\endgroup$ – user5232061 Feb 15 '16 at 19:45
  • $\begingroup$ You are a bit confused and you have to study a bit more theory. For the first one: since X is a G-martingale, X is adapted to G. Since F is the smallest filtration to which X is adapted you have $\mathcal{F}_n \subseteq \mathcal{G}_n$ for all n. This suggests how you need to proceed on how to show X is F-martingale: tower property. $\endgroup$ – Kolmo Feb 15 '16 at 20:29

For the second point, by choosing $H_n = \sigma\{ \emptyset \}$, you have that $H_n \subseteq G_n$ but $X_n$ is constant for all $n$. Hopefully, there are some nonconstant martingales.


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