Square integrable stochastic process

Question

Suppose that for a stochastic process we have

\begin{align} \mathbb{E}\left[\int_{0}^{T}X^{2}(t)dt \right]<\infty \end{align}

where $T<\infty$. Does it holds that $|X(t)|<M$, where $M$ constant? I am a little confused.

Answer 1

The statement $|X(t)|<M$ can be interpreted in two different ways: Either

$$\exists M: |X(t,\omega)| \leq M \, \quad \text{for all} \, \omega \in \Omega \tag{1}$$

(i.e. the bound is uniform for all $\omega \in \Omega$) or

$$\forall \omega \in \Omega \, \, \exists M=M(\omega): |X_t(\omega)| \leq M. \tag{2}$$

• $(1)$ is in general not correct. To see this, simply consider a Brownian motion $(B_t)_{t \geq 0}$: From $\mathbb{E}(B_t^2) = t$, we get that $$\mathbb{E} \left( \int_0^T X_t^2 \, dt \right) = \frac{T^2}{2}< \infty.$$ On the other hand, there does not exist $M$ such that $(1)$ is satisfied since this would contradict $\mathbb{P}(|B_t|>M)>0$ (which follows from $B_t \sim N(0,t)$).
• $(2)$ usually holds, but it has little to do with condition $$\mathbb{E}\left( \int_0^T X_t^2 \, dt \right)<\infty.$$ Rather it is a consequence of the fact that cádlág (or cáglád) processes are considered - and these are bounded (in the sense of $(2)$) on any finite time interval.