## Sequence of Decompositions (22/365)

I am now at the end of Chapter 1 in with four sections to go on random walks, martingales, and markov chains. I am going to skip the section on random walks for now and come back to it later. I haven’t seen martingales before so I’ll start with that. Since martingales makes use of expectations with respect to decompositions heavily I want to get a little more comfortable with it.

Suppose that $\mathcal{D}_1 \le \mathcal{D}_2$ are two decompositions of the sample space where $\mathcal{D}_2$ is finer than $\mathcal{D}_1$. Finer means that $\forall D \in \mathcal{D}_2, \exists E \in \mathcal{D}_1 \text{ s.t. } D \subseteq E$.

Let $\theta$ be a random variable. First, recall the expectation of a random variable with respect to a decomposition $\mathcal{D}_1$. $\displaystyle E(\theta | \mathcal{D}_1) = \sum_j x_j P(A_j | \mathcal{D}_1) \\ \text{where } A_j = \theta^{-1}(x_j)$

Note the special case when $\mathcal{A} \le \mathcal{D}_1$ (i.e., when $\theta$ is $\mathcal{D}_1$-measurable). $\displaystyle E(\theta | D \in \mathcal{D}_1) = \sum_j x_j P(A_j | D) \\ = \sum_j x_j \frac{P(A_j \cap D)}{P(D)} \\ = \sum_j x_j I(D \subseteq A_j) \\ = x_j \\ \text{hence } E(\theta | \mathcal{D}_1) = \theta$

Next, recall the generalized total probability formula $\displaystyle E \left[ E(\theta | \mathcal{D}) \right] = E \theta$

Suppose we took a conditional expection instead $\displaystyle E \left[ E(\theta | \mathcal{D}) | \mathcal{D}' \right] \\ \text{let } D' \in \mathcal{D}' \\ E \left[ E(\theta | \mathcal{D}) | D' \right] \\ = \sum_i \left[ \sum_j x_j P(A_j | D_i) \right] P(D_i | D')$

This gets simplified if $\mathcal{D}$ is a finer decomposition than $\mathcal{D}'$ because $D'$ is now decomposed by $\mathcal{D}$ $\displaystyle = \sum_i \sum_j x_j P(A_j | D_i) P(D_i | D') \\ = \sum_j x_j \sum_i P(A_j | D_i) P(D_i | D') \\ = \sum_j x_j P(A_j | D')$

Therefore if $\mathcal{D}_1 \le \mathcal{D}_2$ $\displaystyle E \left[ E(\theta | \mathcal{D}_2) | \mathcal{D}_1 \right] = E(\theta | \mathcal{D}_1)$

And in general if $\mathcal{D}_1 \le \mathcal{D}_2 \le \dots \le \mathcal{D}_n$ $\displaystyle E \left[ E(\theta | \mathcal{D}_j) | D_1 \right] = E(\theta | \mathcal{D}_1) \text{ for } j > 1$
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