I'm currently studying some properties of general stochastic processes, and am having some issue understanding how to prove this (probably simple) example.
First, let me introduce the notation & definition:
Let $X$ be a stochastic process, and $\tau$ be a stopping time. We say that a space of processes $\mathcal{X}$ is closed under truncation if, whenever $X \in \mathcal{X}$, then $X^\tau \in \mathcal{X}$, where $X^\tau$ is the process X stopped at $\tau$, i.e. $$X^\tau(t,\omega) = X(\tau(\omega) \wedge t, \omega).$$ Now, there are several examples of spaces of stochastic process closed under truncation.
- $\mathcal{M}$, the space of uniformly integrable martingales
- $\mathcal{H}^2$, the space of square-integrable martingales
- $\mathcal{K}$, the space of bounded processes
Of course, I wanted to very that these are actually closed under truncation. For the space of uniformly integrable martingales, the result follows directly from the Optional Stopping/Sampling theorem. For the square-integrable martingales, it follows from the inequality $$\mathbb{E}\left[ \left( \sup_{t \geq 0} |X^\tau (t)|\right)^2\right] \leq \mathbb{E} \left[ \left( \sup_{t\geq 0}|X(t)|\right)^2\right] < \infty,$$ where the right part of the inequality is just the definition of square-integrable martingales (although square-integrable martingale seems like a misnomer to me, it looks more like $L^p$ boundedness).
However, I'm not sure how to prove it for the third case, i.e. when $X$ is a bounded process. The proof is probably trivial, but somehow I can't see it right now.