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

[jimmy7430]

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. Homework 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.

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.

 

[Ray Vickson]

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. Homework 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.

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.

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

 

[jimmy7430]

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]

This is n choose [tex] \frac{1}{2}(n+k)[/tex]

Appreciate your comments.

 

[Ray Vickson]

This is n choose [tex] \frac{1}{2}(n+k)[/tex]

Appreciate your comments.

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 T_z 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