2
$\begingroup$

Let $(\Omega, \mathcal{F}, P)$ be a probability space, and let $X_1,X_2,\dots$ be a sequence of i.i.d. random variables. Let $$S_n:=X_1+\cdots+ X_n,$$ and define $\mathcal{G}:=\sigma(S_n)$, the $\sigma$-field generated by the sum $S_n$. Compute $E(X_1\vert S_n)$.

For reference, by definition, $E(X_1\vert S_n):= E(X_1\vert\sigma(S_n))$, where the latter is the a.s. unique $\sigma(S_n)$-measurable random variable such that $$\int_A E(X_1\vert\sigma(S_n))\,dP=\int_A X_1\,dP\quad\text{for all }A\in\sigma(S_n).$$

My question is if the independence of the $X_i$ is unnecessary? This seems to be because my first attempt only requires the fact that they're identically distributed:

$$S_n = E(S_n\vert S_n) = E(X_1\vert S_n) +\cdots + E(X_n\vert S_n) =: Z_1+\cdots Z_n.$$

The first equality is immediate from the definition and the second is from linearity. Then for any $1\le i< j\le n$ and any $A\in\sigma(S_n)$ we have that $Z_i$ and $Z_j$ are $\sigma(S_n)$-measurable and:

$$\int_A Z_i\,dP = \int_A X_i\,dP = \int_A X_j\,dP = \int_A Z_j\,dP,$$

since the $X_i$ are identically distributed and $\sigma(S_n)\subset \mathcal{F}$. Thus by definition we see that $Z_i=Z_j$, so that $E(X_1\vert S_n) = E(X_i\vert S_n)$ for all $i=1,\dots, n$. Hence

$$S_n = n\cdot E(X_1\vert S_n) \implies E(X_1\vert S_n) =\frac{1}{n} S_n.$$

Of course, in most of the equalities I mean almost sure equality. It doesn't seem like independence is necessary, but I want to verify.

$\endgroup$
  • $\begingroup$ To decide that $E(X_i\mid S_n)$ does not depend on $i$, one needs that the distribution of $(X_i,S_n)$ does not depend on $i$. The hypothesis that $(X_i)$ is identically distributed does not suffice for this. The hypothesis that $(X_i)$ is identically distributed and independent suffices. $\endgroup$ – Did Nov 15 '16 at 22:38
  • $\begingroup$ @Did I see my mistake, but I am still a little confused. It would suffice to show that $X_i \mathbb{1}_A$ has the same distribution as $X_j \mathbb{1}_A$ for any $A\in\sigma(S_n)$. However, I couldn't get very far because we don't necessarily have $X_i$ independent of $S_n$. Could you elaborate? $\endgroup$ – Satana Nov 16 '16 at 2:50
  • $\begingroup$ Sorry but is your question actually to understand why $E(X_i\mid S_n)$ does not depend on $i$ when $(X_i)$ is i.i.d.? $\endgroup$ – Did Nov 16 '16 at 9:31
  • $\begingroup$ @Did Yes. Perhaps I wasn't clear in my previous post, but to do precisely this I was trying to prove that $E[X_i \mathbb{1}_A] = E[X_j\mathbb{1}_A]$ for $A\in\sigma(S_n)$, for which it will then suffice to prove that they have the same distribution. But perhaps there's a better way. $\endgroup$ – Satana Nov 16 '16 at 16:29
  • $\begingroup$ The answer to that is that $(X_i,S_n)$ and $(X_j,S_n)$ are identically distributed and that conditional expectations depend only on joint distributions, in the sense that if $(X,Y)=(X',Y')$ in distribution and if $E(X\mid Y)=u(Y)$ then $E(X'\mid Y')=u(Y')$. $\endgroup$ – Did Nov 16 '16 at 16:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.