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A Submartingale has uniformly Bounded First Moment iff it has uniformly Bounded Positive Moment

Last updated Nov 1, 2022

# Statement

Let (Ω,B,P)(\Omega, \mathcal{B}, \mathbb{P}) be a Probability Space

. Let $(X_{n}){n \geq 1}$ be a discrete-time

Discrete-Time Process

Definition A is a for which the totally ordere is N\mathbb{N}. Other Outlinks ...

11/7/2022

Submartingale

Submartingale

Definition 1 Let (Ω,A,P)(\Omega, \mathcal{A}, \mathbb{P}) be a . Let X:ΩRTX: \Omega \to \mathbb{R}^{T} be an wrt $\mathcal{F}{*} :=...

11/7/2022

wrt Discrete-Time Filtration

Discrete-Time Filtration

Definition Let (Ω,A,P)(\Omega, \mathcal{A}, \mathbb{P}) be a . The Collectio F={Bn:nN}\mathcal{F} = \{\mathcal{B}_{n} : n \in \mathbb{N}\} is a if...

11/7/2022

$\mathcal{F}
{*} := (\mathcal{F}{n} \subset \mathcal{B}){n \in \mathbb{N}}.Then . Then \sup\limits_{n \geq 1} \mathbb{E}(X_{n})^{+} < \infty$ If and Only If

If and Only If

...

11/7/2022

supn1EXn<\sup\limits_{n \geq 1} \mathbb{E}|X_{n}| < \infty.

# Proof

(\Rightarrow): Because $(X_{n}){n \geq 1}$ is a Submartingale

Submartingale

Definition 1 Let (Ω,A,P)(\Omega, \mathcal{A}, \mathbb{P}) be a . Let X:ΩRTX: \Omega \to \mathbb{R}^{T} be an wrt $\mathcal{F}{*} :=...

11/7/2022

and Submartingales have Non-Decreasing Expectation, n1\forall n \geq 1, $$\begin{align*} &\mathbb{E}(X{n}) \geq \mathbb{E}(X_{1})\\ \Rightarrow&\mathbb{E}(X_{n})^{+} - \mathbb{E}(X_{n})^{-} \geq \mathbb{E}(X_{1})\\ \Rightarrow&\mathbb{E}(X_{n})^{+} - \mathbb{E}(X_{1}) \geq \mathbb{E}(X_{n})^{-} \\ \Rightarrow&\infty > \sup\limits_{n \geq 1} \mathbb{E}(X_{n})^{+} - \mathbb{E}(X_{1}) \geq \sup\limits_{n \geq 1} \mathbb{E}(X_{n})^{-} \end{align*}$$

Thus, supn1EXn=supn1(E(Xn)++E(Xn))supn1(E(Xn)+)+supn1(E(Xn))< \begin{align*} \sup\limits_{n \geq 1} \mathbb{E}|X_{n}| &= \sup\limits_{n \geq 1} (\mathbb{E}(X_{n})^{+} + \mathbb{E}(X_{n})^{-})\\ &\leq \sup\limits_{n \geq 1} (\mathbb{E}(X_{n})^{+}) + \sup\limits_{n \geq 1} (\mathbb{E}(X_{n})^{-})\\ &< \infty \end{align*} \checkmark

(\Leftarrow): This follows because >supn1EXnsupn1(E(Xn)++E(Xn))supn1E(Xn)+\infty > \sup\limits_{n \geq 1} \mathbb{E}|X_{n}| \geq \sup\limits_{n \geq 1} (\mathbb{E}(X_{n})^{+} + \mathbb{E}(X_{n})^{-}) \geq \sup\limits_{n \geq 1} \mathbb{E}(X_{n})^{+} \checkmark \blacksquare