Doob's Maximal Inequality/Discrete Time/Proof 2

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Theorem

Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \mathop \ge 0}, \Pr}$ be a filtered probability space.

Let $\sequence {X_n}_{n \mathop \ge 0}$ be a non-negative $\sequence {\FF_n}_{n \mathop \ge 0}$-submartingale.

Let:

$\ds X_n^\ast = \max_{0 \mathop \le k \mathop \le n} X_k$

where $\max$ is the pointwise maximum.

Let $\lambda > 0$.


Then:

$\lambda \map \Pr {X_n^\ast \ge \lambda} \le \expect {X_n}$


Proof

Let $\lambda > 0 $ and $n \ge 0$.

Let:

$E := \set {X^\ast _n \ge \lambda}$.

That is, $E$ is a disjoint union:

$(1):\quad \ds E = \bigsqcup _{0 \mathop \le k \mathop \le n} E_k$

where:

$\ds E_k := \set {X_k \ge \lambda} \cap \bigcap _{0 \mathop \le j \mathop \le k-1} \set {X_j < \lambda}$

By construction, we have:

$\forall k \in \set {0, \ldots, n} : E_k \in \FF_k$

In particular:

\(\ds \expect {X_n \chi_{E_k} \mid \FF_k }\) \(=\) \(\ds \expect {X_n \mid \FF_k } \chi_{E_k}\) Rule for Extracting Random Variable from Conditional Expectation of Product
\(\text {(2)}: \quad\) \(\ds \) \(\ge\) \(\ds X_k \chi_{E_k}\) Definition of Submartingale

where $\chi_A$ denotes the characteristic function of $A \subseteq \Omega$.

Therefore:

\(\ds \expect {X_n}\) \(\ge\) \(\ds \expect {X_n \chi_E}\) as $X_n \ge 0$
\(\ds \) \(=\) \(\ds \sum_{0 \mathop \le k \mathop \le n} \expect {X_n \chi_{E_k} }\) by $(1)$
\(\ds \) \(=\) \(\ds \sum_{0 \mathop \le k \mathop \le n} \expect { \expect {X_n \chi_{E_k} \mid \FF_k } }\) Tower Property of Conditional Expectation
\(\ds \) \(\ge\) \(\ds \sum_{0 \mathop \le k \mathop \le n} \expect { X_k \chi_{E_k} }\) by $(2)$
\(\ds \) \(\ge\) \(\ds \sum_{0 \mathop \le k \mathop \le n} \expect { \lambda \chi_{E_k} }\) as $X_k \ge \lambda$ on $E_k$
\(\ds \) \(=\) \(\ds \lambda \sum_{0 \mathop \le k \mathop \le n} \expect {\chi_{E_k} }\) Expectation is Linear
\(\ds \) \(=\) \(\ds \lambda \sum_{0 \mathop \le k \mathop \le n} \map \Pr {E_k}\) Integral of Characteristic Function
\(\ds \) \(=\) \(\ds \lambda \map \Pr E\) by $(1)$

$\blacksquare$