For $t\ge 0$ define the stochastic process
$$Y_t:[0,1]\rightarrow\mathbb{R},\quad Y_t(x)=\begin{cases}0,&\text{if }t-x\notin\mathbb{Q}\\
1,&\text{if }t-x\in\mathbb{Q}
\end{cases}$$
on the filtration $(F_t)_{t\ge0}$, $F_t=\mathcal{B}(\mathbb{R})$ for all $t\ge 0$. I want to show, that this is a martingale.
Of course $Y_t\le1$, such that $Y_t$ is integrable. For $s,t\in\mathbb{Q}_+$ I find $Y_t=Y_s$. So we have the martingale property for this case. But how do you find the solution for the other cases? And what can you say about the continuity of the paths of this process? Thanks in advance for any help!
Best Answer
We may write $$ Y_t(x) = 1_{(t+\mathbb{Q})\cap [0,1]}(x) $$ and it follows $Y_t$ is $F_t=\mathcal{B}(\mathbb{R})$ measurable. And also, we can see that $ Y_t(x)= 0 $ holds almost surely, thus giving $$ E[Y_t|F_s]=E[0|F_s]=0=Y_s $$ holds almost surely for all $s<t$. This establishes that $\{Y_t,F_t\}$ is a martingale. Finally, we can observe that for all $x\in [0,1]$, the path $$ t\mapsto Y_t(x)=1_{x+\mathbb{Q}}(t) $$ of $Y_t$ is everywhere discontinuous.