For $X_n$ iid, $S_n=\sum X_j$, $\mathscr{G}_n=\sigma(S_n,S_{n+1}, \dots)$, show $E(X_j \mid \mathscr{G}_n)=E(X_1 \mid S_n)$

Question

If $(X_n)$ is iid in $L^1$, and $S_n = \sum_{i=1}^{n} X_i$ and $\mathscr{G_n} = \sigma(S_n, S_{n+1},...)$,

then show that $E[X_1 \mid \mathscr{G_n}]=E[X_1 \mid S_n]$,

and that $E[X_j \mid \mathscr{G_n}]=E[X_j \mid S_n]$, for $1 \leq j \leq n$,

and that $E[X_j \mid \mathscr{G_n}]=E[X_1 \mid S_n]$, for $1 \leq j \leq n$.

So, I'm thinking that $\mathscr{G}_n$ = $\sigma(S_n, S_{n+1}, ...) = \sigma(S_n, \mathscr{G_{n+1}})$ and intuitively it seems to me that $\mathscr{G}_{n+1}$ is independent of $\sigma (X_j, S_n)$ since knowing $S_m$ for any $m \leq n$ should have no impact on $X_j$ with $j \leq n$.

That would mean that $E[X_j \mid \mathscr{G}_n] = E[X_j \mid \sigma(S_n,\mathscr{G}_{n+1})]=E[X_{j} \mid S_n]$ for any $j \leq n$, including $j=1$, by the following Theorem:

Theorem: $X \in L^1, \mathscr{G}, \mathscr{H} \subset \mathscr{F}$, and $\mathscr{H }$ independent of $\sigma(X, \mathscr{G})$ means that $E(X \mid \sigma(\mathscr{G},\mathscr{H})) = E(X \mid \mathscr{G})$

That would take care of the first two parts. So, my question is -- how exactly do I show those events are independent? If I take one event from each field $A, B$, I don't really see how to show $P(A\cup B)=P(A)P(B)$.

The third part, I see intuitively also. The information in $\mathscr{G}_n$ that is relevant is contained in $S_n$, and from there since $X_n$ are iid, it makes no difference if we are looking at $X_1$ or $X_j$.

Answer 1

1. Regarding $\mathbb{E}(X_j \mid \mathcal{G}_n) = \mathbb{E}(X_j \mid S_n)$. Note that $$\mathcal{G}_n := \sigma(S_n,S_{n+1},\ldots,) = \sigma(S_n,X_{n+1},X_{n+2},\ldots).$$ Since, by assumption, $(X_{n+1},X_{n+2},\ldots)$ are independent from $(X_n,S_n)$ you can apply the mentioned theorem.
2. Regarding $\mathbb{E}(X_j \mid S_n) = \mathbb{E}(X_1 \mid S_n)$: You are right, intuitively this is clear. In order to prove this rigorously, we note that the $\sigma$-algebra $\mathcal{\sigma}(S_n)$ is generated by sets of the form $$G := \{S_n = X_1+\ldots+X_n \in B\}$$ where $B$ is a Borel set. Therefore, it suffices to show $$\int_{G} X_1 \, d\mathbb{P} = \int_G X_j \, d\mathbb{P}$$ for $G$ of the form $(1)$, see the Lemma below. To this end, write $$\begin{align*} \int_G X_j \, d\mathbb{P} &= \int 1_B(X_1+\ldots+X_n) X_j \, d\mathbb{P} \\ &= \int \ldots \int 1_B(x_1+\ldots+x_n) x_j \, d\mu(x_1) \ldots d\mu(x_n) \end{align*}$$ where $\mu$ denotes the distribution of $X_j$.

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Lemma: Let $X \in L^1(\mathbb{P})$ and $\mathcal{F}$ a $\sigma$-algebra. Then $Y \in L^1(\mathbb{P})$ equals $\mathbb{E}(X \mid \mathcal{F})$ if, and only, if $Y$ is $\mathcal{F}$-measurable and $$\int_F X \, d\mathbb{P} = \int_F Y \, d\mathbb{P} \qquad \text{for all} \, F \in \mathcal{F}.$$

If $\mathcal{G}$ is a $\cap$-stable generator of $\mathcal{F}$ (i.e. $G_1,G_2 \in \mathcal{G} \Rightarrow G_1 \cap G_2 \in \mathcal{G}$ and $\sigma(\mathcal{G})=\mathcal{F}$), then $Y=\mathbb{E}(X \mid \mathcal{F})$ if, and only if, $Y$ is $\mathcal{F}$-measurable and $$\int_G X \, d\mathbb{P} = \int_G Y \, d\mathbb{P} \qquad \text{for all} \, G \in \mathcal{G}.$$