Stochastic Processes – Problem 31: prove that

The following is the grading rubric based on the official solution:

1. Checkpoints (Key Scoring Points, Total 7 Points)

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

Chain A: Official Reverse Martingale and Inequality Proof Path

• Construction of the $\sigma$-field (1 point)
• Explicitly write the definition of $\mathcal{F}_{-m}$, which must include $S_m$ and the future order statistics (e.g., $\mathcal{F}_{-m} = \sigma(S_m, U_{m+1}, U_{m+2}, \dots, U_{n+1})$).
• *Note: Any $\sigma$-field that makes $R_{-m}$ adapted and satisfies the reverse Markov property is acceptable.*
• Conditional expectation of the random walk part (1 point)
• Correctly state or derive $E(S_{m-1} \mid S_m) = \frac{m-1}{m}S_m$.
• *Justification: symmetry of i.i.d. sums.*
• Coefficient cancellation from order statistics (2 points)
• [additive]
• 1 point: State the relationship between $U_m$ and $U_{m+1}$ (e.g., $U_m$ viewed as the maximum order statistic of a uniform on $(0, U_{m+1})$, or invoke the proportional independence property).
• 1 point: Correctly handle the coefficient cancellation. This may involve explicitly computing $E(U_m^{-1} \mid U_{m+1}) = \frac{m}{m-1}U_{m+1}^{-1}$, or stating that the distributional property produces the coefficient $\frac{m}{m-1}$ which exactly cancels the $\frac{m-1}{m}$ from the random walk part.
• *If the coefficient $\frac{m}{m-1}$ is not shown but "the constants cancel" is claimed to yield the correct martingale conclusion, award 0.5 points.*
• Verification of the reverse martingale property (1 point)
• Using the independence of the $S$ and $U$ sequences, explicitly write the verification:

$E(R_{-(m-1)} \mid \mathcal{F}_{-m}) = \frac{1}{U_{m+1}} E(S_{m-1} \mid S_m) \times (\text{cancellation factor}) = R_{-m}$.

• Deriving the upper bound using the inequality (2 points)
• [additive]
• 1 point: Invoke and apply the maximal inequality for reverse martingales (or Doob's inequality), i.e., $P(\sup_{m \le n} R_{-m} \ge 1 \mid S_n=y) \le E(R_{-n} \mid S_n=y)$.
• 1 point: Substitute the terminal value $R_{-n} = S_n/U_{n+1} = y/t$, and combine with the trivial bound ($\le 1$) to obtain the final conclusion $\min(y/t, 1)$.

Total (max 7)

2. Zero-credit items

• Only copying the given conditions from the problem (e.g., the definition of $R_{-m}$).
• Only proving that $S_m$ is a martingale without addressing $R_{-m}$.
• Only stating "this is a martingale" without any conditional expectation computation or property invocation.
• When proving the inequality, using the unconditional expectation $E(R_{-m})$ instead of the conditional expectation $E(\cdot|S_n=y)$, without explanation.

3. Deductions

Apply at most the single largest deduction.

• Logical gap (-1): When computing the conditional expectation, not implicitly or explicitly using the independence between the $X$ sequence and the $U$ sequence (i.e., directly splitting the expectation of a product into a product of expectations without justification).
• Missing trivial bound (-1): Having computed the bound $y/t$ but not stating that a probability must be $\le 1$ or not writing $\min(\cdot, 1)$ (especially when $y > t$ the result is obviously wrong).
• Conditioning confusion (-1): When substituting the final value, failing to correctly use the conditions $S_n=y$ and $U_{n+1}=t$, leaving random variables in the result.