Ryan McCorvie
home
academic
Williams Exercises
Extra gumption
Chapter 1
Chapter 4
Chapter 5
Chapter 7
Chapter 9
Chapter 10
Chapter 12
Chapter 13
Chapter 14
Chapter 16
Chapter 18
Appendix 13


Probability with Martingales by David Williams

\[ \newcommand{\Q}{\mathbb Q} \newcommand{\R}{\mathbb R} \newcommand{\C}{\mathbb C} \newcommand{\Z}{\mathbb Z} \newcommand{\N}{\mathbb N} \newcommand{\abs}[1]{\lvert #1 \rvert} \newcommand{\norm}[1]{\lVert #1 \rVert} \newcommand{\abs}[1]{\lvert #1 \rvert} \newcommand{\Norm}[1]{\left \lVert #1 \right \rVert} \newcommand{\Abs}[1]{\left \lvert #1 \right \rvert} \newcommand{\pind}{\bot \!\!\! \bot} \newcommand{\probto}{\buildrel P\over \to} \newcommand{\vect}[1]{\boldsymbol #1} \DeclareMathOperator{\EE}{\mathbb E} \DeclareMathOperator{\PP}{\mathbb P} \DeclareMathOperator{\E}{E} \DeclareMathOperator{\dnorm}{\mathcal N} \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Leb}{Leb} \DeclareMathOperator{\Bin}{Bin} \newcommand{\wto}{\buildrel w\over \to} \]

Conditional Expectation

Problem 9.1

Prove that if \(\mathcal G\) is a sub \(\sigma\)-algebra of \(\mathcal F\) and if \(X\in \mathcal L^1(\Omega,\mathcal F, \Pr)\) and if \(Y\in \mathcal L^1(\Omega, \mathcal G, \Pr)\) and

\[ \begin{equation} \E(X;G) = E(Y;G) \label{eq:gcond} \end{equation} \]

for every \(G\) in a \(\pi\)-system which contains \(\Omega\) and generates \(\mathcal G\), then \(\eqref{eq:gcond}\) holds for every \(G\in \mathcal G\).

We make a standard monotone class argument. We will show that the collection of sets satisfying \(\eqref{eq:gcond}\) is a \(d\)-system, so by Dynkin's lemma and the hypothesis the \(\pi\)-system \(\mathcal G\) has property implies \(\sigma( \mathcal G)\) has the property.

  • By assumption \(\Omega \in \mathcal G\)

  • Suppose \(A,B\in \mathcal G\). Then \(L = A\cap B \in \mathcal G\) because \(\mathcal G\) is a \(\pi\)-system. Now since \(B = (B\setminus A) \coprod L\) is the union of these disjoint sets. Therefore \(I_{B} = I_{B \setminus A} + I_L\). Thus by the linearity of expectations

\[ \begin{equation} \E(X;B\setminus A) = \E X I_{B\setminus A} = \E X(I_B - I_L) = \E(X;B) - \E(X;L) \end{equation} \]

Since the same is true for \(Y\) we get

\[ \begin{equation} \E(X;B\setminus A) = \E(X;B) - \E(X;L) = \E(Y;B) - \E(Y;L) = \E(Y;B\setminus A) \end{equation} \]

and the class of sets satisfying \(\eqref{eq:gcond}\) is closed under taking differences.

  • Suppose \(A_n \uparrow A\). Then \(X I_{A_n}\) is dominated by \(\abs{ X I_{A}}\) and also \(X I_{A_n} \to X I_{A}\). Thus by dominated convergence

\[ \begin{equation} \E(X;A_n) \to \E(X;A) \end{equation} \]

Mutatis mutandis, the same is true for \(Y\) hence

\[ \begin{equation} \E(X;A) = \lim_{n\to \infty} \E(X;A_n) = \lim_{n\to \infty} \E(Y;A_n)= \E(Y;A) \end{equation} \]

This shows that \(A\) is in the class of sets which satisfies \(\eqref{eq:gcond}\)

Thus, the class of sets which satisfies \(\eqref{eq:gcond}\) is a \(d\)-system as well as a \(\pi\)-system, so it includes \(\sigma(\mathcal G)\).

I guess another approach is to note that \(G\to E[X;G]\) is a (signed) measure, so we can use something like theorem 1.6 on the uniqueness of measures defined on \(\pi\)-systems?

Problem 9.2

Suppose that \(X,Y\in \mathcal L^1(\Omega, \mathcal F, \Pr)\) and that

\[ \begin{equation} \E(X|Y) = Y \text{ a.s.}\qquad \E(Y|X) = X \text{ a.s.} \label{eq:cond exp hypothesis} \end{equation} \]

Prove that \(\Pr(X=Y) = 1\)

Recall the defining property of of conditional expectation. First, \(\E[Y\mid X]\) is \(\sigma(X)\)measurable, and second and for \(G\in \sigma(X)\),

\[ \begin{equation} \E[ \E[ Y\mid X]; G] = \E[ Y; G ] \label{eq:conditional expectation property} \end{equation} \]

Let \(A\in \sigma(X)\). From \(\eqref{eq:cond exp hypothesis}\) and \(\eqref{eq:conditional expectation property}\)

\[ \begin{equation} \E[X; A] = \E[ E[Y\mid X] ; A ] = \E[Y; A] \end{equation} \]

Also, mutatis mutandis, \(\E[Y;B] = \E[X;B]\) for any \(B\in \sigma(Y)\). For \(c\in \R\), taking \(A= \{X\leq c\}\) and \(B=\{Y\leq c\}\) gives

\[ \begin{equation} \begin{split} &\E( X; X\leq c) = \E(Y; X\leq c)\Rightarrow \\ &\qquad \qquad \E(X-Y; X\leq c, Y>c ) + \E(X-Y; X\leq c, Y\leq c) = 0\\ &\E( X; Y\leq c) = \E(Y; Y\leq c)\Rightarrow \\ &\qquad \qquad \E(X-Y; X>c , Y\leq c) + \E(X-Y; X\leq c, Y\leq c) = 0 \end{split} \end{equation} \]

Therefore

\[ \begin{equation} \E(X-Y; X\leq c, Y>c ) = \E(X-Y; X>c , Y\leq c) \end{equation} \]

since both equal \(\E(Y-X; X\leq c, Y\leq c)\). Since the left hand side is non-positive, and the right hand side is non-negative, both sides must be zero.

I think this clinches it. Note \( \{X>Y\} \subset \bigcup_{c\in \Q} \{X>c, Y\leq c \}\). However

\[ \begin{equation} \E(X-Y; X>Y) \leq \sum_{c\in \Q} \E(X-Y; X>c, Y\leq c) = 0 \end{equation} \]

However if \(\Pr( X>Y)>0\) then \(\E(X-Y;X>Y) > 0\). Thus \(X\leq Y\) almost surely. By symmetry, \(Y\leq X\) almost surely, and the intersection of these almost sure events is almost sure. So \(X=Y\) almost surely.

Contact

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