Suppose we want to show that $$P[|S_j - S_i| \wedge |S_k - S_j| \ge \lambda] \le \frac{1}{\lambda}^{4\beta}C_{i,k}$$ for some $\lambda>0$, $\beta \ge 0$ and some $C_{i,k}$. Then how is this implied by the following moment inequality? $$E[|S_j - S_i|^{2\beta}|S_k-S_j|^{2\beta}]\le C_{i,k}$$
From Markov's inequality we have $P[||S_j - S_i| \wedge |S_k - S_j| \ge \lambda] \le \frac{E[|S_j-S_i| \wedge |S_k - S_j|]^{4\beta} }{\lambda^{4\beta}}$. But I don'tknow how to bound this by the expectation above. I would greatly appreciate any help.
Note the inclusion $$(|S_j - S_i| \wedge |S_k - S_j| \ge \lambda) = (|S_j - S_i| \ge \lambda) \cap (|S_k - S_j| \ge \lambda) \subset (|S_j - S_i||S_k - S_j| \geq \lambda^2)$$
Thus $$\begin{aligned}P(|S_j - S_i| \wedge |S_k - S_j| \ge \lambda) &\leq P(|S_j - S_i||S_k - S_j| \geq \lambda^2) \\ &=P(|S_j - S_i|^{2\beta}|S_k - S_j|^{2\beta}\geq \lambda^{4\beta}) \\ &\leq \frac{C_{i,k}}{\lambda^{4\beta}} \end{aligned}$$