In the book of Rennie & Baxter "Financial Calculus" a martingale with respect to a measure $P$ is defined as a stochastic process $M_{t}$ which satisfies some boundedness condition and: $$\mathbb{E}_{P}(M_{t}|F_{s})=M_{s} $$

where $ F_{s}$ is the filtration (or history) of the process up until time $s$.

Now clearly the filtration is a random variable. $1)$ Does this mean that in the definition of a martingale we condition upon the random variable $F_{s}$ or upon a given/known outcome of the history say $F_{s}=f_{s}$. Rephrasing the question, is $M_{s}$ a random variable or a particular known outcome of the stochastic process at time $s$ in the equation above?

$2)$ Surely conceptionally there is a difference between an expectation conditional on a random variable or an expectation conditional on a fixed outcome but does it matter for the definition of a martingale (does one definition imply the other)?

The book is somewhat confusing to me because sometimes it seems that they use the fact that $F_{s}$ is actually known (as in $F_{s}=f_{s}$) in this conditional expectation while the definition seems to imply that it is just a random variable.

Edit: I thought the filtration was a random variable but I was told it is not. Now I am wondering about the following:

In the book a filtration is explained by means of a non-recombinant binomial tree with times $t=0,1,2.$ At any node the tree can either move up or down to some particular value of a stock. Let the nodes of the tree be numbered from $1$ to $7$ where at time 2, node $7$ has the highest value of the stock, then node 6, then node 5 and node $4$ the lowest. At time 1 node 3 has the higest value and node 2 the lowest and at time 0 there is only one node, node 1.

Now it says, the filtration at time 1, $F_{1}$, is {1,3} if the first move was up or {1,2} if the first move was down. This 'splitting' of filtrations conditional on the path taken (or in particular the move on the tree at time 0) on the tree is what is bothering me; it seems that the information is not really contained in the filtration itself but rather in the filtration conditional on the path taken (moving up or down from any node).

So if you then take an expectation conditional on some filtration $F_{s}$ lets say$$\mathbb{E}_{P}(M_{t}|F_{s})=M_{s} $$ then it seems you really dont have any information because you dont know whether the 'moves' of the process were such and such.

Also, this definition of a filtration doesn't seem to strike me as a $\sigma$-algebra

  • 1
    $\begingroup$ Usually $F_s$ is a $\sigma$-algebra, not a random variable. $\endgroup$ Sep 23, 2016 at 20:27
  • 2
    $\begingroup$ The filtration isn't a random variable, it's a $\sigma$-algebra. It's usually interpreted as including all of the information observed when watching $M_t$ up to time $s$ $\endgroup$ Sep 23, 2016 at 20:27
  • $\begingroup$ what does it mean to condition upon a $\sigma$-algebra in this setting? $\endgroup$
    – Joogs
    Sep 23, 2016 at 20:29
  • $\begingroup$ If you think of $M_t$ as a stock price, $\mathbb{E}(M_t \vert \mathcal{F}_s)$ is essentially "the expected future price of the stock, $M_t$, given what we've observed up until time $s$". $\endgroup$ Sep 23, 2016 at 20:32
  • $\begingroup$ @measure_theory I have provided some extra details on how the book defines a filtration but it doesn't seem to be a $\sigma$-algebra to me. $\endgroup$
    – Joogs
    Sep 24, 2016 at 9:36

1 Answer 1


Given a probability space $(\Omega,\mathcal F,\mathbb P)$ and a totally ordered set $I$, a filtration is a collection of $\sigma$-algebras $\{\mathcal F_\alpha:\alpha\in I\}$ such that $\alpha<\beta$ implies $\mathcal F_\alpha\subset\mathcal F_\beta$ and $$\bigcup_{\alpha\in I}\mathcal F_\alpha\subset \mathcal F.$$ In the case where $I=\mathbb N_0:=\{0,1,2,\ldots\}$, we would have a sequence of $\sigma$-algebras $\{\mathcal F_n:n=0,1,2,\ldots\}$ with $\mathcal F_n\subset\mathcal F_{n+1}$ and $\bigcup_{n=0}^\infty \mathcal F_n\subset\mathcal F$. Suppose now $\{X_n:n=0,1,2,\ldots\}$ is a stochastic process defined on the same probability space as $\{\mathcal F_n\}$. Then $\{X_n\}$ is a martingale with respect to $\{\mathcal F_n\}$ if for all $n=0,1,2,\ldots$,

  • $\mathbb E[|X_n|]<\infty$
  • $\mathbb E[X_{n+1}\mid \mathcal F_n] = X_n.$

The interpretation of a filtration as "information about a process" is most clearly seen when we take $\{\mathcal F_n\}$ to be the natural filtration $$\mathcal F_n = \sigma\left(\bigcup_{i=0}^n \sigma(X_i) \right),$$ with $$\sigma(X_i) = \sigma\left(\{X^{-1}(B):B\in\mathcal B \right) $$ the $\sigma$-algebra generated by $X_i$.


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