Prove that $\tau$ is a stopping time

Question

Let $X$ be a random variable with the following distribution: $P\big(X=1\big)=P\big(X=-1\big)=\frac{1}{2}$.

Assume that: $F_{t}=\sigma\big\{sX\colon\ s\leq t\big\}$

Task:

Check if $\tau(\omega)=\inf\big\{t>0\colon\ tX(\omega)>0 \big\}$ is stopping time with respect to $F_{t}$.

Solution:

We must check whether: $\big\{\omega\in\Omega\colon\ \tau(\omega)\leq t\big\}\in F_{t}$, for every $t>0$.

Thanks to Kavi Rama Murthy comment:

1. If $X(\omega)=-1$, then $\tau(\omega)=+\infty$ since infimum of empty set is eual to $+\infty$.

2. If $X(\omega)=1$, then $\tau(\omega)=0$.

How to formally show that $\tau$ is stopping time?

Answer 1

Let $0<s<t$, then \begin{align} \{\tau\leqslant t\}&= (\{\tau\leqslant t\}\cap\{X=-1\}) \cup (\{\tau\leqslant t\}\cap\{X=1\} )\\ &= \{X=1\}\\ &= \{sX=s\}\in\mathcal F_t, \end{align} and hence $\tau$ is a stopping time with respect to $\{\mathcal F_t\}$.