Let $T$ is a Markov time such that $T= \min \{ n : X[n] = 1\}$ , $X[n]$ is the number of $h$ (heads) in coin tossing for $n$ times. Let's say I will toss the coin 3 times, so the event collection is $\{hhh,hht,hth,htt,thh,tht,tth,ttt\}$.
Let $T\wedge n$ denote the minimum between $n$ and $T$ i.e., $$T\wedge n=\begin{cases} n & \mbox{ if }n< T,\\ T &\mbox{ if }n\geq T.\end{cases}$$
what is the value of $X[T\wedge 2] (tth)$?
the value of $X[T\wedge 2] (hhh) = 1$.
Please help since I am getting confused. I think $T$ is always $1$, which means
$$X[1\wedge 2] (tth) = X[1] (tth) = 0,$$
However when you look at it the following way :
$$X[(T\wedge 2)(tth)](tth) = X[3\wedge 2](tth)=X[2](tth)=0.$$
Although the answer is the same, but $X[1]$ is not the same as $X[2]$.
If $\omega$ is some sample point, then the expression $X[T(\omega)\wedge 2](\omega)$ means that you should evaluate the time point $T(\omega)\wedge 2$ at $\omega$ and then substitute that time point into the $X$ process evaluated at the same $\omega$.
For example, if $\omega=tth$ then $T(\omega)=3$ so that $T(\omega)\wedge 2=3\wedge 2=2$. Therefore $$X[T(tth)\wedge 2](tth) =X[2](tth)=0.$$ It is zero since no heads have been observed up to and including the second coin toss.