Wilmott Forums - Previsible Vs. Martingale Process

Forum Navigation:

login

FORUMS > Student Forum

Topic Title: Previsible Vs. Martingale Process
Created On Wed Feb 05, 03 03:59 AM

Topic View:

View thread in raw text format

asd
Senior Member

Posts: 660
Joined: Aug 2002

Wed Feb 05, 03 03:59 AM

User is offline

Could someone please tell the difference between a Previsible process and a Martingale process?

Thanks,
asd

-------------------------
Life is a journey and not a destination
quant code directory

Stefanone
Senior Member

Posts: 332
Joined: Aug 2002

Wed Feb 05, 03 10:10 AM

User is offline

Basically, by def, a stochastic process X is a Ft-martingale iff:
1) is adapted to the filtration Ft (adapted means that the sigma algebra of (X0,X1,X2,..,Xn) is a subset of the filtration Ft)
2) E(|Xt|)<infinity for all t
3) E(Xt|Fs)=Xs

Therefore you can see that a martingale is an Ft measurable process for all t. Instead, broadly speaking, we say that a process X is previsible if it is F(t-1) measurable, i.e. E(Xt|Ft-1)=Xt (this property descends from the definition of conditional expectation).
That means that, based on the information u get at time t-1 (which are embedded in the filtration Ft-1), you can say what the value of the random variable Xt will be like. That is the reason why it is called previsible.

Hope it helps,

regards,

S.

-------------------------
Life is complex, partially because it has a real part and an imaginary part.

sam
Senior Member

Posts: 703
Joined: Dec 2001

Wed Feb 05, 03 10:19 AM

User is offline

A martingale process is just process who's expected future value at any time is just the present value. Using notation from measure theory,

if M(t) = E [ M(T) | F(t) ] then M(s) s> 0, is a martingale, and F(s) is the filtration.

A previsible process is just a process that satisfies certain properties relating to the adaptablility of the process. In discrete time, if
M(t+1) is F(t) measurable, then M(t) is a previsible process. Hence, this can never be a martingale, since a martingale must be F(t) measurable.

In continuous time, this becomes more complex.... I'll leave it to someone more qualified (Martingale... your help is needed here!!!

I guess previsibilty just means that a process that is less random that a purely stochastic one like BM. Less random in the sense that the value of the process in known 'slightly' in advance.

Sam

sam
Senior Member

Posts: 703
Joined: Dec 2001

Wed Feb 05, 03 10:21 AM

User is offline

How's that for efficiency in the Wilmott Forums???

... Two nearly identical answers in the space of ten mins!

Edited: Wed Feb 05, 03 at 10:21 AM by sam

chin
Junior Member

Posts: 12
Joined: Jul 2002

Wed Feb 05, 03 10:29 AM

User is offline

How about predictable process, is it another name for it?
I have another related question: what's the difference between markov chain and markov process?

sam
Senior Member

Posts: 703
Joined: Dec 2001

Wed Feb 05, 03 11:35 AM

User is offline

A markov process is just a process who's expected future value depends only on the current state of the system. We can be talking about a continuous or discrete process.

A Markov chain is first of all a DISCRETE time process, where at each discrete time point, the process can only take one out of a set of finite values. And it is still a markov process. Since it is discrete, we can more easily attach a probability for each state and this is usually done using a probability transition matrix. This is the basic description, although there are some other properties like time homogenity of the transition matrix.

Sam

chin
Junior Member

Posts: 12
Joined: Jul 2002

Wed Feb 05, 03 01:03 PM

User is offline

Sam, thanks for the reply.

Note that Markov chain could be continous-time, in which case the jump time is random with inter-jump time being exponentially distributed. Aslo, it can also take infinite number of values, e.g. Poisson process (from one to infinite).

My questions are: what is the difference between the two? Possible answer - uncountable number of values, discontinuity (jumps) ? Is Markov chain also a Markov process? (probably true). What do you (and other gurus) think about this?

Edited: Wed Feb 05, 03 at 01:06 PM by chin

BeautifulMind
Member

Posts: 146
Joined: Mar 2002

Wed Feb 05, 03 01:39 PM

User is offline

Quote
Is Markov chain also a Markov process? (probably true).

As far as I know Markov chains are discrete version of Markov processes, depending on discrete or continuous state space issues: anyone can confirn this?

-------------------------
Presque toute la vie humaine dépend des probabilités - Voltaire

sam
Senior Member

Posts: 703
Joined: Dec 2001

Wed Feb 05, 03 02:39 PM

User is offline

Hi,

I also think that markov chains are more concerned with discrete time processes... i.e. processes which change only at specific points in time, t = 0, 1, 2.... At each of these times, the process can only take on one out of a fixed number of possible states.

Contrast this with a general markov process, where the variable defining the process can change continuously at any time and by an amount that comes from an infinitly large list.

What is the relavence of all this?

Well, Markov chains can be seen as a simplification of the continuous markov process case. Both hold the markov property, but in one, we simplify the process by only considering those outcomes that we deem significant, and looking at small increments of time to mimic continuous time.

