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How Prove a process is Markov?

by vale
Tags: markov, process, prove
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vale
#1
Aug3-05, 11:15 AM
P: 4
This is my first time here, so... Hi everybody!

I've very little time to figure out the following problem ... and Im wandering if some of you can give me any help or just suggest me any good reading material...

The question is how you can prove a process [tex] P_t[/tex], given the dynamics, is Markov.
In short my process is on alternate intervals, a mean reverting brownian bridge [tex]dP_t = \frac{\alpha}{G-t}(Q-P_t)dt + \sigma dW_t [/tex], and a mean reverting proportional volatility process : [tex]dP_t = K(\theta -P_t)dt + \nu dW_t [/tex]. The length of the intervals and their occurence is determined by an exogenous bootstrap procedure, which I believe, doesn't give any problems, being a resampling procedure with replacement, it doesn't generate any dependence with the past history...

How should I procede on your opinion? Any hints ?

Thank you very much in advance,
Vale
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EnumaElish
#2
Aug3-05, 03:38 PM
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Could this be of any help? When you say prove, do you mean empirically or mathematically?
vale
#3
Aug4-05, 04:44 AM
P: 4
Thank you for the reference and the reply!

Actually I meant a mathematical proof....
I think I should show somehow the transition probabilities are independent from the past realizations... but I don't Know how to retrieve them from the dynamic...

Many thanks...
Vale

EnumaElish
#4
Aug4-05, 09:54 AM
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How Prove a process is Markov?

I guess I'd argue W(t) is independent of the past. Then equate Eq. (1) to Eq. (2) and solve for P(t). It'll be a function of t, W(t) and some constants. Since W is independent of the past, so's P(t).

{P.S. Oh, whoops! You said "on alternating intervals." Does that mean the two Eq's do not hold simultaneously?}

{P.P.S. In that case:

P(t+1) = P(t) + dP(t) = P(t) + a(dt) P(t) + b dW(t) = [a(dt)+1] P(t) + b dW(t).

Et+1[P(t+1)|P(t),P(t-1)...,P(0)] = (a+1) Et+1[P(t)|P(t),P(t-1)...,P(0)] + b Et+1[dW(t)|P(t),P(t-1)...,P(0)] = (a+1) P(t) + b Et+1[dW(t)]. QED

The last step is based on two premises: (i) E[X|X,Y,Z,...] = X, and (ii) dW(t) is independent of past history so E[dW(t)|P(t),P(t-1)...,P(0)] = E[dW(t)].}


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