If $Y_1, Y_2, \ldots$ are iid, with $S_n:= Y_1+\cdots+Y_n$ why do we have that $E(Y_1\mid\sigma(S_n, S_{n+1}, \ldots))=S_n/n$ for all $n \geq 1$?

Durrett's textbook asserts this is true when trying to give a simpler proof of the Strong Law of Large Numbers from the backwards Martingale theorem. However, he merely says it is "by symmetry" that we have that $E(Y_1\mid\sigma(S_n, S_{n+1}, \ldots))=E(Y_k\mid \sigma(S_n, S_{n+1}, \ldots))$ whenever $k\leq n$. But I'd like to see an actual proof of this.

  • $\begingroup$ Please replace trying to give a simpler proof by giving a simpler proof. $\endgroup$ – Did Feb 20 '13 at 7:26

It really is "by symmetry". Because the Y_i are iid, the distribution of $$ (Y_1,\ldots,Y_n, S_n, S_{n+1},\ldots) $$ is unchanged when $Y_i$ and $Y_j$ are exchanged. It follows that $$ \tag{1}E(Y_i|\sigma(S_n,S_{n+1},\ldots)) = E(Y_j|\sigma(S_n,S_{n+1},\ldots)) $$ for $i,j\in\{1,\ldots,n\}$. Also, $$ S_n = E(S_n |\sigma(S_n,S_{n+1},\ldots)) = \sum_{i=1}^n E(Y_i | \sigma(S_n,S_{n+1},\ldots)), $$ so $$ S_n = n E(Y_1| \sigma(S_n,S_{n+1},\ldots)). $$

EDIT How does one deduce (1) above? Observe that for any constants $s_n,s_{n+1},\ldots$, $$ E(Y_i 1_{S_n\le s_n,S_{n+1}\le s_{n+1},\ldots}) = E(Y_j 1_{S_n\le s_n,S_{n+1}\le s_{n+1},\ldots}) $$ for all $i,j\in\{1,\ldots,n\}$. This follows from the symmetry of the law of $(Y_1,\ldots,Y_n,S_n,S_{n+1},\ldots)$ with respect to exchanging the $Y_i$. We conclude that $$ E(Y_i 1_B) = E(Y_j 1_B) $$ for all $B\in\sigma(S_n,S_{n+1},\ldots)$. Staring at the definition of conditional expectation, we see $$ E(Y_i | \sigma(S_n,S_{n+1},\ldots)) = E(Y_j | \sigma(S_n,S_{n+1},\ldots)). $$

  • $\begingroup$ Well, where I can't fill in for a proof like this is how you get from the fact that $(Y_1, ..., Y_n, S_n, S_n+1, ...)$ is distributionally equal to the same thing with i and j exchanged, to the fact that the conditional expectations are equal. Are you viewing conditional expectation as expectation against a random measure? Cause if not that, the definition I'm using is the unique (up to a.s.) RV that is measurable in the smaller sigma algebra with all the same integrals as the original random variable when taken over sets in the smaller sigma algebra. $\endgroup$ – Jeff Feb 20 '13 at 5:34
  • $\begingroup$ Forgot to tag you in the last comment. $\endgroup$ – Jeff Feb 20 '13 at 6:36
  • $\begingroup$ @Jeff - I edited my answer to clarify that step. I hope my answer helps. Also, I believe the conditional expectation $E(X|Y_1,Y_2,\ldots)$ is determined by the joint law of $(X,Y_1,Y_2,\ldots)$. That's a useful fact when dealing with the iid assumption. $\endgroup$ – Will Nelson Feb 20 '13 at 7:10
  • $\begingroup$ I tried a similar argument as this, but couldn't get it to work. So what I did was to recognize that I have the desired equality of integrals over sets in $\sigma(S_k)$ for any $k \geq n$ But I couldn't extend it to the sigma closure of all such sets, because the integral relation isn't closed under countable unions. I tried using $\pi-\lambda$ but $\cup_{k\geq n} \sigma(S_k)$ is not a $\pi$ system. Here, what you have done is pointed out that the desired integral relation holds on slightly more, which is actually a generating pi-system. $\endgroup$ – Jeff Feb 20 '13 at 8:01
  • $\begingroup$ For the additional statement added in your latest comment, is this what you mean: Let $X$ and $Z$ be such that $(X, Y_1, ...)$ is distributionally equal to $(Z, Y_1, ...)$. Then the conclusion holds. I clarify this to say that the $Y_1$ cannot be changed to something else that makes distributional equality hold. It must be literally the same random variables I'm conditioning against, correct? (That is obeyed here) $\endgroup$ – Jeff Feb 20 '13 at 8:05

Note that by the independence of $(Y_j)_{j \geq 1}$ we have $$\mathbb{E}(Y_j \mid \sigma(S_n,S_{n+1},S_{n+2},\ldots)) = \mathbb{E}(Y_j \mid \sigma(S_n,Y_{n+1},Y_{n+2},\ldots))= \mathbb{E}(Y_j \mid \sigma(S_n))$$ for all $j \leq n$.


$$\mathbb{E}(Y_1 \cdot 1_B(S_n)) = \int \ldots \int_{y_1+\ldots+y_n \in B}y_1 \, \underbrace{d\mathbb{P}_{Y_1}(y_1)}_{d\mathbb{P}_{Y_j}(y_1)} \, d\mathbb{P}_{Y_2}(y_2) \ldots \underbrace{d\mathbb{P}_{Y_j}(y_j)}_{d\mathbb{P}_{Y_1}(y_j)} \ldots d\mathbb{P}_{Y_n}(y_n) \\ = \int \ldots \int_{y_1+\ldots+y_n \in B} y_1 \, d\mathbb{P}_{Y_j}(y_1) \, d\mathbb{P}_{Y_2}(y_2) \ldots d\mathbb{P}_{Y_1}(y_j) \ldots d\mathbb{P}_{Y_n}(y_n) = \mathbb{E}(Y_j \cdot 1_B(S_n))$$

for all Borel sets $B \in \mathcal{B}(\mathbb{R})$, $j \leq n$. Thus $$\mathbb{E}(Y_1 \mid \sigma(S_n)) = \mathbb{E}(Y_j \mid \sigma(S_n))$$ for all $j \leq n$. Therefore

$$S_n = \mathbb{E}(S_n \mid \sigma(S_n)) = n \cdot \mathbb{E}(Y_1 \mid \sigma(S_n)) = n \cdot \mathbb{E}(Y_1 \mid \sigma(S_n,S_{n+1},\ldots))$$


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