Dob's Inequality Proof
Theorem
Given a probability space $( \Omega , \mathcal{F} , P)$ and a submartingale $\left\{ ( X_{n} , \mathcal{F}_{n} ) \right\}$.
If for some $N \in \mathbb{N}$ and $p>1$, $X_{n} \ge 0 (n \le N)$, $E X_{N}^{p} < \infty$ then $$ E \left( \max_{n \le N} X_{n}^{p} \right) \le \left( {{ p } \over { p-1 }} \right)^{p} E X_{N}^{p} \text{ a.s.} $$
Explanation
The form of the equation can be seen as extracting the $\displaystyle \left( {{ p } \over { p-1 }} \right)^{p}$ resulting from $\displaystyle \max_{n \le N} \cdot_{n} ^{p}$ and calculating its upper bound. Since $\displaystyle \left( {{ p } \over { p-1 }} \right)>1$, if $p$ is too large, the inequality’s value decreases, and it is crucial that the final number $N$ can be large. In practical applications to statistics beyond theory, this $N$ can be considered quite large, so there is no particular concern that $N$ is limited.
Proof
Strategy: First, for a constant $L$, choose $\displaystyle \left( L \land \max_{n \le N} X_{n} ^{p} \right)$ and cut off the value based on $L$ to develop the equation. Here, $\land$ means $\displaystyle (f \land g) (x):= \min \left\{ f(x) , g(x) \right\}$ for two functions $f,g$. After obtaining a satisfactory expression through various tricks, sending $L$ to infinity will eventually leave only $\displaystyle \infty \land \max_{n \le N} X_{n} ^{p} = \max_{n \le N} X_{n} ^{p}$.
Part 1.
Consider an increasing sequence $\left\{ L_{n} \right\}$ that satisfies $L_{n+1} > L_{n}$ for all $n \in \mathbb{N}$, where $\displaystyle Y:= \max_{n \le N} X_{n}$.
$$ EX = \int_{0}^{\infty} P(X>t) dt $$
By considering $t:= \lambda^{p}$, according to another expression of expectation, $$ \begin{align*} E \left[ \left( Y \land L_{n} \right)^{p} \right] =& \int_{0}^{\infty} P \left[ \left( Y \land L_{n} \right)^{p} > t \right] dt \\ =& \int_{0}^{\infty} P \left[ \left( Y \land L_{n} \right)^{p} > \lambda^{p} \right] p \lambda^{p-1} d \lambda \\ =& \int_{0}^{\infty} P \left( Y \land L_{n} > \lambda \right) p \lambda^{p-1} d \lambda \\ =& \int_{0}^{L_{n}} P \left( Y \land L_{n} > \lambda \right) p \lambda^{p-1} d \lambda \end{align*} $$ The reason why the integration interval can be reduced from $[0,\infty]$ to $[0,L_{n}]$ is because if $0 \le \lambda \le L_{n}$, then there is no need to consider up to $\infty$ since it deals with $P \left( Y \land L_{n} > \lambda \right)$ anyway. Meanwhile, since $L_{n}$ will be sent to infinity at the end of the proof, cases where $L_{n} < \lambda $ is even less of a concern.
Part 2.
If $\left\{ (X_{n} , \mathcal{F}_{n}) \right\}$ is a submartingale, for all $\lambda > 0$ $$ \lambda P \left( \max_{n \le N} X_{n} \ge \lambda \right) \le \int_{(\max_{n \le N} X_{n} \ge \lambda)} X_{N} dP $$
If $0 \le \lambda \le L_{n}$, since it’s essentially $Y \ge L_{n}$ and according to the inequality for submartingales [3], $\displaystyle P \left( Y \ge \lambda \right) \le \int_{(Y \ge \lambda)} X_{N} dP$, thus $$ \begin{align*} E \left[ \left( Y \land L_{n} \right)^{p} \right] \le & \int_{0}^{L_{n}} P \left( Y \land L_{n} > \lambda \right) p \lambda^{p-1} d \lambda \\ \le & \int_{0}^{L_{n}} P \left( Y \ge \lambda \right) p \lambda^{p-1} d \lambda \\ \le & \int_{0}^{L_{n}} {{ 1 } \over { \lambda }} \int_{(Y \ge \lambda)} X_{N} dP p \lambda^{p-1} d \lambda \\ \le & p \int_{0}^{L_{n}} \int_{(Y \ge \lambda)} X_{N} \lambda^{p-2} dP d \lambda \end{align*} $$
Part 3.
