MathIsimple

Stochastic Processes – Problem 38: Prove:

Question

Consider the simple symmetric random walk {Sn,n0}\{S_n, n \geq 0\}, Sn=x+k=1nξkS_n = x + \sum_{k=1}^{n} \xi_k, {ξk}k=1\{\xi_k\}_{k=1}^{\infty} i.i.d., with P(ξk=1)=P(ξk=1)=12P(\xi_k = 1) = P(\xi_k = -1) = \frac{1}{2}. (Note S0=xS_0 = x.) Let N=inf{n0:Sn(a,b)}N = \inf\{n \geq 0 : S_n \notin (a,b)\}, and x(a,b)x \in (a,b). Prove:

(i) P(SN=a)=bxba(ii) E[N]=(bx)(xa).\begin{array}{l}{(i)\ P(S_N = a) = \frac{b-x}{b-a}} \\ {(ii)\ E[N] = (b-x)(x-a).}\end{array}

Step-by-step solution

Step 1. Prove part (i). First show {Sn}\{S_n\} is a martingale with respect to the natural filtration Fn=σ(ξ1,,ξn)\mathcal{F}_n = \sigma(\xi_1, \dots, \xi_n). Since ξk\xi_k are i.i.d. with E[ξk]=0E[\xi_k]=0: E[Sn+1Fn]=Sn+E[ξn+1]=Sn.E[S_{n+1} | \mathcal{F}_n] = S_n + E[\xi_{n+1}] = S_n. So {Sn}\{S_n\} is a martingale. Define the stopping time N=inf{n0:Sn(a,b)}N = \inf\{n \ge 0 : S_n \notin (a,b)\}. Since the random walk on a finite interval must exit, N<N < \infty a.s. and E[N]<E[N] < \infty. Before time NN, SnS_n is bounded in (a,b)(a,b), so SnNmax(a,b)+1|S_{n \wedge N}| \le \max(|a|, |b|) + 1. By the optional stopping theorem: E[SN]=E[S0]=x.E[S_N] = E[S_0] = x.

Step 2. Use the distribution of SNS_N to solve for the probability. SNS_N can only take values aa or bb. Let P(SN=a)=paP(S_N = a) = p_a, then P(SN=b)=1paP(S_N = b) = 1 - p_a. Substituting into the expectation: E[SN]=apa+b(1pa)=x.E[S_N] = a \cdot p_a + b \cdot (1 - p_a) = x. Solving: pa(ba)=bxp_a(b - a) = b - x, so pa=bxbap_a = \frac{b-x}{b-a}. Hence P(SN=a)=bxbaP(S_N = a) = \frac{b-x}{b-a}. Part (i) is proved.

Step 3. Prove part (ii). Construct the process Mn=Sn2nM_n = S_n^2 - n. Compute: E[Sn+12Fn]=E[(Sn+ξn+1)2Fn]=Sn2+2SnE[ξn+1]+E[ξn+12]=Sn2+1.E[S_{n+1}^2 | \mathcal{F}_n] = E[(S_n + \xi_{n+1})^2 | \mathcal{F}_n] = S_n^2 + 2S_n E[\xi_{n+1}] + E[\xi_{n+1}^2] = S_n^2 + 1. Therefore E[Sn+12(n+1)Fn]=Sn2n=MnE[S_{n+1}^2 - (n+1) | \mathcal{F}_n] = S_n^2 - n = M_n, so {Sn2n}\{S_n^2 - n\} is also a martingale.

Step 4. Apply the optional stopping theorem to MnM_n to find E[N]E[N]. Since E[N]<E[N] < \infty and increments are bounded, OST applies: E[SN2N]=E[S02]=x2.E[S_N^2 - N] = E[S_0^2] = x^2. So E[N]=E[SN2]x2E[N] = E[S_N^2] - x^2. Using the probabilities from Step 2: E[SN2]=a2bxba+b2xaba.E[S_N^2] = a^2 \cdot \frac{b-x}{b-a} + b^2 \cdot \frac{x-a}{b-a}. Simplifying: E[N]=a2(bx)+b2(xa)bax2=ab+x(b+a)x2=(bx)(xa).E[N] = \frac{a^2(b-x) + b^2(x-a)}{b-a} - x^2 = -ab + x(b+a) - x^2 = (b-x)(x-a). Part (ii) is proved.

