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If It Is a Regular Martingale, It Is a Uniformly Integrable Martingale 📂Probability Theory

If It Is a Regular Martingale, It Is a Uniformly Integrable Martingale

Definition

Let there be a given probability space (Ω,F,P)( \Omega , \mathcal{F} , P). When a set of random variables Φ\Phi is given, if for all ε>0\varepsilon>0 there exists a k∈Nk \in \mathbb{N} that satisfies sup⁡X∈Φ∫(∣X∣≥k)∣X∣dP<ε \sup_{ X \in \Phi } \int_{ \left( \left| X \right| \ge k \right) } \left| X \right| dP < \varepsilon , then Φ\Phi is said to be uniformly integrable. If a stochastic process {Xn}\left\{ X_{n} \right\} is uniformly integrable, then a martingale {(Xn,Fn)}\left\{ ( X_{n} , \mathcal{F}_{n} ) \right\} is said to be uniformly integrable.

Theorem

If a martingale {(Xn,Fn)}\left\{ ( X_{n} , \mathcal{F}_{n} ) \right\} is regular, it is uniformly integrable.

Explanation

Understanding why we consider ∣X∣≥k|X| \ge k makes it easier to accept the definition. Asking whether something is uniformly integrable in probability theory is the same as asking whether a stochastic process always possesses a first moment. In other words, it’s about checking E∣Xn∣<∞E |X_{n}| <\infty, and if a natural number k∈Nk \in \mathbb{N} is fixed, E∣Xn∣=∫(∣Xn∣≥k)∣Xn∣dP+∫(∣Xn∣<k)∣Xn∣dP E |X_{n}| = \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| X_{n} \right| dP + \int_{ \left( \left| X_{n} \right| < k \right) } \left| X_{n} \right| dP then ∫(∣Xn∣<k)∣Xn∣dP<∫(∣Xn∣<k)kdP<∫ΩkdP<kP(Ω)<∞ \begin{align*} \int_{ \left( \left| X_{n} \right| < k \right) } \left| X_{n} \right| dP <& \int_{ \left( \left| X_{n} \right| < k \right) } k dP \\ <& \int_{ \Omega } k dP \\ <& k P ( \Omega ) \\ <& \infty \end{align*} and there’s no need to think about ∫(∣Xn∣<k)∣Xn∣dP\displaystyle \int_{ \left( \left| X_{n} \right| < k \right) } \left| X_{n} \right| dP, only needing to check if ∫(∣Xn∣≥k)∣Xn∣dP\displaystyle \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| X_{n} \right| dP is finite.

Proof

It suffices to show that for all ε>0\varepsilon > 0 there exists a k∈Nk \in \mathbb{N} that satisfies: sup⁡n∈N∫(∣Xn∣≥k)∣Xn∣dP<ε \sup_{ n \in \mathbb{N} } \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| X_{n} \right| dP < \varepsilon

Part 1. ∫(∣Xn∣≥k)∣Xn∣dP≤∫(∣η∣>M)∣η∣dP+ME∣Xn∣k\displaystyle \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| X_{n} \right| dP \le \int_{ \left( \left| \eta \right| > M \right) } | \eta| dP + M {{ E |X_{n} | } \over {k}}

Properties of conditional expectation:

  • [3]: If XX is F\mathcal{F}-measurable, then E(X∣F)=X a.s.E(X|\mathcal{F}) =X \text{ a.s.}
  • [10]: ∣E(X∣G)∣≤E(∣X∣∣G) a.s.\left| E( X | \mathcal{G} ) \right| \le E ( | X | | \mathcal{G} ) \text{ a.s.}
  • [11]: For all sigma fields G\mathcal{G}, E[E(X∣G)]=E(X)E \left[ E ( X | \mathcal{G} ) \right] = E(X)

