Proof of the Monotone Convergence Theorem for Conditional Cases
Theorem
Let’s assume that a probability space $( \Omega , \mathcal{F} , P)$ is given.
Considering the sequence of random variables $\left\{ X_{n} \right\}_{n \in \mathbb{N}}$ and $X \in \mathcal{L}^{1} (\Omega)$, we have $$ X_{1} \le X_{2} \le \cdots \le X \\ X_{n} \to X \text{ a.s.} $$ then $$ \lim_{n \to \infty} E( X_{n} | \mathcal{G} ) = E( \lim_{n \to \infty} X_{n} | \mathcal{G} ) \text{ a.s.} $$
- $\text{a.s.}$ means almost surely.
Explanation
The conditional monotone convergence theorem simply states that the monotone convergence theorem applies to conditional expectations just as well. Its role in probability theory is the same as that of MCT.
Proof
Strategy: Use the monotone convergence theorem to sandwich $\displaystyle \int$ around $\displaystyle \lim_{n \to \infty}$, and by tweaking the definition of expectation to add and remove $E$, make the integrands the same.
Part 1. $X_{1} \ge 0$
According to the monotone convergence theorem, for all $A \in \mathcal{G}$ $$ \begin{align*} \int_{A} \lim_{n \to \infty} E( X_{n} | \mathcal{G} ) dP \color{red}{=}& \lim_{n \to \infty} \int_{A} E( X_{n} | \mathcal{G} ) dP \\ =& \lim_{n \to \infty} \int_{A} X_{n} dP \\ \color{red}{=}& \int_{A} \lim_{n \to \infty} X_{n} dP \\ =& \int_{A} E( \lim_{n \to \infty} X_{n} | \mathcal{G} ) d P \end{align*} $$ So, $ $\displaystyle \forall A \in \mathcal{F}, \int_{A} f dm = 0 \iff f = 0 \text{ a.e.}$ 이므로 $$ \lim_{n \to \infty} E( X_{n} | \mathcal{G} ) = E( \lim_{n \to \infty} X_{n} | \mathcal{G} ) \text{ a.s.} $$
Part 2. $X_{1} < 0$
$Y_{n} := X_{n} - X_{1}$ 이 확률 변수 $Y = X - X_{1}$ 에 대해 $Y_{n} \nearrow Y$ 라고 하면 $Y_{1} \ge 0$, therefore, according to Part 1. $$ \lim_{n \to \infty} E( Y_{n} | \mathcal{G} ) = E( \lim_{n \to \infty} Y_{n} | \mathcal{G} ) \text{ a.s.} $$ Then, by the linearity of conditional expectation, we obtain the following. $$ \begin{align*} \lim_{n \to \infty} E( X_{n} | \mathcal{G} ) =& \lim_{n \to \infty} E( Y_{n} + X_{1} | \mathcal{G} ) \\ =& \lim_{n \to \infty} E( Y_{n} | \mathcal{G} ) + E( X_{1} | \mathcal{G} ) \\ =& E( X - X_{1} | \mathcal{G} ) + E( X_{1} | \mathcal{G} ) \\ =& E( X - X_{1} + X_{1} | \mathcal{G} ) \\ =& E( \lim_{n \to \infty} X_{n} | \mathcal{G} ) \text{ a.s.} \end{align*} $$
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