CONTENTS

Exercise: Expected Exit Time from a Bounded Interval

Problem

A symmetric random walk SnS_n starts at k(0,N)k \in (0, N) and exits the interval (0,N)(0, N) at the stopping time τ=inf{n:Sn{0,N}}\tau = \inf\{n : S_n \in \{0, N\}\}.

  1. Show that Yn=Sn2nY_n = S_n^2 - n is a martingale. (Hint: expand (Sn+Xn+1)2(S_n + X_{n+1})^2 and use E[Xn+1]=0\mathbb{E}[X_{n+1}] = 0, E[Xn+12]=1\mathbb{E}[X_{n+1}^2] = 1.)

  2. Verify that OST applies to YnY_n at the stopping time τ\tau. Use condition (3): bounded increments and E[τ]<\mathbb{E}[\tau] < \infty.

  3. Apply OST to YY at τ\tau and derive E[τ]=k(Nk)\mathbb{E}[\tau] = k(N - k). (You will use the symmetric walk's gambler's-ruin result P(Sτ=N)=k/N\mathbb{P}(S_\tau = N) = k/N.)

  4. Concrete case. Compute E[τ]\mathbb{E}[\tau] for (k,N)=(50,100)(k, N) = (50, 100). If each step takes 11 second, how many seconds on average until the walk exits? Plot E[τ]\mathbb{E}[\tau] as a function of kk for N=100N = 100 and observe the parabola.

Hint

For part 2, that E[τ]<\mathbb{E}[\tau] < \infty can be proved by a geometric tail bound — e.g. the probability of exiting in any block of NN consecutive steps is at least some ϵ>0\epsilon > 0, so τ\tau has an exponential tail.

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