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What is the difference between a random process that is adapted to a filteration and one that had the martingale property. It seems the two notions are quite similar and would be helpful to construct examples when they are the same and when they differ.

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A very interesting question. Nevertheless such a mathematical question could be improved using formulas or at least references. –  Richard May 16 '14 at 6:17
@Richard Point taken; can't edit the question anymore; Suppose $(\Omega,\mathcal{F},\mathcal{F}_{t},\mathcal{P})$ is a filtered probability space, and suppose $X$ is process on this space. In this setup $X$ is adapted to $\mathcal{F}_{t}$ iff for each $t$, $X_t$ is measurable w.r.t the sigma algebra $\mathcal{F}_{t}$. $X$ is has the martingale property if $E(X_{t+s}|F_t) = X_t$ -- the martingale property says that the expected value of the $X$ at time $t+s$ is only dependent on data available at $t$. This is similar to the notion of adaptation, so when is a process adapt but not a martingale. –  Don Shanil May 19 '14 at 3:53
Hi Don, you really don't have the edit button right below the question? I just thought that a clear definition (with formulas) would already reveal the answer partially. But giving some examples is for sure more valuable. –  Richard May 19 '14 at 6:23
@Richard hm...I can't really find this button! Yes I agree with you Richard, my initial post definitely needed some improvement. Also, I feel its good to edit the question since it might be useful for others if presented properly. Yes you are right about the clear definitions, and I have got the answer -- I have also accepted quasi's answer as he gives good examples. Thanks for the feedback and appreciate the help. –  Don Shanil May 19 '14 at 6:44

3 Answers 3

up vote 2 down vote accepted

Let's consider a random process $X$. If $X$ is an adapted process, then we know, without any uncertainty, what its value is at the present time. This idea is formalized with measure theory.

For $X$ to be a martingale, it needs to have the following property: at any given time, our best estimate of the value at some point in the future (i.e. forecast), is the present value. In other words, the present and previous values never make us believe there's a local trend in time.

Implicit in this belief is the idea that we know what the present value is. In other words, every martingale is necessarily adapted for the concept to make sense.


  1. Let $Y_t = t$ be deterministic and increasing. It is adapted but not a martingale. This is the only real counterexample, since martingale implies adapted. But one other.
  2. Let $\Omega = \{a,b\}$. let $t \in [0,1,2]$, and let $Y_0 = 0, Y_1(a) = 1, Y_2(a) = 0, Y_1(b) = -1, Y_2(b) = 0$. Let the filtration be the one generated by $Y$, so that $Y$ is by default adapted. But, $Y$ is not a martingale.
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I am not sure we need to know the present value. It is a stochastic process, so we would only know one realization, i.e. one trajectory up till now. But the definition is on random variables, so it suffices to know "all the possible values the r.v. can be" in and their probability. The r.v. tomorrow could be allowed to be have values in more states as now, but if we condition an all the now allowed states we "project" the future random variable to the present r.v. It is also nice to think of best forecast, but for this you might need L^2 and not only L^1 r.v., to say something about "best"? –  Marco Breitig May 18 '14 at 9:18
@quasi Thanks this was exactly what I was looking for! Appreciate the insight. –  Don Shanil May 18 '14 at 13:10

A random process that is adapted to a filtration is measurable (ie X_t is F_t-measurable) but not necessarily a martingale. X_t is a martingale if E(X_t | F_s) = X_s for s < t.

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For a stochastic process $\left(X_{t}\right)$ to be adopted to a filtration $\left(\mathcal{F}_{t}\right)_{t\in T}$ the random variable $X_{t}$ must be $\mathcal{F}_{t}$-measurable for each $t\in T$. A stochastic process is a collection of random variables $X_{t}$, indexed by some set $T$. Each random variable is a mapping from a probability space into a measurable space. A measurable space is a set together with a collection of subsets of this set which fulfill certain properties. This collection is called a $\sigma$-algebra. It is nearly the same as a topology, but for a slight technical difference (countable vs finite intersection of sets belong to the collection). A topology is essentially a collection of open sets. For two topological space one can define a mapping from one into the other. If this function respects the openness of sets, i.e. the preimage of an open set is open, we say this mapping is continuous. In much the same light a $\sigma$-algebra includes (sub)sets which are measurable, i.e. it makes sense to give that subset a measure (a volume). Measurable functions are functions where the preimage of a measurable set is measurable. In Probability Theory we call a measurable function a random variable.

As a stochastic process is a collection of (different) random variables, i.e. measurable functions, a filtration is the sequence of the different $\sigma$-algebras of these random variables. So a stochastic process is adapted to a filtration, if this filtration makes every single random variable measurable according to its $\sigma$-algebra. I hope this was not too technical.

On the other hand a martingale is a stochastic process with the martingale property: $$ \textrm{for all}\space s,t\in T \textrm{ with } s\leq t\textrm{ it holds that }\mathbb{E}_{\mathbb{P}}\left[X_{t}\mid\mathcal{F}_{s}\right] = X_{s}\space\left(\mathbb{P}\textrm{-almost surely}\right) $$ This defining property depends on the probability measure $\mathbb{P}$ and the filtration $\left(\mathcal{F}_{t}\right)$. So a stochastic process is only a martingale with respect to a filtration $\left(\mathcal{F}_{t}\right)$ and a probability measure $\mathbb{P}$ if it fulfills the martingale property.

So the adapted filtration is a building block of the definition of a martingale.

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just curious, does Steven Shreve ring a bell for you? –  Matt Wolf May 17 '14 at 12:44
Continuity and measurability are defined slightly differently than what you said. For a continuous function, inverse image of an open set is open, and similarly with measurable. If you unpack the $\epsilon-\delta$ definition of continuity, you'll see that this is what it's logically equivalent to. –  quasi May 18 '14 at 3:19
Also, this is something I think about from time to time. You can definite martingality intrinsically, by taking the filtration to be the natural one generated by $X$. Does this mean that martingality is an intrinsic property? –  quasi May 18 '14 at 3:20
@quasi: You are right, the exact definition for a function to be continuous or measurable is that preimages of open resp. measurable is open resp. measurable. It is always difficult on how to find the right balance in explaining something without making mistakes. –  Marco Breitig May 18 '14 at 8:37
@quasi, second comment: Of course a stochastic process is always adapted to the filtration it generates. But that gives you only the minimal collection of events this stochastic process is allowed to be in. The main ingredience for me is the probabiliy measure. It connects all the measure spaces and ensures the martingale property. And yes, Matt Wolf, at university we worked with Karatzas/Shreve and Revuz/Yor. –  Marco Breitig May 18 '14 at 8:51

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