Stochastic Processes – Problem 18: Let nonneg

Let nonneg. random variables $\zeta_{1},\zeta_{2},\cdots$ be independent and identically distributed. Define $X_{n}=\zeta_{1}\zeta_{2}\cdots\zeta_{n}$.

(1) What is the necessary and sufficient condition for $\{X_{n}\}$ to be a martingale? In that case, what can you conclude from the martingale convergence theorem?

Let $\mathcal{F}_n = \sigma(\zeta_1, \dots, \zeta_n)$. Then $E[X_{n+1} | \mathcal{F}_n] = E[X_n \zeta_{n+1} | \mathcal{F}_n] = X_n \cdot E[\zeta_{n+1}]$, since $X_n$ is $\mathcal{F}_n$-measurable and $\zeta_{n+1}$ is independent of $\mathcal{F}_n$.

For $\{X_n\}$ to be a martingale, we need $E[X_{n+1} | \mathcal{F}_n] = X_n$ a.s., i.e., $X_n \cdot E[\zeta_{n+1}] = X_n$ a.s. Since $X_n \ge 0$, this is equivalent to $E[\zeta_1] = 1$.

If $E[\zeta_1] = 1$, then $X_n$ is a nonneg. martingale. By Doob's martingale convergence theorem, there exists $X_\infty$ such that $X_n \to X_\infty$ a.s. with $E[X_\infty] \le E[X_0] = 1$.

By Kakutani's theorem for product martingales: - If $\prod_{k=1}^\infty E[\sqrt{\zeta_k}] > 0$ (equiv. $\sum(1 - E[\sqrt{\zeta_k}]) < \infty$), then $X_n$ converges in $L^1$ with $E[X_\infty] = 1$. - Otherwise $X_n \to 0$ a.s. with $E[X_\infty] = 0$.

Since $\zeta_k$ are i.i.d., $E[\sqrt{\zeta_1}] \le \sqrt{E[\zeta_1]} = 1$ by Jensen's inequality. If $E[\sqrt{\zeta_1}] < 1$, then $\prod E[\sqrt{\zeta_k}] = 0$, so $X_n \to 0$ a.s. If $E[\sqrt{\zeta_1}] = 1$, then $\mathrm{Var}(\sqrt{\zeta_1}) = 0$, i.e., $\zeta_1 = 1$ a.s., so $X_n \equiv 1$.

Conclusion: $\{X_n\}$ is a martingale iff $E[\zeta_1] = 1$. In that case, $X_n$ is a nonneg. martingale converging a.s. to some $X_\infty \ge 0$, with: - If $\zeta_1 = 1$ a.s., then $X_\infty = 1$; - If $E[\sqrt{\zeta_1}] < 1$, then $X_\infty = 0$ a.s.

$\{X_n\}$ is a martingale if and only if $E[\zeta_1] = 1$. Conclusion: $X_n$ is a nonneg. martingale converging a.s. to $X_\infty \ge 0$, with: - If $\zeta_1 = 1$ a.s., then $X_\infty = 1$; - If $E[\sqrt{\zeta_1}] < 1$, then $X_\infty = 0$ a.s.