Existence of nonnegative and nonconstant martingales

Question

Are there martingales that are nonnegative and nonconstant? If so, are their any intuitive examples for such?

Answer 1

Example 1: Let $(X_j)_{j \in \mathbb{N}}$ be a sequence of independent random variables such that $$\mathbb{P}(X_j = - 2^{-j})=\frac{1}{2} \quad \text{and} \quad \mathbb{P}(X_j=2^{-j})=\frac{1}{2}$$ and define $$M_n := 1 + \sum_{j=1}^n X_j.$$ As $\mathbb{E}(X_j)=0$ for all $j \in \mathbb{N}$, the independence of the random variables $(X_j)_{j \in \mathbb{N}}$ implies that $(M_n)_{n \in \mathbb{N}}$ is a martingale. Moreover, the process is clearly nonconstant. As $$\sum_{j=1}^{\infty} |X_j| = \sum_{j=1}^{\infty} 2^{-j} = 1$$ we have $$M_n = 1+ \sum_{j=1}^n X_j \geq 1- \sum_{j=1}^{\infty} |X_j| \geq 0,$$ i.e. $M_n \geq 0$ for all $n \in \mathbb{N}$.

Example 2: Let $(\xi_j)_{j \in \mathbb{N}}$ be independent identically distributed random variables such that $$\mathbb{P}(\xi_j = 3/2) = \frac{1}{2} \quad \text{and} \quad \mathbb{P}(\xi_j = 1/2) = \frac{1}{2}.$$ Since $\mathbb{E}(\xi_j)=1$ for all $j \geq 1$, it follows from the independence of the random variables that $$M_n := \prod_{j=1}^n \xi_j$$ is a martingale. Clearly, $(M_n)_{n \in \mathbb{N}}$ is non-constant and non-negative.

Remark: It is not difficult to show that that the only non-negative martingale $(M_n)_{n \in \mathbb{N}}$ with $M_0=0$ is the trivial martingale.

Answer 2

Another standard example is to take simple random walk started at $1$ and stopped when it hits 0. That is, let $S_n = 1 + \xi_1 + \xi_2 + \dots + \xi_n$ where $\xi_i$ are iid $\pm1$, $T = \min\{n : S_n = 0\}$, and $X_n = S_{n \wedge T}$.

This is nonconstant in the sense that none of the $X_n$, $n \ge 1$, is a constant random variable. But it is "eventually constant" in the sense that for almost every $\omega$, there is an $N(\omega)$ such that $X_n(\omega) = 0$ for all $n \ge N(\omega)$. In other words, almost surely, it is zero from some time onwards, but that time is random and unbounded.