MathIsimple

Stochastic Processes – Problem 50: Prove:

Question

Consider the simple symmetric random walk {Sn,n0}\{S_{n},n\geqslant0\}, Sn=x+k=1nξkS_{n}=x+\sum_{k=1}^{n}\xi_{k}, where {ξk}k=1\{\xi_{k}\}_{k=1}^{\infty} are i.i.d. with P(ξk=1)=P(ξk=1)=12P\left(\xi_{k}=1\right)=P\left(\xi_{k}=-1\right)=\frac{1}{2}. (Note S0=xS_{0}=x.) Let N=inf{n0,Sn(a,b)}N=\inf\{n\geqslant0,S_{n}\notin(a,b)\}, with x(a,b)x\in(a,b). Prove:

(i) P(SN=a)=bxba(ii) E[N]=(bx)(xa).\begin{array}{l}{(i)\ P\left(S_{N}=a\right)=\displaystyle\frac{b-x}{b-a}}\\ {(ii)\ E\left[N\right]=(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 Fn=σ(ξ1,,ξn)\mathcal{F}_n = \sigma(\xi_1, \dots, \xi_n). Since E[ξk]=0E[\xi_k]=0, E[Sn+1Fn]=SnE[S_{n+1} | \mathcal{F}_n] = S_n. Define the stopping time N=inf{n0,Sn(a,b)}N = \inf\{n \ge 0, S_n \notin (a, b)\}. Since the random walk must exit the finite interval, N<N < \infty a.s. and SnNmax(a,b)+1|S_{n \wedge N}| \le \max(|a|, |b|) + 1. By the optional stopping theorem: E[SN]=E[S0]=xE[S_N] = E[S_0] = x.

Step 2. Since SN{a,b}S_N \in \{a, b\}, let P(SN=a)=paP(S_N=a) = p_a. Then apa+b(1pa)=xa p_a + b(1 - p_a) = x, giving pa=bxbap_a = \frac{b-x}{b-a}.

Step 3. Prove part (ii). Construct Mn=Sn2nM_n = S_n^2 - n. Verify E[Mn+1Fn]=MnE[M_{n+1} | \mathcal{F}_n] = M_n using E[ξn+1]=0E[\xi_{n+1}] = 0 and E[ξn+12]=1E[\xi_{n+1}^2] = 1. So {Sn2n}\{S_n^2 - n\} is also a martingale.

Step 4. Apply the optional stopping theorem to MnM_n: E[SN2N]=x2E[S_N^2 - N] = x^2, so E[N]=E[SN2]x2E[N] = E[S_N^2] - x^2. Using the probabilities from part (i): E[SN2]=a2bxba+b2xabaE[S_N^2] = a^2 \frac{b-x}{b-a} + b^2 \frac{x-a}{b-a}. Simplifying: E[N]=(bx)(xa)E[N] = (b-x)(x-a).

Final answer

(i) QED. (ii) QED.

Marking scheme

The following is the complete rubric for this problem.

1. Checkpoints (max 7 pts total)

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

Chain A: Martingale Approach (Official Solution)

*Part (i): Probability proof (3 pts)*

  • Verify {Sn}\{S_n\} is a martingale (show E[Sn+1Fn]=SnE[S_{n+1}|\mathcal{F}_n] = S_n). (1 pt)
  • Apply the optional stopping theorem to get E[SN]=xE[S_N] = x (implicitly or explicitly noting N<N<\infty and boundedness). (1 pt)
  • 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 (ii): Expectation proof (4 pts)*

  • Construct and verify {Sn2n}\{S_n^2 - n\} is a martingale (showing E[ξ2]=1E[\xi^2]=1). (1 pt)
  • Apply the optional stopping theorem to get E[SN2N]=x2E[S_N^2 - N] = x^2. (1 pt)
  • Substitute the probability distribution from part (i) to compute E[SN2]E[S_N^2]. (1 pt)
  • Complete the algebra to obtain (bx)(xa)(b-x)(x-a). (1 pt)

Chain B: Difference Equation / Gambler's Ruin Approach

*Part (i) (3 pts)*

  • 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)
  • Identify the general solution as linear u(x)=C1x+C2u(x)=C_1 x + C_2. (1 pt)
  • Solve for constants and obtain the final formula. (1 pt)

*Part (ii) (4 pts)*

  • Set up the non-homogeneous 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)
  • Determine the general solution structure (particular solution x2-x^2 plus homogeneous C1x+C2C_1 x + C_2). (1 pt)
  • Solve for constants from boundary conditions. (1 pt)
  • Substitute back and simplify to (bx)(xa)(b-x)(x-a). (1 pt)

Total (max 7)


2. Zero-credit items

  • Citing standard Gambler's Ruin formulas or E[N]=Var(SN)E[N]=Var(S_N) without derivation.
  • Merely copying definitions or target formulas.
  • Only verifying specific numerical cases.

3. Deductions

  • -1: Logical jumps (e.g., claiming martingale without verification).
  • -1: Missing key qualifiers (e.g., not mentioning finiteness of NN).
  • -1: Boundary condition errors.
  • -1: Algebra errors leading to mismatched final result.
  • -2: Conceptual confusion (e.g., treating SNS_N as constant or assuming E[SN2]=(E[SN])2E[S_N^2] = (E[S_N])^2).
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