Bounded expectation of a martingale process

Question

Suppose $N(t)$ is a rate $\lambda$ Poisson process. I define $X(t) = N(t) - \lambda t$. I need to argue that $\mathbb{E}[|X(t)|]< \infty$. I would proceed as follows \begin{align} \mathbb{E}[|X(t)|] = \mathbb{E}[|N(t) - \lambda t|] \le \mathbb{E}[ N(t) + \lambda t] = 2 \lambda t, \end{align} So it is clearly bounded in expectation for finite times $t<\infty$. But is this argument sufficient to establish boundedness of $|X(t)|$ in expectation? I would appreciate any insights.

Answer 1

Yes, it suffices. The target $\mathbb{E}\left|X_t\right|<\infty$ means exactly that, for each given $t$, $\mathbb{E}\left|X_t\right|$ is a finite value; it does not mean that $\mathbb{E}\left|X_t\right|$ is uniformly bounded by all $t\ge 0$.

In fact, note that $$ \mathbb{E}\left|N_t-\lambda t\right| $$ is the mean deviation of $N_t$.

Provided that $$ N_t\sim\text{Poiss}(\lambda t), $$ we have, by referring to this table if you do not want to carry out the details of the calculation, $$ \mathbb{E}\left|N_t-\lambda t\right|=\frac{2}{\lfloor\lambda t\rfloor!}\left(\lambda t\right)^{\lfloor\lambda t\rfloor+1}e^{-\lambda t}, $$ where $\lfloor x\rfloor$ denotes the largest integer smaller than or equal to $x$.

With this exact value of $\mathbb{E}\left|X_t\right|$, you may see that as $t\to\infty$, $\mathbb{E}\left|X_t\right|\to\infty$.