I have been practising a question set by my lecturer and try to verify the answer, unfortunately I am unable to understand the following question and answer.
$\textbf{Question:}$ Let $\Omega=\{1,2,3\}, \mathcal{A}=\mathcal{P}(\Omega)$ and $P(\{\omega\})=\tfrac{1}{3}$ for each $\omega \in \Omega$. Define a stochastic process $(X(t):t\ge 0)$ by $X(t)(\omega) = \max\{t-\omega,0\}.$ Then the filtration generated by the stochastic process $X$ computes as \begin{align} \mathcal{F} = \begin{cases} \{0,\Omega\}, \qquad \qquad \qquad \text{if $t\in[0,1],$} \\ \{0,\Omega,\{1\},\{2,3\}, \phantom{xx}\text{if $t \in (1,2]$,}\\ \mathcal{P}(\Omega), \qquad \qquad \qquad \phantom{.}\text{if $t>2$.} \end{cases} \end{align}
Would anyone be able to briefly explain how the filtration was generated and how the $t$ interval was selected?
Many thanks,
John
We need to check that the random variable $X(t)$ is measurable with respect to $\sigma$-algebra $\mathcal F_t$ for each $t\ge0$.
If $t\in[0,1]$, the only value that $X(t)$ can take is $0$. So we need to find the smallest $\sigma$-algebra that contains $X^{-1}(t)(\{0\})=\Omega$. Such $\sigma$-algebra is $\{\emptyset,\Omega\}$.
If $t\in(1,2]$, the range of $X(t)$ is $\{0, t-1\}$. Then $X^{-1}(t)(\{0\})=\{2,3\}$ and $X^{-1}(t)(\{t-1\})=\{1\}$. The smallest $\sigma$-algebra is $\{\emptyset, \Omega,\{1\},\{2,3\}\}$.
If $t\in(2,3]$, the range of $X(t)$ is $\{0,t-1,t-2\}$, so that the $\sigma$-algebra must contain the sets $\{1\}$,$\{2\}$ and $\{3\}$. The smallest $\sigma$-algebra is $\mathcal P(\Omega)$.