I have just encountered this question
Suppose $({X_n}, n \in \mathbb N)$ is a martingale. Let $n \ge 1$ and $\alpha > 1$ such that $E\left[|X_n|^\alpha\right]<\infty$. Then
$$
E\left[\max_{0\leq k \leq n}|X_k| \right]\leq \frac{\alpha}{\alpha-1} E[|X_n|^{\alpha}]^{\frac{1}{\alpha}}.
$$
Hint:
$$
E\left[\max_{0\leq k \leq n}|X_k| \right]=\int _{0}^{\infty} P \left [\max_{0\leq k \leq n}|X_k|>t \right ] dt.
$$
Now use the maximal inequality on the submartingale $|X_n|^{\alpha}$ (Let ${X_n}$ be a submartingale for which $X_n\geq 0$ for all n. Then for any positive $\lambda>0$ we have $\lambda P[\max_{0\leq k\leq n} X_k>\lambda] \leq E[X_n]$).
and its answer
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
$$
\int_0^{\infty} P [Y>t] dt \leq 1 + \int_1^{\infty} P[Y>t] dt \leq 1+ \frac 1 {\alpha -1}=\frac {\alpha} {\alpha -1}.
$$
My understanding I assume $Y =\max_{0\leq k \leq n}|Y_k|$. By maximal inequality for the sub-martingale $(Y_n, n \in \mathbb N)$,
$$
\int_1^{\infty} P[Y>t] dt \le \int_1^{\infty} \frac{E[Y_n]}{t} dt = E(|X_n|^{\alpha})^{1-\frac 1 {\alpha}}\int_1^{\infty} \frac{1}{t} dt \color{red}{= +\infty}.
$$
Could you please elaborate on how $\int_1^{\infty} P[Y>t] dt \leq \frac 1 {\alpha -1}$ in the answer?
Best Answer
Let $X=(X_n)_{n \in \mathbb{N}}$ be a $\mathscr{F}_n$-martingale s.t. $0<E[|X_n|^\alpha]<\infty$ for some $\alpha >1$, $n\geq 1$. Then, define for this fixed $n$ $$Y_k=\frac{|X_k|}{(E[|X_n|^\alpha])^{1/\alpha}}$$ We claim $Y=(Y_k)_{1\leq k\leq n}$ is a nonnegative $\mathscr{F}_k$-submartingale. Indeed, by Jensen's inequality, for any $\ell < k\leq n$ $$E[Y_k|\mathscr{F}_\ell]=\frac{E[|X_k||\mathscr{F}_\ell]}{(E[|X_n|^\alpha])^{1/\alpha}}\geq \frac{|E[X_k|\mathscr{F}_\ell]|}{(E[|X_n|^\alpha])^{1/\alpha}}=\frac{|X_\ell|}{(E[|X_n|^\alpha])^{1/\alpha}}=Y_\ell$$ By using the nonnegative submartingale maximal inequality with $\alpha$ we obtain for $t \geq 0$: $$P\bigg(\max_{1\leq k \leq n}Y_k\geq t\bigg)\leq t^{-\alpha}E[|Y_n|^\alpha]=t^{-\alpha}\frac{E[|X_n|^\alpha]}{(E[|X_n|^\alpha])^{\frac{1}{\alpha}\cdot \alpha}}=t^{-\alpha}$$ So we get $$E\bigg[\max_{1\leq k \leq n}Y_k\bigg]=\int_0^\infty P\bigg(\max_{1\leq k \leq n}Y_k\geq t\bigg)dt\leq 1+\int_1^\infty t^{-\alpha}dt=\frac{\alpha}{\alpha-1}$$