Let be nonnegative i.i.d. random variables. Define . (2) What is the necessary and sufficient condition for to be a uniformly integrable martingale?

Stochastic Processes Solution – Problem 27: Let be nonnegative i.i.d

Question

Let $\zeta_1, \zeta_2, \ldots$ be nonnegative i.i.d. random variables. Define $X_n = \zeta_1 \zeta_2 \cdots \zeta_n$ .

(2) What is the necessary and sufficient condition for $\{X_n\}$ to be a uniformly integrable martingale?

Step-by-step solution

First, determine when $\{X_n\}$ is a martingale. By the martingale definition, we need $E[X_{n+1} | \mathcal{F}_n] = X_n$ , where $\mathcal{F}_n = \sigma(\zeta_1, \dots, \zeta_n)$ . Computing: $E[X_{n+1} | \mathcal{F}_n] = E[X_n \zeta_{n+1} | \mathcal{F}_n] = X_n E[\zeta_{n+1}] = X_n E[\zeta_1]$ . For $\{X_n\}$ to be a martingale, we must have $E[\zeta_1] = 1$ .

Analyze the almost sure convergence of $\{X_n\}$ . Since $\{X_n\}$ is a nonneg martingale, by the martingale convergence theorem it converges a.s. to some $X_\infty$ . Taking logarithms (assuming $\zeta_i > 0$ a.s.): $\ln X_n = \sum_{i=1}^n \ln \zeta_i = n \cdot \frac{1}{n} \sum_{i=1}^n \ln \zeta_i.$ By the strong law of large numbers, $\frac{1}{n} \sum \ln \zeta_i \to E[\ln \zeta_1]$ a.s. By Jensen's inequality, since $\ln$ is strictly concave (unless $\zeta_1$ is constant), $E[\ln \zeta_1] < \ln E[\zeta_1] = 0$ . Setting $\mu = E[\ln \zeta_1] < 0$ , we get $\ln X_n \to -\infty$ , hence $X_n \to 0$ a.s. The degenerate case $\zeta_1 \equiv 1$ gives $X_n \equiv 1$ , which is trivially UI.

Apply the equivalence condition for uniform integrability. For a nonneg martingale, UI is equivalent to $L^1$ convergence, i.e., $E[X_n] \to E[X_\infty]$ . In the non-degenerate case, $X_\infty = 0$ a.s., so $L^1$ convergence requires $E[X_n] \to 0$ . But as a martingale, $E[X_n] = E[X_0] = 1$ for all $n$ . This is a contradiction: $1 \to 0$ is impossible.

Summary of the necessary and sufficient condition. If $\zeta_1 \equiv 1$ a.s., then $X_n \equiv 1$ , which is UI. If $\zeta_1$ is not identically 1 and $E[\zeta_1] = 1$ , then $X_n \to 0$ a.s. but $E[X_n] = 1$ , so it is not UI. Therefore, $\{X_n\}$ is a uniformly integrable martingale if and only if $P(\zeta_1 = 1) = 1$ .

Final answer

$\zeta_1 \equiv 1$ a.s.

Marking scheme

The following is the grading rubric for this probability theory problem.

1. Checkpoints (max 7 pts total)

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

Chain A: Analysis via the Law of Large Numbers and Jensen's Inequality

Necessary condition for the martingale [additive]
State that $E[\zeta_1] = 1$ is necessary for $\{X_n\}$ to be a martingale. (1 pt)
Convergence analysis [additive]
Introduce the log transform $\ln X_n = \sum \ln \zeta_i$ and use SLLN to state $\frac{1}{n}\ln X_n \to E[\ln \zeta_1]$ a.s. (1 pt)
Use Jensen's inequality: if $\zeta_1$ is non-degenerate, then $E[\ln \zeta_1] < \ln E[\zeta_1] = 0$ (must show strict inequality or discuss non-degeneracy). (1 pt)
Conclude: in the non-degenerate case, $\ln X_n \to -\infty$ , i.e., $X_n \to 0$ a.s. (1 pt)
Uniform integrability (UI) criterion [additive]
Cite the theorem: a nonneg martingale is UI iff it converges in $L^1$ (or equivalently $E[X_n] \to E[X_\infty]$ ). (1 pt)
Identify the contradiction: if $\zeta_1$ is non-degenerate, $E[X_\infty] = 0$ contradicts $E[X_n] \equiv 1$ , so not UI. (1 pt)
Final conclusion: The necessary and sufficient condition is $P(\zeta_1 = 1) = 1$ . (1 pt)

Chain B: Direct invocation of product martingale properties

Necessary condition for the martingale [additive]: State $E[\zeta_1] = 1$ . (1 pt)
Limit property invocation [additive]: Directly cite a known result on nonneg product martingales (e.g., Kakutani dichotomy): if $E[\zeta_1]=1$ and $\zeta_1 \not\equiv 1$ , then $X_n \to 0$ a.s. (3 pts)
UI criterion [additive]: State $L^1$ convergence requires $E[X_n] \to E[X_\infty]$ . (1 pt)
Identify the $1 \to 0$ contradiction. (1 pt)
Final conclusion: $P(\zeta_1 = 1) = 1$ . (1 pt)

Total (max 7)

2. Zero-credit items

Only copying the definition of UI or martingale without connecting to the specific variables.
Only stating $\zeta_1 \equiv 1$ is a UI martingale without proving it is the only case.
Incorrectly asserting: "since $E[X_n]$ is bounded, it is UI" (this is a basic martingale property, not a sufficient condition).
Incorrectly asserting: "since $X_n$ converges, it is UI" (ignoring mass loss).

3. Deductions

Jensen's inequality logic gap: When using $E[\ln \zeta] \le \ln E[\zeta]$ , not stating that equality holds iff $\zeta$ is constant. (-1 pt)
Sufficiency/necessity confusion: Only proving sufficiency or only necessity, not both. (Cap at 4/7)
Confusing $L^1$ and a.s. convergence: Believing $X_n \to 0$ a.s. implies $E[X_n] \to 0$ . (-2 pts)

Stochastic Processes – Problem 27: Let be nonnegative i.i.d