Probability with Martingales by David Williams

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Uniform Integrability

Assume the conditions are true. Let

\[ \begin{equation} B_K = \sup_{X \in \mathcal C} \{ \Pr( \abs{X} > K ) \} \end{equation} \]

Now for any \(X\in \mathcal C\), Markov's inequality implies

\[ \begin{equation} \Pr( \abs{X} > K ) \leq \frac{ \E \abs{X}} K \leq \frac A K \end{equation} \]

Taking the supremum over all \(X\in \mathcal C\), this shows \(B_K \leq A/K\). Therefore \(B_K\to 0\) as \(K\to \infty\). For any \(\epsilon>0\), there is a \(\delta\) satisfying condition (ii). Take \(K\) large enough that \(B_K < \delta\). From the definition of \(B_K\), all of the sets \(F=\{\abs{X} > K \}\) satisfy \(\Pr( F )<\delta\), and hence condition (ii) implies \(\E(\abs{X}; \abs{X}>K) < \epsilon\) for each \(X\in \mathcal C\). This verifies that \(\mathcal C\) is UI.

Conversely assume \(\mathcal C\) is UI. For any \(K>0\) and \(F \in \mathcal F\), note

\[ \begin{equation} I_F \leq I_F ( I_{\abs{X}>K} + I_{\abs{X}\leq K}) \leq I_{\abs{X}>K} + I_{\abs{X}\leq K} I_F \end{equation} \]

Multiplying both sides by \(\abs{X}\) and taking expectatins gives

\[ \begin{equation} \begin{split} \E(\abs{X};F) &\leq \E(\abs{X}; \abs{X}>K ) + \E(\abs{X}; F \cap \{ \abs{X}\leq K \})\\ &\leq \E(\abs{X}; \abs{X}>K ) + K \Pr( F ) \end{split} \label{eq:UI tricky bound} \end{equation} \]

Hence, given \(\epsilon>0\) choose \(K\) such that \(\E(\abs{X}; \abs{X}>K ) < \frac \epsilon 2\) for all \(X\). Then choose \(\delta = \frac{\epsilon}{2K}\). Equation \(\eqref{eq:UI tricky bound}\) shows that for this choice of \(\delta\), condition (ii) is satisfied. Furthermore with \(F=\Omega\) and \(K\) chosen so \(\E(\abs{X}; \abs{X}>K ) \leq 1\), \(\eqref{eq:UI tricky bound}\) gives a uniform bound of \(1+K\) for \(\E(\abs{X})\), so condition (i) is satisfied.

We'll verify the conditions in 13.1. First note that for \(Z_0 \in \mathcal C + \mathcal D\) with \(Z_0=X_0+Y_0\)

\[ \begin{equation} \E \abs{Z_0} \leq \E \abs{X_0} + \E \abs{Y_0} \leq \sup_{X\in \mathcal C} \E \abs{X} + \sup_{Y\in \mathcal D} \E \abs{Y} \end{equation} \]

The bound on the right hand side is independent of \(Z\), so its a bound for \(\sup_{Z\in \mathcal C+\mathcal D} \E \abs{Z}\), and condition (i) is satisfied.

For \(\epsilon>0\), choose \(\delta\) such that for any \(X\in \mathcal C\) and \(Y\in \mathcal D\) and \(F\in \mathcal F\) with \(\Pr( F ) < \delta\), then \(\E( \abs{X};F) < \frac \epsilon 2\) and \(\E(\abs{Y};F) < \frac \epsilon 2\) (Take \(\delta\) to be the minimum of the \(\delta\) needed for \(\mathcal C\) or \(\mathcal D\) alone). Then for \(Z=X+Y\)

\[ \begin{equation} \E( \abs{Z}; F) \leq \E(\abs{X};F) + \E(\abs{Y};F) < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon \end{equation} \]

Thus condition (ii) is satisfied for \(\mathcal C + \mathcal D\)

Let \(G\in \mathcal G\) and \(Y=\E( X\mid \mathcal G)\). Then

\[ \begin{equation} \begin{split} \E(\abs{Y}; G) &= \E(\Abs{\E(X\mid \mathcal G)}I_G ) = \E(\Abs{\E(X I_G \mid \mathcal G)}) \\ &\leq \E(\E(\abs{X I_G} \mid \mathcal G)) = \E(\abs{X} I_G) = \E(\abs{X};G) \end{split} \end{equation} \]

The inequality in the middle is the conditional Jensen's inequality for the convex function \(\abs{X}\). When \(G=\Omega\), this gives a bound \(\E(\abs{Y}) \leq \E(\abs{X})\). So \(A=\sup_{X\in \mathcal C} \E(\abs{X})\) is a bound for \(\sup_{Y\in \mathcal D} \E(\abs{Y})\), and condition (i) of 13.1 is satisfied. Given \(\epsilon>0\), choose \(\delta>0\) to satisfy condition (ii) of 13.1 for \(\mathcal C\). Then for any \(G\in \mathcal G \subset \mathcal F\) with \(\Pr(G)<\delta\) have for any \(Y\in \mathcal D\)

\[ \begin{equation} \E(\abs{Y};G) \leq \E(\abs{X};G) <\epsilon \end{equation} \]

Thus \(\mathcal D\) also satisfies condition (ii) of 13.1. Satisfying both conditions, it is UI.

Contact

For comments or corrections please contact Ryan McCorvie at ryan@martingale.group