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#1 Re: Exercises » solve the matrix » 2014-02-21 01:43:25

thanks for your effort. Yes finally E is based on four values. So if i change v i'll get a different E value for the same u. What i need is an another simpler equation for E, not with all these matrices. I know it from simulations that  E=2(n-v) if v > u and E=2(n-u) if v < u. But is it possible to get these results from the equation given in the question?  I saw a paper with similar problem. They've solved it using recurrence relations.
Title: "the random walk between a reflecting and an absorbing barrier". (I couldn't use the url)

#2 Re: Exercises » solve the matrix » 2014-02-20 17:17:01

Yes thats right.
If xp > u then E(xp) linearly decreases as xp increases.

#3 Re: Exercises » solve the matrix » 2014-02-20 00:04:26

sorry about that... i was confused while typing. I've changed it now.

#4 Re: Exercises » solve the matrix » 2014-02-19 21:37:22

I've modified my question. Hope its clear now.

#5 Re: Exercises » solve the matrix » 2014-02-19 21:02:52


let say u=2 and xp=1;
then

   will give a column matrix with first three elements constant and the rest will be in linear form.


Q will be like this,


G(1) depends on 'u'; It has 0.5 in u-1 and u+1 th position. (count starts from 0);
P(xp)  depends on xp ; It has 0.5 in xp-1 and xp+1 th position.

#6 Exercises » solve the matrix » 2014-02-19 17:46:04

soni85
Replies: 13

hi, i have this equation


where,
I- identity matrix. (n*n)
Q-(n*n) matrix

so,

If u=2 and xp=1;
then,


(G(1) - (n*1) matrix ,depends on 'u'; It has 0.5 in u-1 and u+1 th position. (count starts from 0); P(xp)  - (1*n) matrix ,depends on xp ; It has 0.5 in xp-1 and xp+1 th position.)

If xp < u then the E(xp) is constant, if xp > u then E(xp) is linearly decreasing. But is it possible to get a final simple equation with notations u , xp and n(matrix size)?

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