I would like to solve the question of the title. My ideas:

  1. $\sigma(S_n,S_{n+1},\ldots)=\sigma(S_n,X_{n+1},X_{n+2},\ldots)$.

  2. $E[X_1|S_n,X_{n+1},X_{n+2},\ldots]=E[X_1|S_n]$ by independence.

  3. Since $X_1,\ldots,X_n$ have the same distribution, $E[X_1|S_n]=E[X_2|S_n]=\ldots=E[X_n|S_n]$, therefore $E[X_1|S_n]=\frac{1}{n}\sum_{j=1}^n E[X_j|S_n]=S_n/n$.

I could prove 1 and 2, but not 3: $E[X_1|S_n]=E[X_j|S_n]$ for $j=2,\ldots,n$. Could you provide a hint?

And just as a curiosity: can we deduce the strong law of large numbers from here? For instance, something like $$\lim_n S_n/n=\lim_n E[X_1|S_n,S_{n+1},\ldots]=E[X_1|\emptyset]=E[X_1].$$

  • $\begingroup$ Concerning 3: the vector$(X_1,X_2,X_3,\dots,X_n)$ has the same distribution as $(X_2,X_1,X_3,\dots,X_n)$. Does this help? $\endgroup$ – zhoraster Dec 14 '16 at 17:10
  • $\begingroup$ @zhoraster I don't exactly know what to do with that... Is that related to $S_n=X_1+X_2+X_3+\ldots+X_n=X_2+X_1+X_3+\ldots+X_n$ or something like that? $\endgroup$ – user39756 Dec 14 '16 at 17:32
  • $\begingroup$ Related it is! Try to play with this equality, the equality in distribution and with the definition of conditional expectation to deduce $E[X_1|S_n] = E[X_2|S_n]$. $\endgroup$ – zhoraster Dec 14 '16 at 17:36
  • $\begingroup$ @zhoraster I think I got it. By definition, we need to prove that $$E[X_11_{\{X_1+\ldots+X_n\in B\}}]=E[X_21_{\{X_1+\ldots+X_n\in B\}}]$$ for every Borel set $B$ in $\mathbb{R}$. We have $$E[X_11_{\{X_1+\ldots+X_n\in B\}}]=\int_{\mathbb{R}^n}x_11_{\{x_1+\ldots+x_n\in B\}}\,P_{(X_1,\ldots,X_n)}(dx_1,\ldots,dx_n).\quad (*)$$ Now we use $P_{(X_1,\ldots,X_n)}=P_{(X_2,X_1,X_3,\ldots,X_n)}$ (from the i.i.d. condition) and $x_1+\ldots+x_n=x_2+x_1+x_3+\ldots+x_n$: $$(*)=\int_{\mathbb{R}^n}x_11_{\{x_2+x_1+x_3+\ldots+x_n\in B\}}\,P_{(X_2,X_1,X_3,\ldots,X_n)}(dx_1,\ldots,dx_n).\quad (**)$$ $\endgroup$ – user39756 Dec 14 '16 at 18:33
  • $\begingroup$ ... Change names to see things clearer: $x_1$ by $y_2$, $x_2$ by $y_1$, $x_j$ by $y_j$ for $j\geq3$: $$(**)=\int_{\mathbb{R}^n}y_21_{\{y_1+\ldots+y_n\in B\}}\,P_{(X_2,X_1,X_3,\ldots,X_n)}(dy_2,dy_1,dy_3,\ldots,dy_n)=\int_{\mathbb{R}^n}y_21_{\{y_1+\ldots+y_n\in B\}}\,P_{(X_1,\ldots,X_n)}(dy_1,\ldots,dy_n)=E[X_21_{\{X_1+\ldots+X_n\in B\}}].$$ $\endgroup$ – user39756 Dec 14 '16 at 18:34

Your last observation is correct! But it is not written properly: in fact $$\lim_n \mathsf E[\cdot \mid S_n,S_{n+1},\dots]\neq \mathsf E[\cdot \mid\varnothing]. $$ The limit of conditioning on $S_n,S_{n+1},\dots$ should be understood as the limit of sigma-algebras $$ \mathcal F_n = \sigma(S_n,S_{n+1},\dots). $$ Why $\mathcal F_\infty = \lim_n \mathcal F_n$ is not empty? Because it consists of so-called exchangeable events$-$those which do not change when you permute any finite number of $X_1,X_2,\dots$ Indeed, under any such permutation the sums $S_n$ do not change for $n$ large enough.

There is a so-called backwards martingale convergence theorem: if $\{\mathcal F_n,n\ge 1\}$ is a decreasing sequence of sigma-algebras, $\mathcal F_\infty = \lim_n \mathcal F_n := \bigcap_n\mathcal F_n$, and $X$ is an integrable random variable, then $$ \lim_n \mathsf E[X \mid \mathcal F_n]= \mathsf E[X \mid\mathcal F_\infty] $$ almost surely. This is still not enough to prove the strong law of large numbers. Fortunately, there is a 0-1 Hewitt-Savage law: if, as in your question, $\mathcal F_\infty$ is the sigma-algebra of exchangeable events, corresponding to a sequence of iid random variables, then each event in $\mathcal F_\infty$ has probability $0$ or $1$. This implies, of course, that $$ \mathsf E[X \mid\mathcal F_\infty] = \mathsf E[X] $$ almost surely, finishing the proof of the strong law of large numbers.

It is also possible to appeal to Kolmogorov's 0-1 law, not that of Hewitt and Savage: $\lim_n S_n/n$ is measurable with respect to the tail sigma-algebra. (And it is better to proceed in this manner, since my argument that $\mathcal F_\infty$ consists of exchangeable random events was rather handwaving.)


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