2
$\begingroup$

I would be very gratefull if someone could help me with my question below. Intuitivly I can see that it is correct but I am unsure of how to prove it.

Let T be a stopping time in $\mathcal{F}_t$ for $t \geq 0$. $Y$ is a random variable that is $\mathcal{F}_T$ measurable.

The process $(X_t)_{t\geq0}$ is defined by $(X_t)=0$ for $0\leq t\leq T$ and $(X_t)=Y$ when $t \gt T$.

I want to show that $(X_t)_{t\geq0}$ is a predictable process. I can see that the process is left continous and therefore should also be predictable but i am unsure how to prove this. Would be very thankfull for any advise on where to start.

$\endgroup$
3
  • $\begingroup$ @ Jare : What is the definition of the predictable sigma algebra ? Hint : wouldn't it be defined by involving left continuous process ? $\endgroup$
    – TheBridge
    Oct 21, 2014 at 22:03
  • 1
    $\begingroup$ Yes, I started in that end but unsure if the random variable Y maps to a predictable rectangle in the sigma algebra. Is it enough to say that $Y$ is $\mathcal{F}_T$ measurable? @TheBridge $\endgroup$
    – Jaure
    Oct 21, 2014 at 22:48
  • $\begingroup$ What do you mean by "predictable rectangle"? $\endgroup$
    – saz
    Oct 22, 2014 at 5:50

1 Answer 1

Reset to default
3
$\begingroup$

By definition, a stochastic process $(X_t)_{t \geq 0}$ is predictable if it is adapted (i.e. $X_t$ is $\mathcal{F}_t$-measurable) and measurable with respect to the $\sigma$-algebra generated by the left-continuous adapted processes. In particular, any left-continuous adapted process is predictable.

It follows obviously from the definition that the given process $(X_t)_{t \geq 0}$ is left-continuous. Hence, it remains to prove that it is adapted. To this end, use the identity

$$\begin{align*} \{X_t \in B\} &= \{X_t \in B, t \leq T\} \cup \{X_t \in B, t>T\} \\ &= \begin{cases} \{t \leq \tau\} \cup (\{Y \in B\} \cap \{T<t\}) & 0 \in B \\ \{Y \in B\} \cap \{t>T\} & 0 \notin B \end{cases} \end{align*}$$ for any $B \in \mathcal{B}(\mathbb{R})$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .