# question about martingale about maximal inequality for submartigales

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 $$\int_0^{\infty} P\{Y>t\}dt \leq 1+\int_1^{\infty} P\{Y>t\}dt \leq 1+ \frac 1 {\alpha -1}=\frac {\alpha} {\alpha -1}$$.