From the Book of Jacod-Protter (Probability Essentials).
Theorem: Let $(X_j)$ be an independent and identically distributed sequence with $E\{|X_j|\} < \infty$. Then
\begin{equation*}
\lim_{n \to \infty} \frac{X_1+ \cdots + X_n}{n} = E\{X_1\} \text{ a.s.}
\end{equation*}Some part of the proof: Let $S_n = X_1 + \cdots + X_n$, and $\mathcal{F}_{-n} = σ(S_n, S_{n+1}, S_{n+2}, . . .)$. Then $\mathcal{F}_{-n} \subseteq \mathcal{F}_{-m}$ if $n ≥ m$, and the process
$$M_{-n} = E\{X_1|\mathcal{F}_{−n}\}$$
is a backwards martingle. Note that $E\{M_n\} = E\{X_1\}$, each $n$. Also note that by symmetry for $1 ≤ j ≤ n:$
$$E\{X_1|\mathcal{F}_{−n}\} = E\{X_j|\mathcal{F}_{−n}\} ~~~~ a.s.$$ Therefore
$$M_{-n} = E\{X_1|\mathcal{F}_{−n}\} = E\{X_2|\mathcal{F}_{−n}\} = \cdots = E\{X_n|\mathcal{F}_{−n}\},$$
hence
$$M_{−n} =\frac{1}{n} \sum_{j=1}^n E\{X_j|\mathcal{F}_{−n}\} = E\left\{ \frac{S_n}{n}|\mathcal{F}_{-n}\right\} = \frac{S_n}{n} ~~~ a.s.$$
Query:
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How can this process $M_{-n} = E\{X_1|\mathcal{F}_{−n}\}$ be backwards martingle? Is this because we have our $(\mathcal{F}_{-n})$ is an increasing sequence and $M_{-n} = E\{X_1|\mathcal{F}_{−n}\}$ is somewhat look like on the definition of the backwards martingle but in this case $M_{-n}$ is $\mathcal{F}_{-n}-$measurable?
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In this part $$E\{X_1|\mathcal{F}_{−n}\} = E\{X_j|\mathcal{F}_{−n}\} ~~~~ a.s.,$$ the book gave the hint that use the result in Exercise 23.17 .
Exercise 23.17 Let $(X_n)_{n \geq 1}$ be an i.i.d. and in $L^p$. Let $S_n = \sum_{j=1}^n X_J$ and $\mathcal{F}_ = \sigma(S_n, S_{n+1}, S_{n+2}, \ldots)$. Show that $E\{X_1|\mathcal{F}_n\} = E\{X_1|S_n\}$ and also $E\{X_j|\mathcal{F}_n\} = E\{X_j|S_n\}$ for $1\leq j \leq n$. Also show that $E\{X_j|\mathcal{F}_n\} = E\{X_1|S_n\}$ for $1 \leq j \leq n$.
My attempt: $E\{X_1|\mathcal{F}_{-n}\} = E\{X_1|S_n\} = E\{X_j|\mathcal{F}_{-n}\}$ for $1 \leq j \leq n$.
However, in the said problem, it does not say about its almost surely but on the proof, it is equal almost surely. Could you please show why it is almost surely?
- And lastly, from the proof, we have
$$E\left\{ \frac{S_n}{n}|\mathcal{F}_{-n}\right\} = \frac{S_n}{n} ~~~ a.s.$$
Could you please show me how is this equal?
Best Answer
1- First, note that a backward martingale is a martingale for which indices start from $-\infty$ towards 0. The filtration $(\mathcal{F}_{n})_{n\in\mathbb{Z}^{-}}$ is also increasing in the sense: $\mathcal{F}_{n}\subset\mathcal{F}_{n+1}$. We also have $M_{n} = \mathbb{E}[M_{n+1}\mid \mathcal{F_{n}}]$ (n is negative) which corresponds exactly to the definition of martingale. So what you described is correct.
2- Note that the conditional expectation is always defined almost surely. So what is meant by equality in the hint is rather equality a.e.
3- By definition of $\mathcal{F}_{-n}$, $S_{n}$ is $\mathcal{F}_{-n}-$measurable, so its conditional expectation is equal to itself (a.e of course).