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I'm not sure, if I understand the definition of backwards martingales. The definition I have says that it is a martingale $(X_n)_{n\in -\mathbb{N}}$ adapted to an increasing $\sigma$-Field $\mathcal{F}_n$.

So does that mean that we have also negative indexes on the $\sigma$-Algebras:

$\mathcal{F}_{-1}\subseteq \mathcal{F}_{-2}\subseteq \cdots \subseteq \mathcal{F}_{n}\quad n\in -\mathbb{N}$

And $\mathcal{F}_{n}=\sigma(X_{n})=\sigma(X_{-1},X_{-2}\dots,X_{n})$

So we only have changed the index $\mathbb{N}\to -\mathbb{N}$ but the real difference is that $\mathbb{E}(X_{n+1}\mid \mathcal{F}_{n})=X_{n}$ means for example $\mathbb{E}(X_{-2}\mid \mathcal{F}_{-3})=X_{-3}$

Is this correct? If not, could someone make an example and writing indexes instead of abbreviations?

1 Answers 1

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The term refers to two equivalent situations. One of these is the usual martingale notion, except that the "infinity" in the index set extends to the left rather than to the right. That is, we have a filtration $$ \cdots\mathcal F_{-n}\subset\mathcal F_{-n+1}\subset\cdots\subset\mathcal F_{-2}\subset\mathcal F_{-1} $$ indexed by $-\Bbb N$, and an adapted sequence $(X_n)_{n\in-\Bbb N}$ of integrable random variables such that $\Bbb E[X_{n}\mid\mathcal F_{n-1}]=X_{n-1}$ for each $n\in -\Bbb N$.

Alternatively, by setting $\mathcal G_n:=\mathcal F_{-n}$ and $Y_n:=X_{-n}$ for $n\in\Bbb N$, we can stick with the index set $\Bbb N$, but now envision a "reverse filtration" $$ \mathcal G_1\supset\mathcal G_2\supset\cdots\supset\mathcal G_n\supset\mathcal G_{n+1}\supset\cdots $$ and the associated "backward" martingale property: $$ \Bbb E[Y_n\mid\mathcal G_{n+1}]=Y_{n+1}, \quad n\in\Bbb N. $$

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    Thank you very much for your answer. What I still not get is whether the process starts at $X_{-1}$ or at $X_{-\infty}$. If it starts at $X_{-1}$ then our $\sigma$-Algebra $\mathcal{F}_{-1}$ already contains all the future information since $\mathcal{F}_{-1}$ is the biggest upper set? Is this correct? – 2017-01-11
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    What do you mean by "starts"? Notice that time ends at time $-1$, so $\mathcal F_{-1}$ does indeed contain all the relevant information. – 2017-01-11
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    With "starts" I mean which of the $X_i$ is the first value of the process (a normal process starts usually at $X_0$ or $X_1$). Since time ends at $-1$ I assume that $X_{-\infty}$ is the first value of the process since $X_{-1}$ is the last? Is this correct? – 2017-01-11
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    With a backward martingale like $(X_n)_{n\in\Bbb -N}$, there a last value (namely $X_{-1}$) but no first value. However, the backward martingale convergence theorem tells you that $X_{-\infty}:=\lim_{n\to-\infty} X_n$ exists a.s and in $L^1$, and that $X_{-\infty}=\Bbb E[X_{-1}\mid\mathcal F_{-\infty}]$, thereby producing a "first value" for the martingale. (Here $\mathcal F_{-\infty}:=\cap_{n\in-\Bbb N}\mathcal F_n$.) The key is that a backward martingale is always uniformly integrable, because $X_{-n}=\Bbb E[X_{-1}\mid\mathcal F_{-n}]$ for each $n\in\Bbb N$. – 2017-01-11
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    Thank you so much, now everything is very clear! – 2017-01-11