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I'm currently try to understand the properties of the conditional expectation. Consider a probability space $(\Omega, \mathcal{A}, \mathbb{P})$ and an iid sequence of random variables $(\varepsilon_j)_{j \in \mathbb{Z}}$ on this probability space and let $\xi_j = (\dots,\varepsilon_{j-1},\varepsilon_j)$. Further let $g$ be a measurable function such that $X_j = g(\xi_j)$ describes a well defined sequence of random variables with mean zero. Next let $F = (\mathcal{F}_k)_{k \in \mathbb{Z}} = (\sigma(\xi_k))_{k \in \mathbb{Z}}$ be the filtration of sigma-algebras generated by $\xi_k$.

I want to prove, that the following representation is valid:

$X_k = \sum_{l \in \mathbb{Z}} \mathbb{E}(X_k \mid F_l) - \mathbb{E}(X_k \mid F_{l-1})$.

So far, I was able to obtain so far is that $\mathbb{E}(X_k \mid F_l) - \mathbb{E}(X_k \mid F_{l-1}) = 0$ for all $l > k$ and hence we can write

$\sum_{l \in \mathbb{Z}} \mathbb{E}(X_k \mid F_l) - \mathbb{E}(X_k \mid F_{l-1}) \\= \sum_{l=-\infty}^{k} \mathbb{E}(X_k \mid F_l) - \mathbb{E}(X_k \mid F_{l-1}) \\= \lim_{n\to \infty} \sum_{l=-n}^{k} \mathbb{E}(X_k \mid F_l) - \mathbb{E}(X_k \mid F_{l-1}) \\ = \lim_{n\to \infty} (\mathbb{E}(X_k \mid F_k) - \mathbb{E}(X_k \mid F_{n-1}))$.

We know that $\mathbb{E}(X_k \mid F_k) = X_k$. Hence it only remains to treat the remainder term. I was wondering, if I could use Levy's downwards theorem to obtain

$\lim_{n\to \infty} \mathbb{E}(X_k \mid F_{n-1}) = \mathbb{E}(X_k \mid F_{-\infty})$, where $F_{-\infty}$ is the intersection of all $F_k$. Then – at least I assume – we have $F_{-\infty} = \{\emptyset, \Omega\}$ the trivial sigma algebra und thus $\mathbb{E}(X_k \mid F_{-\infty}) = \mathbb{E}(X_k) = 0$ and thus the proof would be complete.

Would someone share his/her thoughts on this?

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1 Answer 1

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$F_{-\infty}$ need not be $\{\emptyset, \Omega\}$ but any set in this $\sigma-$ field has probability $0$ or $1$ and this is good enough for your conclusion. Apply Kolmogorobv's $0-1$ law to the sequence $\epsilon_{-1},\epsilon_{-2},\epsilon_{-3},\cdots$ to see that any set in $F_{-\infty}$ has probability $0$ or $1$.

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  • $\begingroup$ But I still need Levy's downwards theorem for the convergence of the sequence $\lim_{n\to \infty} \mathbb{E}(X_k \mid F_{n-1}) = \mathbb{E}(X_k \mid F_{-\infty})$, don't I? $\endgroup$ Apr 23, 2022 at 11:45

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