Showing that the r-th moment of $u_t$ is finite

Question

I am a reading a paper, in which $u_t = \pi(L)\epsilon_t$, where $L$ is the lag operator and $\pi(z)=\sum_{k=0}^{\infty}\pi_kz^k$. Also,

It is said that the Marcinkiewicz-Zygmund inequality for martingales implies that the r-th moment of $u_t$ exists and is uniformly bounded in t. However, $u_t$ is not a martingale, thus I wonder how I could derive this result. Any comment would be appreciated even if it would not be about the MZ inequality.

Answer 1

You have $$ u_t=\sum_{k=0}^{\infty} \pi_k \epsilon_{t+k} $$ The partial sums of the right hand side are the martingale that you apply MZ to, separately for each $t$.

EDIT: More details: Define $u_{t, s}:= \sum_{k=0}^{s}\pi_k \epsilon_{t+k}$. Then $(u_{t, s})_{s}$ is a martingale because $$ E[u_{t, s+1} | \mathcal{F}_s] = u_{t,s} + E[\pi_{s+1}\epsilon_{t+s+1} | \mathcal{F}_{s}] = u_{t,s} $$