Is this product is a martingale?

Question

Let $\tau$ be a random variable with exponentiel distribution with parameter $1$ and Let U be a random variable with law $P(U=1)=P(U=-1)=1/2$ independent of $\tau$. We put $X_t=U\Bbb{I}_{(t\geq \tau)}$. Then X is a martingale.

Thanks

Answer 1

Here is an outline of an argument which I believe works. There may be a more elegant way to do it, but I could not think of one.

Let $\mathcal F_t = \sigma\{X_s : s\leq t\}$. The variable $X_s$ takes on three values: it is 0 on $A_s:= \{s < \tau\}$, 1 on $B_s:= \{s\geq \tau\}\cap \{U = 1\}$, and $-1$ on $C_s:= \{s\geq \tau\} \cap \{U=-1\}$. Hence $\mathcal F_s$ is generated by the sets $A_s, B_s, C_s$ for $s\leq t$. Notice that the collection of $A_s$, $B_s$, $C_s$ for $s \leq t$ is closed under finite intersections, so is a $\pi$-system.

Furthermore, if $r > t$, the collection of sets $E$ such that $\mathbb E(1_E X_r) = \mathbb E(1_E X_t)$ is a $\lambda$-system. By the Dynkin $\pi-\lambda$ theorem (http://en.wikipedia.org/wiki/Dynkin_system), it will then suffice to show that the collection of sets with this property contains $A_s , B_s , C_s$ for each $s\leq t$. For $E=A_s$, $\mathbb E(1_{E} X_r) = \mathbb E(1_{E} X_t)$ (since $U$ is symmetric on these sets). For $E = B_s$, both sides are 1. For $E = C_s$, both sides are $-1$. Hence $\mathbb E(1_E X_r) = \mathbb E(1_E X_t)$ for each $E \in \sigma(A_s , B_s , C_s : s\leq t) = \mathcal F_t$. This proves $\mathbb E(X_r | \mathcal F_t) = X_t$.