Let $M_t$ be a martingale whose quadratic variation is finite (and also any higher moments).
Then for any $s<t$ and $k > 1$, is it true that \begin{align*} \mathbb{E}|M_s|^k \le \mathbb{E}|M_t|^k? \end{align*}
My intuition says that due to the quadratic variation of any martingale is increasing, so $\mathbb{E}|M_s|^2 \le \mathbb{E}|M_t|^k$ is quite trivial due to the Ito's isometry. But is this still hold for any arbitrary $k>1$?
Thanks,
$|M_s| \leq E(|M_t| \, |\mathcal F_s)$ and $f(x)=x^{k}$ is an increasing convex function on $[0,\infty)$ Hence, $|M_s|^{k} \leq (E(|M_t|\mathcal F_s))^{k} \leq E(|M_t|^{k}\mathcal F_s)$ by Jensen's inequality. Take expectation on both sides.