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Mean and Variance of first hitting (entrance, passage) time

  1. Aug 2, 2012 #1
    Let [itex]Pr(X_{i} = +1) =\frac{2}{3} = 1 - Pr(X_{i} = -1) [/itex], and [itex] S_{n} = \sum{X_{i}}[/itex], For each k≥1, define [itex]T_{k}\ =\ min \left\{n≥1: S_{n} = k \right\}[/itex]. Calculate [itex]E[T_k][/itex], and [itex]Var[T_k][/itex].

    2. Relevant equations

    [itex]E[T_k] = \sum_{n=1}^{n=∞}{n Pr(T_k = n)} [/itex], and [itex]E[T_k]^2 = \sum_{n=1}^{n=∞}{n^2 Pr(T_k = n)} [/itex] will be used.

    3. The attempt at a solution

    k is given as an known constant. A theorem says [itex] Pr(T_k = n ) = Pr(S_{1} S_{2} ... S_{n} ≠ 0, S_{n} = k) = \frac{|b|}{n}Pr(S_n = k)[/itex], and
    [itex] Pr(S_n = k) = \left( \stackrel{n}{\frac{1}{2}(n+k)} \right) p^{\frac{1}{2}(n+k)}q^{\frac{1}{2}(n-k)} [/itex]

    IMHO, problem should be asolved. However, It doesn't seem to have a nice expression. My guess will be use Sterling approximation, or Mathematica. Any comments are welcome.
     
  2. jcsd
  3. Aug 2, 2012 #2

    Ray Vickson

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    In your expression for P(T_k = n) (an expression that I don't believe), what is [itex]|b|? [/itex] What is the meaning of
    [tex] \left( \stackrel{n}{\frac{1}{2}(n+k)} \right)?[/tex]

    Anyway, using Stirling's (not Sterling's) approximation would not help you to get an *exact* expression, although it might help you to determination an approximate answer. Rather than going to Mathematica (or Maple), my first inclination would be to consult a good book, such as W. Feller, "An Introduction to Probability Theory and Its Applcications", Vol. I (Wiley). It has whole chapters on this material.

    Alternatively, you can recognize that to get E(T_k) you need the expected first passage-time in an infinite-state Makov chain, and you can try to apply the usual equations for expected first-passage times to get an expression. (This will result from the solution of infinitely many coupled linear equations in infinitely many unknows, but the special structure makes it tractable.)

    RGV
     
  4. Aug 2, 2012 #3
    This is n choose [tex] \frac{1}{2}(n+k)[/tex]

    Appreciate your comments.
     
  5. Aug 3, 2012 #4

    Ray Vickson

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    In Chapter 14, page 351 of the Feller book I cited before, there is a simple formula for the generating function [itex] U_z(s) = \sum_{n=0}^{\infty} s^n P\{T_z = n\}[/itex] of the absorption time Tz in an absorbing random walk on (0,∞), with absorption at 0 and starting from z > 0. The walk has probabilities p and q = 1-p of unit steps up and down, respectively. The problem is essentially the same as yours: you have the interval (-∞,k) with absorption at k and you start from 0, while Feller's case has the interval (0,∞) with absorption at 0 and starts from z>0. Feller's expression is
    [tex]U_z(s) = \left[ \frac{1-\sqrt{1-4pqs^2}}{2ps}\right]^z,[/tex]
    and you can get the moments [itex] E(T_z^m)[/itex] from this in the usual way:
    [tex] E\,T_z = U_z ^{\prime}(1),\; E\, T_z^2 = U_z ^{\prime \prime} (1) + U_z ^{\prime}(1) .[/tex] For Feller's case you need to assume p < q in order to have finite mean and variance of absorption times, and when evaluating the derivatives using a computer package (I used Maple) that assumption must be passed to the system. Of course, you could do the derivatives manually, but this is one case where using a CAS really saves hours of work and a tree's worth of paper. The final results are amazingly simple.

    In your case you want to go up instead of down, and you have p > q, so it is essentially equivalent.

    RGV
     
    Last edited: Aug 3, 2012
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