Doob's Maximal Inequality/Discrete Time/Proof 1

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:

$\map T \omega = \inf \set {k \ge 0 : \map {X_k} \omega \ge \lambda} \wedge n$

for each $\omega \in \Omega$.

From:

Least Time at which Discrete-Time Adapted Stochastic Process equals or exceeds Real Number is Stopping Time

Constant Function is Stopping Time

Pointwise Minimum of Stopping Times is Stopping Time

we have:

$T$ is a stopping time with respect to $\sequence {\FF_n}_{n \mathop \ge 0}$.

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Further:

$T \le n$

Note that $X_T \ge \lambda$ if and only if:

$X_k \ge \lambda$ for some $0 \le k \le n$.

This is equivalent to:

$\ds \sup_{0 \mathop \le k \mathop \le n} X_k = X_n^\ast \ge \lambda$

Let $\FF_T$ be the stopped $\sigma$-algebra associated with $T$.

By Doob's Optional Stopping Theorem for Stopped Sigma-Algebra of Bounded Stopping Time: Discrete Time: Submartingale, we have:

$\expect {X_n \mid \FF_T} \ge X_T$ almost surely.

From Adapted Stochastic Process at Stopping Time is Measurable with respect to Stopped Sigma-Algebra:

$X_T$ is $\FF_T$-measurable.

From Conditional Expectation of Measurable Random Variable, we have:

$X_T = \expect {X_T \mid \FF_T}$ almost surely.

So, by Conditional Expectation is Linear we have:

$\expect {X_n - X_T \mid \FF_T} \ge 0$ almost surely.

So from Condition for Conditional Expectation to be Almost Surely Non-Negative, we have:

$\expect {X_n \cdot \chi_A} \ge \expect {X_T \cdot \chi_A}$

for all $A \in \FF_T$.

We can now calculate:

\(\ds \map \Pr {X_n^\ast \ge \lambda}\)

\(=\)

\(\ds \map \Pr {X_T \ge \lambda}\)

\(\ds \)

\(=\)

\(\ds \expect {\chi_{\set {X_T \mathop \ge \lambda} } }\)

Integral of Characteristic Function

\(\ds \)

\(\le\)

\(\ds \expect {\frac {X_T} \lambda \chi_{\set {X_T \mathop \ge \lambda} } }\)

since if $X_T \ge \lambda$, we have $\dfrac {X_T} \lambda \ge 1$ and can apply Expectation is Monotone

\(\ds \)

\(=\)

\(\ds \frac 1 \lambda \expect {X_T \chi_{\set {X_T \mathop \ge \lambda} } }\)

Expectation is Linear

\(\ds \)

\(\le\)

\(\ds \frac 1 \lambda \expect {X_n \chi_{\set {X_T \mathop \ge \lambda} } }\)

since $X_T$ is $\FF_T$-measurable and so $\set {X_T \ge \lambda} \in \FF_T$

\(\ds \)

\(\le\)

\(\ds \frac 1 \lambda \expect {X_n}\)

Expectation is Monotone

Multiplying through $\lambda > 0$ allows us to conclude.

$\blacksquare$