Final answer

(i) QED. (ii) QED.

Marking scheme

The following is the complete grading rubric for this problem and official solution.

1. Checkpoints (max 7 pts total)

Score exactly one chain | take the maximum subtotal among chains; do not add points across chains.

Chain A: Martingale Approach (Based on Official Solution)

*Part 1: Probability proof (3 pts)*

  • [additive] Verify {Sn}\{S_n\} is a martingale (show E[Sn+1Fn]=SnE[S_{n+1}|\mathcal{F}_n] = S_n or use the independent zero-mean increment property). (1 pt)
  • [additive] Apply the optional stopping theorem (OST) to get E[SN]=xE[S_N] = x (implicitly or explicitly noting N<N<\infty and boundedness). (1 pt)
  • [additive] Set up aP(SN=a)+b(1P(SN=a))=xa P(S_N=a) + b(1-P(S_N=a)) = x and solve P(SN=a)=bxbaP(S_N=a) = \frac{b-x}{b-a}. (1 pt)

*Part 2: Expectation proof (4 pts)*

  • [additive] Construct and verify {Sn2n}\{S_n^2 - n\} is a martingale (must show E[ξ2]=1E[\xi^2]=1 and key computations). (1 pt)
  • [additive] Apply OST again to get E[SN2N]=x2E[S_N^2 - N] = x^2 (or equivalently E[N]=E[SN2]x2E[N] = E[S_N^2] - x^2). (1 pt)
  • [additive] Substitute the probability distribution from part 1 to compute E[SN2]=a2pa+b2(1pa)E[S_N^2] = a^2 p_a + b^2(1-p_a). (1 pt)
  • [additive] Complete the algebraic simplification from E[SN2]x2E[S_N^2]-x^2 to the final form (bx)(xa)(b-x)(x-a). (1 pt)

Chain B: Difference Equation / Gambler's Ruin Approach

*Part 1: Probability proof (3 pts)*

  • [additive] Set up the difference equation u(x)=12u(x+1)+12u(x1)u(x) = \frac{1}{2}u(x+1) + \frac{1}{2}u(x-1) with boundary conditions u(a)=1,u(b)=0u(a)=1, u(b)=0. (1 pt)
  • [additive] Identify the general solution as linear: u(x)=C1x+C2u(x)=C_1 x + C_2. (1 pt)
  • [additive] Apply boundary conditions to solve for constants and obtain the final probability formula. (1 pt)

*Part 2: Expectation proof (4 pts)*

  • [additive] Set up the nonhomogeneous difference equation v(x)=1+12v(x+1)+12v(x1)v(x) = 1 + \frac{1}{2}v(x+1) + \frac{1}{2}v(x-1) with v(a)=0,v(b)=0v(a)=0, v(b)=0. (1 pt)
  • [additive] Determine the general solution structure (particular solution x2-x^2 plus homogeneous solution C1x+C2C_1 x + C_2). (1 pt)
  • [additive] Apply boundary conditions to set up and solve the system for C1,C2C_1, C_2. (1 pt)
  • [additive] Substitute back and simplify to the final form (bx)(xa)(b-x)(x-a). (1 pt)

Total (max 7)


2. Zero-credit items

  • Citing Gambler's Ruin formulas or E[N]=Var(SN)E[N]=\mathrm{Var}(S_N) without derivation.
  • Only copying definitions or target formulas without substantive derivation.
  • Only verifying specific numerical cases rather than a general proof.

3. Deductions

  • -1: Logical gap, e.g., claiming "this is a martingale" without verification, or writing the general solution without the difference equation.
  • -1: Missing key qualifiers (e.g., not mentioning finiteness/boundedness of NN as a prerequisite for applying the theorem).
  • -1: Boundary condition error (e.g., swapping the probability values of a,ba, b).
  • -1: Algebraic error leading to a mismatched final result.
  • -2: Conceptual confusion, e.g., treating SNS_N as a constant, or assuming E[SN2]=(E[SN])2E[S_N^2] = (E[S_N])^2.
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