Moments of a random walk

  • Thread starter humenghu
  • Start date
  • #1
4
0

Main Question or Discussion Point

Hi, can someone who is familiar with the analysis of random walks (statistical mechanics, condensed matter physics etc.) help me on solving a particular problem?

We define the following random walk, the random variable w(t) is evolved as

w(t+1)=w(t), with probability of p;

w(t+1)=w(t)+C*(1-w(t))*f, with probability 1-p.

C is a constant, f is a random variable with known distribution, and f is not correlated with w.

Following the way of analyzing a Brownian motion, what I have tried is to write down
E(w(t)|w(t-1),p), then show the E(w(t)|p) using the induction, for a given w(0).

Next, at least I need to find the Var(w(t)) , or equivalently E(w(t)^2) since I have known (E(w(t)))^2 for a given w(0).

I hope that someone can help me find the moment generating function of this random walk

(E(w(t)^n)), or alternatively just the mean and the variance of w(t). Any hint on this is badly needed.
 

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
33,810
5,592
You can obtain E(w(t)) as something of the form A+Bλt, right?
You should also get a recurrence relation on E(w(t)2) which involves E(w(t)). This should have a solution like C+Dλt+Eμt.
Plugging in the coefficients from the recurrence relations and the initial conditions should allow you to determine the 7 constants.
 
  • #3
4
0
You can obtain E(w(t)) as something of the form A+Bλt, right?
You should also get a recurrence relation on E(w(t)2) which involves E(w(t)). This should have a solution like C+Dλt+Eμt.
Plugging in the coefficients from the recurrence relations and the initial conditions should allow you to determine the 7 constants.

Thanks for the hint.

Yeah the E(W[itex]_{n}[/itex])=K[itex]^{n}[/itex]W[itex]_{0}[/itex]+K[itex]^{n-1}[/itex]L+K[itex]^{n-2}[/itex]L+...+KL+L

Where K=p-C(1-p)E(f), L=C(1-p)E(f).

However I found that the E(Wt[itex]^{2}[/itex]) is a little bit hard to show since the terms grow somehow fast with n. I could write down:

E(W[itex]_{n}[/itex][itex]^{2}[/itex])=A[itex]^{n-1}[/itex]M+ A[itex]^{n-2}[/itex][BE(W[itex]_{1}[/itex])+C]+ A[itex]^{n-3}[/itex][BE(W[itex]_{2}[/itex])+C]+...A[BE(W[itex]_{n-2}[/itex])+C]+[BE(W[itex]_{n-1}[/itex])+C]

where A,B,C are terms consisting of p, C and random variable f.M=AE(W[itex]_{0}[/itex][itex]^{2}[/itex])+BE(W[itex]_{0}[/itex])+C

so do you think this is what you meant? or?

Addtionally I am still curious about the moments of the Wt, in a general sense. Is it possible to show them in closed form here?
 
  • #4
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
33,810
5,592
Yeah the E(W[itex]_{n}[/itex])=K[itex]^{n}[/itex]W[itex]_{0}[/itex]+K[itex]^{n-1}[/itex]L+K[itex]^{n-2}[/itex]L+...+KL+L

Where K=p-C(1-p)E(f), L=C(1-p)E(f).
I don't think it should be that bad.
K[itex]^{n}[/itex]W[itex]_{0}[/itex]+K[itex]^{n-1}[/itex]L+K[itex]^{n-2}[/itex]L+...+KL+L
= K[itex]^{n}[/itex]W[itex]_{0}[/itex]+L(K[itex]^n[/itex]-1)/(K-1)
= K[itex]^{n}[/itex](W[itex]_{0}[/itex]+L/(K-1))-L/(K-1)
So as I said, it is possible to represent the first moment as A+BKn, and the second moment will have only two additional constants.
 
  • #5
4
0
Now I see what you mean. Thanks a lot!

Did you magically predict the first two moments just by looking at the evolution rules? or did you actually write down the terms like I did, any additional tips on this ?
 
  • #6
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
33,810
5,592
I wrote out the recurrence relation for w(t) and saw it took the form
w(t+1) = Aw(t) + B.
Clearly this is basically multiplying by A at each step, so the answer will be something with At. To cover the inhomogeneous term (the B), look for a stationary value, i.e. w(t+1) = w(t) = w:
w = Aw+B
w = B/(1-A)
So now we have a general solution of w(t+1) = Aw(t), and a specific solution of w(t+1) = Aw(t) + B. Since the relationship is linear, we can add any multiple of the homogeneous solution to the inhomogeneous one:
w(t) = cAt + B/(1-A)

The recurrence relation for w2 is of the form
w2(t+1) = Dw2(t) + Ew(t) + F
= Dw2(t) + E(cAt + B/(1-A)) + F
Collecting up constants, this reduces to
w2(t+1) = Dw2(t) + E'At + F'
Again, the homogeneous part gives w2(t) = gDt. What to do with E'At? Iterating the recurrence will result in adding terms like that for successive values of t, i.e. the sum of a geometric series. So we know this will add a term like At to the answer. So just assume an answer of the form gDt + hAt + k, plug it into the recurrence relation and initial conditions, and see what you get for g, h and k.
 
  • #7
4
0
I see. Thanks!
 

Related Threads on Moments of a random walk

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
11
Views
19K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
5
Views
14K
Replies
7
Views
2K
  • Last Post
Replies
0
Views
1K
Top