The following computations are based on Lemma 5.9 of Karatzas and Shreve. Suppose $X$ is a continuous-time martingale and that $\left|X_t\right|\leq K$ $\mathbb{P}$-a.s. . Let $0=t_{0,n}<t_{1,n}<\dots<t_{n,n}=t$ be a partition of $[0,t]$. Thanks to the martingale property and a telescopic sum it is pretty easy to verify that, for any $k=0,\dots,n-1$, $$ \mathbb{E}\left[\sum_{j=k+1}^n\left(X_{t_{j,n}}-X_{t_{j-1,n}}\right)^2\left|\mathcal{F}_{t_{k,n}}\right.\right] = \mathbb{E}\left[\left(\sum_{j=k+1}^n\left(X_{t_{j,n}}-X_{t_{j-1,n}}\right)\right)^2\left|\mathcal{F}_{t_{k,n}}\right.\right] =\mathbb{E}\left[\sum_{j=k+1}^n\left(X^2_{t_{j,n}}-X^2_{t_{j-1,n}}\right)\left|\mathcal{F}_{t_{k,n}}\right.\right]\leq \mathbb{E}\left[X^2_{t_{n,n}}\left|\mathcal{F}_{t_{k,n}}\right.\right]\leq K^2 $$ from which it follows that \begin{eqnarray} \mathbb{E}\left[\sum_{k=1}^{n-1}\sum_{j=k+1}^n\left(X_{t_{k,n}}-X_{t_{k-1,n}}\right)^2\left(X_{t_{j,n}}-X_{t_{j-1,n}}\right)^2\right] &=& \mathbb{E}\left[\sum_{k=1}^{n-1}\left(X_{t_{k,n}}-X_{t_{k-1,n}}\right)^2\sum_{j=k+1}^n\mathbb{E}\left[\left(X_{t_{j,n}}-X_{t_{j-1,n}}\right)^2\left|\mathcal{F}_{t_{k,n}}\right.\right]\right]\\ &\leq& K^2\,\mathbb{E}\left[\sum_{k=1}^{n-1}\left(X_{t_{k,n}}-X_{t_{k-1,n}}\right)^2\right]\leq K^4 \end{eqnarray} Up to here everything works fine to me. The authors conclude also that $$ \mathbb{E}\left[\sum_{k=1}^{n-1}\left(X_{t_{k,n}}-X_{t_{k-1,n}}\right)^4\right]\leq 4\,K^2\,\mathbb{E}\left[\sum_{k=1}^{n-1}\left(X_{t_{k,n}}-X_{t_{k-1,n}}\right)^2\right] $$ This last inequality it's hard to explain for me. I tired with Holder or expanding the fourth power inside the parenthesis and using the martingale property, but nothing works.
2026-04-15 09:52:36.1776246756
Inequality for expected value of sum of increments of a martingale (from Karatzas and Shreve)
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