Suppose ${X_n}$ is a martingale satisfying, for some $\alpha > 1$,

$E\left[|X_n|^\alpha\right]<\infty$, for all n. Show $$E\left[\max_{0\leq k \leq n}|X_k| \right]\leq \frac{\alpha}{\alpha-1} E[|X_n|^{\alpha}]^{\frac{1}{\alpha}}$$

Hint:$$E\left[\max_{0\leq k \leq n}|X_k| \right]=\int _{0}^{\infty}\{\max_{0\leq k \leq n}|X_k|>t\}dt.$$ Now use the maximal inequality on the submartingale $|X_n|^{\alpha}$

Note the maximal inequality on the submartingale ${X_n}$. Let ${X_n}$ be a submartingale for which $X_n\geq 0$ for all n. Then for any positive $\lambda \Pr\{\max_{0\leq k\leq n} X_k>\lambda\}\leq E[X_n]$.


Hints: apply submartingale inequality to $\{Y_j:1\leq j \leq n\}$ where $Y_j=\frac {|X_j|^{\alpha}} {E(|X_n|^{\alpha})^{\frac 1 {\alpha}}}$. [By Jensen's inequality this is indeed a submartingale]. Note that $\int_0^{\infty} P\{Y>t\}dt \leq 1+\int_1^{\infty} P\{Y>t\}dt \leq 1+ \frac 1 {\alpha -1}=\frac {\alpha} {\alpha -1}$.

  • $\begingroup$ Thank you so much! $\endgroup$ – glorianrvt Dec 27 '18 at 5:45

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