Since {(Xn,Fn)}\left\{ ( X_{n} , \mathcal{F}_{n} ) \right\} is a regular martingale, there exists an integrable random variable η\eta that satisfies Xn=E(η∣Fn)X_{n} = E \left( \eta | \mathcal{F}_{n} \right). According to the properties of conditional expectation [3] and [10], ∫(∣Xn∣≥k)∣Xn∣dP=∫(∣Xn∣≥k)∣E(η∣Fn)∣dP≤∫(∣Xn∣≥k)E(∣η∣∣Fn)dP=∫(∣Xn∣≥k)∣η∣dP \begin{align*} \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| X_{n} \right| dP =& \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| E \left( \eta | \mathcal{F}_{n} \right) \right| dP \\ \le & \int_{ \left( \left| X_{n} \right| \ge k \right) } E \left( | \eta| | \mathcal{F}_{n} \right) dP \\ =& \int_{ \left( \left| X_{n} \right| \ge k \right) } | \eta| dP \end{align*} Now, if we split (∣Xn∣≥k)\left( \left| X_{n} \right| \ge k \right) into two parts, (∣η∣>M)\left( \left| \eta \right| > M \right) and (∣η∣≤M)\left( \left| \eta \right| \le M \right), ∫(∣Xn∣≥k)∣Xn∣dP=∫(∣Xn∣≥k)∩(∣η∣>M)∣η∣dP+∫(∣Xn∣≥k)∩(∣η∣≤M)∣η∣dP≤∫(∣η∣>M)∣η∣dP+∫(∣Xn∣≥k)MdP≤∫(∣η∣>M)∣η∣dP+MP(∣Xn∣≥k) \begin{align*} \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| X_{n} \right| dP =& \int_{ \left( \left| X_{n} \right| \ge k \right) \cap \left( \left| \eta \right| > M \right) } | \eta| dP + \int_{ \left( \left| X_{n} \right| \ge k \right) \cap \left( \left| \eta \right| \le M \right)} | \eta| dP \\ \le & \int_{ \left( \left| \eta \right| > M \right) } | \eta| dP + \int_{ \left( \left| X_{n} \right| \ge k \right) } M dP \\ \le & \int_{ \left( \left| \eta \right| > M \right) } | \eta| dP + M P \left( \left| X_{n} \right| \ge k \right) \end{align*}

Markov’s inequality: P(u(X)≥c)≤E(u(X))c P(u(X) \ge c) \le {E(u(X)) \over c}

By Markov’s inequality, ∫(∣Xn∣≥k)∣Xn∣dP≤∫(∣η∣>M)∣η∣dP+MP(∣Xn∣≥k)≤∫(∣η∣>M)∣η∣dP+ME∣Xn∣k \begin{align*} \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| X_{n} \right| dP \le & \int_{ \left( \left| \eta \right| > M \right) } | \eta| dP + M P \left( \left| X_{n} \right| \ge k \right) \\ \le & \int_{ \left( \left| \eta \right| > M \right) } | \eta| dP + M {{ E |X_{n} | } \over {k}} \end{align*}

Part 2. sup⁡n∈N∫(∣Xn∣≥k)∣Xn∣dP≤∫(∣η∣>M)∣η∣dP+MkE∣η∣ a.s.\displaystyle \sup_{n \in \mathbb{N}} \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| X_{n} \right| dP \le \int_{ \left( \left| \eta \right| > M \right) } | \eta| dP + {{M} \over {k}} E | \eta | \text{ a.s.}

Since Xn=E(η∣Fn)X_{n} = E \left( \eta | \mathcal{F}_{n} \right), according to the properties of conditional expectation [10] and [11], E∣Xn∣=E∣E(η∣Fn)∣≤EE(∣η∣∣Fn)≤E∣η∣ \begin{align*} E |X_{n} | =& E \left| E \left( \eta | \mathcal{F}_{n} \right) \right| \\ \le & E E \left( \left| \eta \right| | \mathcal{F}_{n} \right) \\ \le & E | \eta | \end{align*} thus, continuing from Part 1, the following inequality is obtained. ∫(∣Xn∣≥k)∣Xn∣dP≤∫(∣η∣>M)∣η∣dP+MkE∣η∣ \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| X_{n} \right| dP \le \int_{ \left( \left| \eta \right| > M \right) } | \eta| dP + {{M} \over {k}} E | \eta | This holds for all n∈Nn \in \mathbb{N} and M>0M>0, so sup⁡n∈N∫(∣Xn∣≥k)∣Xn∣dP≤∫(∣η∣>M)∣η∣dP+MkE∣η∣ \sup_{n \in \mathbb{N}} \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| X_{n} \right| dP \le \int_{ \left( \left| \eta \right| > M \right) } | \eta| dP + {{M} \over {k}} E | \eta |