According to Fubini’s theorem, $$ \begin{align*} E \left[ \left( Y \land L_{n} \right)^{p} \right] \le & p \int_{0}^{L_{n}} \int_{(Y \ge \lambda)} X_{N} \lambda^{p-2} dP d \lambda \\ \le & p \int_{\Omega} \int_{0}^{ Y \land L_{n}} X_{N} \lambda^{p-2} d \lambda dP \\ =& p \int_{\Omega} X_{N} \int_{0}^{ Y \land L_{n}} \lambda^{p-2} d \lambda dP \\ =& p \int_{\Omega} X_{N} {{ 1 } \over { p-1 }} \left( Y \land L_{n} \right)^{p-1} dP \\ =& {{ p } \over { p-1 }} \int_{\Omega} X_{N} \left( Y \land L_{n} \right)^{p-1} dP \\ =& {{ p } \over { p-1 }} E \left[ X_{N} \left( Y \land L_{n} \right)^{p-1} \right] \end{align*} $$
Hölder’s inequality: For $p>1$, if $\displaystyle {{1} \over {p}} + {{1} \over {q}} = 1$ and $f \in \mathcal{L}^{p} (E) $, $g \in \mathcal{L}^{q} (E)$ then $fg \in \mathcal{L}^{1} (E)$ and $| fg |_{1} \le | f |_{p} | g |_{q}$
If we set $\displaystyle q: = {{ p } \over { p-1 }}$, then $q (p-1) = p$ and, according to Hölder’s inequality, $$ \begin{align*} E \left[ \left( Y \land L_{n} \right)^{p} \right] \le & {{ p } \over { p-1 }} E \left[ X_{N} \left( Y \land L_{n} \right)^{p-1} \right] \\ \le & {{ p } \over { p-1 }} \left\| X_{N} \left( Y \land L_{n} \right)^{p-1} \right\|_{1} \\ \le & q \left\| X_{N} \right\|_{p} \left\| \left( Y \land L_{n} \right)^{p-1} \right\|_{q} \\ \le & q \left[ E X_{N}^{p} \right]^{1/p} E \left[ \left( Y \land L_{n} \right)^{(p-1) \cdot q} \right]^{1/q} \\ \le & q \left[ E X_{N}^{p} \right]^{1/p} E \left[ \left( Y \land L_{n} \right)^{p} \right]^{1/q} \end{align*} $$ Dividing both sides by $\displaystyle E \left[ \left( Y \land L_{n} \right)^{p} \right]^{1/q}$, $$ E \left[ \left( Y \land L_{n} \right)^{p} \right]^{1-1/q} \le q \left[ E X_{N}^{p} \right]^{1/p} $$ Taking the $p$ power to both sides, $1 - 1/q = 1 - (p-1)/p = 1/p$, therefore $$ E \left[ \left( Y \land L_{n} \right)^{p} \right] \le q^{p} \left[ E X_{N}^{p} \right] $$ Taking the limit to both sides, $$ \lim_{n \to \infty} E \left[ \left( Y \land L_{n} \right)^{p} \right] \le \lim_{n \to \infty} q^{p} \left[ E X_{N}^{p} \right] $$ When $n \to \infty$, since $Y \land L_{n} \nearrow Y$, according to the Monotone Convergence Theorem, $$ E \lim_{n \to \infty} \left[ \left( Y \land L_{n} \right)^{p} \right] \le q^{p} \left[ E X_{N}^{p} \right] $$ Organizing the equation, $$ E Y^{p} \le q^{p} \left[ E X_{N}^{p} \right] $$
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