Stochastic Processes Solution – Problem 10: Prove that there exists a constant such that: where

Question

Let $(X_{n})_{0\leqslant n\leqslant N}$ be a martingale or a nonnegative submartingale. Prove that there exists a constant $C>0$ such that:

$E\left[\sup_{n}\left|X_{n}\right|\right]\leqslant C\left(1+\sup_{n}\,E\left[\left|X_{n}\right|\mathrm{ln}_{+}\left|X_{n}\right|\right]\right)$

where $\ln_{+}=\max\{\ln x,0\}$ .

Step-by-step solution

Step 1. Let $X^* = \sup_{0 \leqslant n \leqslant N} |X_n|.$ Since $(X_{n})_{0\leqslant n\leqslant N}$ is a martingale or a nonnegative submartingale, $|X_n|$ is a nonnegative submartingale. Using the integral formula for the expectation of a nonnegative random variable: $E[X^*] = \int_0^\infty P(X^* > \lambda) d\lambda \leqslant 1 + \int_1^\infty P(X^* \geqslant \lambda) d\lambda.$

Step 2. By Doob's maximal inequality, for any $\lambda > 0$ : $\lambda P(X^* \geqslant \lambda) \leqslant E\left[|X_N|\mathbb{I}_{\{X^* \geqslant \lambda\}}\right].$ Substituting into the integral: $\int_1^\infty P(X^* \geqslant \lambda) d\lambda \leqslant \int_1^\infty \frac{1}{\lambda} E\left[|X_N|\mathbb{I}_{\{X^* \geqslant \lambda\}}\right] d\lambda.$ Using Fubini's theorem to interchange the integral and expectation, and noting that $\ln_+ x = \int_1^x \frac{1}{t} dt$ when $x \geqslant 1$ : $\int_1^\infty \frac{1}{\lambda} E\left[|X_N|\mathbb{I}_{\{X^* \geqslant \lambda\}}\right] d\lambda = E\left[|X_N| \int_1^{X^*} \frac{1}{\lambda} \mathbb{I}_{\{X^* \geqslant 1\}} d\lambda\right] = E\left[|X_N| \ln_+ X^*\right].$ Thus: $E[X^*] \leqslant 1 + E\left[|X_N| \ln_+ X^*\right].$

Step 3. Introduce the inequality: for any $a, b > 0$ , $a \ln b \leqslant a \ln a + \frac{b}{e}.$ Proof: let $f(b) = \frac{b}{e} - a \ln b$ , then $f'(b) = \frac{1}{e} - \frac{a}{b}$ . Setting $f'(b)=0$ gives $b=ae$ , at which $f(b)$ attains its minimum $a - a \ln(ae) = -a \ln a$ . Hence $f(b) \geqslant -a \ln a$ , i.e., $a \ln b \leqslant a \ln a + \frac{b}{e}.$ When $X^* \geqslant 1$ , taking $a = |X_N|, b = X^*$ : $|X_N| \ln_+ X^* = |X_N| \ln X^* \leqslant |X_N| \ln |X_N| + \frac{X^*}{e} \leqslant |X_N| \ln_+ |X_N| + \frac{X^*}{e}.$ When $X^* < 1$ , $\ln_+ X^* = 0$ , so the inequality holds trivially. Taking expectations: $E\left[|X_N| \ln_+ X^*\right] \leqslant E\left[|X_N| \ln_+ |X_N|\right] + \frac{1}{e} E[X^*].$

Step 4. Substituting Step 3 into Step 2: $E[X^*] \leqslant 1 + E\left[|X_N| \ln_+ |X_N|\right] + \frac{1}{e} E[X^*].$ Rearranging (if $E[X^*]$ is finite, or via a truncation argument): $\left(1 - \frac{1}{e}\right) E[X^*] \leqslant 1 + E\left[|X_N| \ln_+ |X_N|\right].$ Solving: $E[X^*] \leqslant \frac{e}{e-1} \left(1 + E\left[|X_N| \ln_+ |X_N|\right]\right).$ Since $E\left[|X_N| \ln_+ |X_N|\right] \leqslant \sup_n E\left[|X_n| \ln_+ |X_n|\right]$ , taking $C = \frac{e}{e-1}$ completes the proof.

Final answer

QED.

Marking scheme

The following is the detailed rubric for this problem:

1. Checkpoints (Total 7)

Score exactly one chain; if multiple paths exist, take the highest score; do not add points across chains.

Chain A: Classical Integration Method (Official Solution Path)

Integral representation and maximal inequality (3 pts total) [additive]
Use the tail probability integral formula to represent the expectation, with a reasonable splitting (e.g., $E[X^*] \le 1 + \int_1^\infty P(X^* \ge \lambda) d\lambda$ ) [1 pt]
Correctly cite and substitute Doob's maximal inequality: $\lambda P(X^* \ge \lambda) \le E\left[|X_N|\mathbb{I}_{\{X^* \geqslant \lambda\}}\right]$ [1 pt]
Use Fubini's theorem to interchange the order of integration, computing the intermediate result with the logarithmic term (i.e., deriving $E[X^*] \le 1 + E[|X_N| \ln_+ X^*]$ or equivalent form) [1 pt]
Decoupling via inequality (3 pts total) [additive]
Core step: Introduce or prove a specific algebraic inequality (such as $a \ln b \leqslant a \ln a + \frac{b}{e}$ ); this inequality must produce a coefficient $k < 1$ for the linear term to enable rearrangement [2 pts]
Note: If only using the ordinary Young inequality leading to a coefficient $\ge 1$ for $E[X^*]$ (which cannot be eliminated), these 2 pts are not awarded.
Substitute the random variables $|X_N|$ and $X^*$ into the above inequality, successfully separating $E[|X_N| \ln_+ |X_N|]$ and $E[X^*]$ [1 pt]
Conclusion and simplification (1 pt total) [additive]
Rearrange by combining the $E[X^*]$ terms (e.g., handling the coefficient $1 - 1/e$ ), solve for the final inequality, and identify the existence of the constant $C$ [1 pt]

2. Zero-credit items

Only writing out the final statement of Doob's $L \log L$ inequality without providing any proof derivation (treated as recitation rather than proof).
Only listing definitions of martingales, submartingales, or variables given in the problem, without substantive derivation.
Attempting to derive via $L^p$ norm ( $p>1$ ) boundedness without proving the limiting behavior or constant control as $p \to 1$ .

3. Deductions

Integration interval error: When handling $\int P(X^* \ge \lambda) d\lambda$ , failing to truncate at $\lambda=1$ (or $\lambda=e$ ), causing $\ln \lambda$ to diverge at $0$ or unclear logic. -1 pt
Missing symbol definition: Not handling the definition of $\ln_+$ (e.g., directly taking logarithm of values less than 1 without noting it equals 0). -1 pt
Coefficient computation error: In the algebraic inequality step, coefficient computation error that, despite correct approach, leads to a negative or meaningless constant $C$ . -1 pt
Circular reasoning: Using the $L \log L$ conclusion to be proved, or a stronger unproved conclusion, in the proof. Deduct to 0 pts.

Stochastic Processes – Problem 10: Prove that there exists a constant such that: where