Part 3. ∫(∣η∣>M)∣η∣dP<ε2\displaystyle \int_{ \left( \left| \eta \right| > M \right) } | \eta| dP < {{\varepsilon} \over {2}}

Dominated convergence theorem: For a measurable set E∈ME \in \mathcal{M} and g∈L1(E)g \in \mathcal{L}^{1} (E), let a sequence of measurable functions {fn}\left\{ f_{n} \right\} almost everywhere satisfies ∣fn∣≤g|f_{n}| \le g in EE. If almost everywhere in EE satisfies f=lim⁡n→∞fn\displaystyle f = \lim_{n \to \infty} f_{n}, then f∈L1(E)f \in \mathcal{L}^{1}(E) and lim⁡n→∞∫Efn(x)dm=∫Efdm \lim_{ n \to \infty} \int_{E} f_{n} (x) dm = \int_{E} f dm

Since ∣η∣1(∣η∣>M)≤∣η∣|\eta| \mathbb{1}_{\left( \left| \eta \right| > M \right) } \le | \eta|, by the dominated convergence theorem, lim⁡M→∞∫(∣η∣>M)∣η∣dP=lim⁡M→∞∫Ω∣η∣1(∣η∣>M)dP=∫Ωlim⁡M→∞∣η∣1(∣η∣>M)dP=0 \begin{align*} \lim_{M \to \infty} \int_{ \left( \left| \eta \right| > M \right) } | \eta | dP =& \lim_{M \to \infty} \int_{ \Omega } | \eta | \mathbb{1}_{\left( \left| \eta \right| > M \right) } dP \\ =& \int_{ \Omega } \lim_{M \to \infty} | \eta | \mathbb{1}_{\left( \left| \eta \right| > M \right) } dP \\ =& 0 \end{align*} In other words, for all ε2>0\displaystyle {{\varepsilon} \over {2}} > 0, there exists a MM that satisfies ∫(∣η∣>M)∣η∣dP<ε2 \int_{ \left( \left| \eta \right| > M \right) } | \eta| dP < {{\varepsilon} \over {2}}

Part 4. MkE∣η∣<ε2\displaystyle {{M} \over {k}} E | \eta | < {{\varepsilon} \over {2}}

Following Part 3, there exists a MM that satisfies sup⁡n∈N∫(∣Xn∣≥k)∣Xn∣dP≤ε2+MkE∣η∣ \sup_{n \in \mathbb{N}} \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| X_{n} \right| dP \le {{\varepsilon} \over {2}} + {{M} \over {k}} E | \eta | This MM for all ε2>0\displaystyle {{\varepsilon} \over {2}} > 0 satisfies MkE∣η∣<ε2 {{M} \over {k}} E | \eta | < {{\varepsilon} \over {2}} Since there exists a k∈Nk \in \mathbb{N} that satisfies the following for all ε>0\varepsilon >0, the regular martingale {(Xn,Fn)}\left\{ ( X_{n} , \mathcal{F}_{n} ) \right\} is uniformly integrable. sup⁡n∈N∫(∣Xn∣≥k)∣Xn∣dP≤ε2+ε2=ε \sup_{n \in \mathbb{N}} \int_{ \left( \left| X_{n} \right| \ge k \right) } \left| X_{n} \right| dP \le {{\varepsilon} \over {2}} + {{\varepsilon} \over {2}} = \varepsilon

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