Section 5.3

Random walk

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Problem 5.3.1

For a simple random walk SS with S0=0S_{0} = 0 and p=1−q<12p = 1-q < \frac{1}{2}, show that the maximum M=max⁡{Sn:n≥0}M = \max \left\{ S_{n}: n \geq 0\right\} satisfies P(M≥r)=(p/q)r\mathbb {P}\left(M \geq r\right) = (p / q)^{r} for r≥0r \geq 0.

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Problem 5.3.2

Use generating functions to show that, for a symmetric random walk,

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(a)

2kf0(2k)=P(S2k−2=0)2 k f_{0}(2 k) = \mathbb {P}\left(S_{2 k-2} = 0\right) for k≥1k \geq 1, and

(b)

P(S1S2⋯S2n≠0)=P(S2n=0)\mathbb {P}\left(S_{1} S_{2} \cdots S_{2 n} \neq 0\right) = \mathbb {P}\left(S_{2 n} = 0\right) for n≥1n \geq 1.

Problem 5.3.3

A particle performs a random walk on the corners of the square ABCDA B C D. At each step, the probability of moving from corner cc to corner dd equals ρcd\rho_{c d}, where

ρAB=ρBA=ρCD=ρDC=α,ρAD=ρDA=ρBC=ρCB=β \rho _{\mathrm{AB}} = \rho _{\mathrm{BA}} = \rho _{\mathrm{CD}} = \rho _{\mathrm{DC}} = \alpha , \quad \rho _{\mathrm{AD}} = \rho _{\mathrm{DA}} = \rho _{\mathrm{BC}} = \rho _{\mathrm{CB}} = \beta

and α,β>0,α+β=1\alpha , \beta > 0, \alpha +\beta = 1. Let GA(s)G_{\mathrm{A}}(s) be the generating function of the sequence (pAA(n):n≥0)\left(p_{\mathrm{AA}}(n): n \geq 0\right), where pAA(n)p_{\mathrm{AA}}(n) is the probability that the particle is at A after nn steps, having started at A. Show that

GA(s)=12{11−s2+11−∣β−α∣2s2} G_{\mathrm{A}}(s) = \frac{1}{2}\left\{ \frac{1}{1-s^{2}}+\frac{1}{1-\left|\beta -\alpha \right|^{2} s^{2}}\right\}

Hence find the probability generating function of the time of the first return to A.

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Problem 5.3.4

A particle performs a symmetric random walk in two dimensions starting at the origin: each step is of unit length and has equal probability 14\frac{1}{4} of being northwards, southwards, eastwards, or westwards. The particle first reaches the line x+y=mx+y = m at the point (X,Y)(X, Y) and at the time TT. Find the probability generating functions of TT and X−YX-Y, and state where they converge.

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Problem 5.3.5

Derive the arc sine law for sojourn times, Theorem (3.10.21), using generating functions. That is to say, let L2nL_{2 n} be the length of time spent (up to time 2n2 n) by a simple symmetric random walk to the right of its starting point. Show that

P(L2n=2k)=P(S2k=0)P(S2n−2k=0) for 0≤k≤n \mathbb {P}\left(L_{2 n} = 2 k\right) = \mathbb {P}\left(S_{2 k} = 0\right) \mathbb {P}\left(S_{2 n-2 k} = 0\right) \quad \text{ for } 0 \leq k \leq n
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Problem 5.3.6

Let {Sn:n≥0}\left\{ S_{n}: n \geq 0\right\} be a simple symmetric random walk with S0=0S_{0} = 0, and let T=min⁡{n>0:Sn=0}T = \min \left\{ n > 0 : S_{n} = 0\right\}. Show that

E[min⁡{T,2m}]=2E[∣S2m∣]=4mP(S2m=0) for m≥0 \mathbb {E}\left[\min \left\{ T, 2 m\right\} \right] = 2 \mathbb {E}\left[\left|S_{2 m}\right|\right] = 4 m \mathbb {P}\left(S_{2 m} = 0\right) \quad \text{ for } m \geq 0
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Problem 5.3.7

Let Sn=∑r=0nXrS_{n} = \sum_{r = 0}^{n} X_{r} be a left-continuous random walk on the integers with a retaining barrier at zero. More specifically, we assume that the XrX_{r} are identically distributed integer-valued random variables with X1≥−1,P(X1=0)≠0X_{1} \geq -1, \mathbb {P}\left(X_{1} = 0\right) \neq 0, and

Sn+1={Sn+Xn+1 if Sn>0Sn+Xn+1+1 if Sn=0 S_{n+1} = \begin{cases} S_{n}+X_{n+1} & \text{ if } S_{n} > 0 \\ S_{n}+X_{n+1}+1 & \text{ if } S_{n} = 0\end{cases}

Show that the distribution of S0S_{0} may be chosen in such a way that E[zSn]=E[zS0]\mathbb {E}\left[z^{S_{n}}\right] = \mathbb {E}\left[z^{S_{0}}\right] for all nn, if and only if E[X1]<0\mathbb {E}\left[X_{1}\right] < 0, and in this case

E[zSn]=(1−z)E[X1]E[zX1]1−E[zX1] \mathbb {E}\left[z^{S_{n}}\right] = \frac{(1-z) \mathbb {E}\left[X_{1}\right] \mathbb {E}\left[z^{X_{1}}\right]}{1-\mathbb {E}\left[z^{X_{1}}\right]}
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Problem 5.3.8

Consider a simple random walk starting at 0 in which each step is to the right with probability p(=1−q)p( = 1-q). Let TbT_{b} be the number of steps until the walk first reaches bb where b>0b > 0. Show that E[Tb∣Tb<∞]=b/∣p−q∣\mathbb {E}\left[T_{b} \mid T_{b} < \infty \right] = b /\left|p-q\right|.

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Problem 5.3.9

Let S={Sk:k=0,1,2,…}S = \left\{ S_{k}: k = 0,1,2, \ldots \right\} be a simple random walk on the non-negative integers, with S0=iS_{0} = i and an absorbing barrier at 0. A typical jump XX has mass function P(X=1)=p\mathbb {P}\left(X = 1\right) = p and P(X=−1)=q=1−p\mathbb {P}\left(X = -1\right) = q = 1-p, where p∈(12,1)p \in \left(\frac{1}{2}, 1\right). Let HH be the event that the walk is ultimately absorbed at 0. Show that, conditional on HH, the walk has the same distribution as a simple random walk W={Wk:k=0,1,2,…}W = \left\{ W_{k}: k = 0,1,2, \ldots \right\} for which a typical jump YY satisfies P(Y=1)=q\mathbb {P}\left(Y = 1\right) = q, P(Y=−1)=p\mathbb {P}\left(Y = -1\right) = p.

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