It's all about simplification. Isn' this what the binomial tree does? To take a continuous stock price process and discretise it? Now imagine a trinomial or 4 x nomial tree or n -nomial tree... This isnt necesssarily a markov chain process, but certainly can be made to be if the right parameters are chosen!

Chin, you are right... u can also pick a jump time to be exponentially distributed, but my point is that those jump times are still specific points in time... between those times what happens to the process? nothing... it stays constant... that's the whole point of a markov chain. It's easier to analyse this than something that changes continuously with every second.
Going back to your post I will accept that in a markov process there are uncountably many values, wheras in a markov chain these are made finite. I won't agree that there is a difference with regard to jumps... a markov chain is by nature a jump process... as it changes suddenly at discrete points of time... but there are many markov jump processes...e.g. poisson process
Finally, a markov chain also incorporates a property that we havent discussed at all... the time homogenous transition matrix...from what i remember this makes the process markov.

Regards,

Sam

Edited: Wed Feb 05, 03 at 02:39 PM by sam

asd
Senior Member

Posts: 660
Joined: Aug 2002

Wed Feb 05, 03 07:16 PM

User is offline

Stefanone,Sam,BeautifulMind & Chin thanks a lot for your replies.

I am not sure if I have properly understood.
Please correct me if I am wrong -

(taking example case of stock movements)

Martingale process implies that stock price expected tomorrow,is the equal to the value of the price today.

Previsible process implies that I already know the value of the price going to be taken tomorrow,or altleast I know the max and min values it could attain from calculating the variance.

Markov Process implies that tomorrow's value will depend only on today's value.

One more confusing thing is Markov implies that tomorrow's value depends on today,while Martingale implies it is equal.

Is not Martingale=>Markov
and
Markov not necassarily =>Martingale?

Thanks once again,
asd

-------------------------
Life is a journey and not a destination
quant code directory

chin
Junior Member

Posts: 12
Joined: Jul 2002

Wed Feb 05, 03 11:21 PM

User is offline

Let me try it first.

Markov process and martingale are talking about two different things.

* Martingale -- the definition is already given by Stefanone.
It is a process with constant mean, i.e. E[X(t)]=E[X(s)] for any t,s.
Therefore, it has orthogonal increaments, i.e. E[X(t) - X(s) | F_s]=0, where F_s is the sigma field (or information) up to time s.

* Markov process -- the process has the property that its conditional distribution p(X(t)|F_s) is only dependent on the your information on X(s) at time s, i.e. p(X(t)|X(s)). Anything else is not necessary.
One type of definiton: E[h(X(t))|F_s]=E[h(X(t))|X(s)] for any function h() such that E[h(X(t))] is finite.
Note here, F_s has way much more information than X(s) provides, e.g in discrete time, F_s = sigma algreba of {X(s), X(s-1), ..., X(0)}.

Now, Markov process is not necessarily martingale (obvious).
It is my belief that that martingale is not necessarily Markov process, e.g. X is margingale, so that E[X(t)|F_s]=X(s), but E[X(t)^2|F_s]= E[ (X(t)-X(s))^2 |F_s] + X(s)^2 where the first (square) term may well be dependent on X(s-1), X(s-2), etc.

Anyone agree on this or can provide an example? asd, maybe you can message "Martingale", he is an expert on this one.

Edited: Wed Feb 05, 03 at 11:22 PM by chin

amgc
Junior Member

Posts: 1
Joined: Jul 2002

Thu Feb 06, 03 01:02 PM

User is offline

I think chin is right about the two being different. An example of a martingale which is not a Markov Process is a simple symmetric random walk which is stopped (i.e. its value remains the same for all future time) on the first occasion that it goes below its past maximum. In this case the distribution is not only dependent on X(t), but X(t-1).

As regards previsibility of continuous time processes, the standard (technical) definition for a process to be previsible is that X(t) is F_t- measurable, where F_t- is the sigma algebra generated by taking the union over sigma algebras F_s where s is less than t. What this means is that any continuous (stochastic) process is previsible. Typical examples on non-previsible processes are Levy-Processes (which are Markov) since these may jump at random times not 'signalled' in the history of the processes. (Of course not all Levy processes are non-previsible, since Brownian motion is also a Levy process)

Hope that helps.

Alex

mj
Senior Member

Posts: 3030
Joined: Dec 2001

Thu Feb 06, 03 02:28 PM

User is offline

If a process is continuous, previsible (or predictable) and adapted are the same thing. This is because

X_t = \lim_{ s \to t-} X_{s}

So the value of X at time t is determined by the info available up to but not including time t.

If the process has jumps this is not the case.

MJ

-------------------------
Quant Job Interview Questions and Answers now available on lulu.com and amazon.com

asd
Senior Member

Posts: 660
Joined: Aug 2002

Fri Feb 07, 03 12:40 AM

User is offline

MJ and Alex, Thanks for your replies.

I am atlast now able to get the idea of a previsible process!

asd

-------------------------
Life is a journey and not a destination
quant code directory

View thread in raw text format

FORUMS > Student Forum

Forum Navigation:

© All material, including contents and design, copyright Wilmott Electronic Media Limited - FuseTalk 4.01 © 1999-2010 FuseTalk Inc.