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I have $(X_i)$ an i.i.d sequence of random variables in $L^1$.

Now I want to compute $\Bbb{E}\left(\sum_{k=1}^n X_k|X_1\right)$. I know that this is equal to $\sum_{k=1}^n\Bbb{E}\left( X_k|X_1\right)$ i.e. the problem reduces to compute $\Bbb{E}(X_k|X_1)$. But somehow I don't see how I can compute this conditional expectation. Could maybe someone help me?

We have the following definition:

If $X\in L^1$ and $B\subset A$ a sub sigma algebra, then $\Bbb{E}(X|B)$ is the unique r.v. $\xi\in L^1$ such that for all $Q\in L^\infty$ $$\Bbb{E}(XQ)=\Bbb{E}(\xi Q)$$

user1294729
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  • Your definition is faulty. Should be: If $X\in L^1(\Omega,A,P)$ and $B\subset A$ a sub sigma algebra, then $\Bbb{E}(X|B)$ is the unique r.v. $\xi\in L^1(\Omega,B,P)$ such that for all $Q\in L^\infty(\Omega,B,P)$ $$\Bbb{E}(XQ)=\Bbb{E}(\xi Q)$$ – GEdgar Oct 01 '22 at 19:37

2 Answers2

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Since the random variables are independent, the conditional expectation equals the expectation for $k\ge 2$, i.e. $\mathbb E[X_k|X_1]=\mathbb E[X_k]$ almost surely. Since the random variables have the same law, we have $\mathbb E[X_k]=\mathbb E[X_1]$, so we have $\mathbb E[\sum_{k=1}^nX_k|X_1]=X_1+(n-1)\mathbb E[X_1]$ almost surely, where we used $\mathbb E[X_1|X_1]=X_1$ (since $X_1$ is $X_1$-measurable).

Matija
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    Why can you do the first equality? – user1294729 Oct 01 '22 at 19:23
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    They are independent, so it follows that $\int_A X_k dP = \int_A (E X_k) dP$ for any $\sigma(X_1)$ measurable set $A$. – copper.hat Oct 01 '22 at 19:29
  • On Wikipedia under Basic Properties, this is the first item Pulling out independent factors. Formally, $\mathbb E[X_k|X_1]:\Omega\rightarrow\mathbb R$ is a function, while $\mathbb E[X_k]\in\mathbb R$ is a number. Sometimes this constant is identified with the function, as pointed out in the proof on Wikipedia. However, to be very precise, I stated that this constant function is one possible candidate for the conditional expectation, hence the almost surely (conditional expectations are allowed to differ on null sets). – Matija Oct 01 '22 at 19:30
  • @Matija ah I see and then $E(X_1|X_1)=X_1$ since $X_1=f(X_1)$ where $f$ is the identity and hence measurable. So this is equivalent to say that $X_1$ is $\sigma(X_1)$ measurable? – user1294729 Oct 01 '22 at 19:34
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    Exactly, when I say $X_1$ is $X_1$-measurable, I mean that $X_1$ is $\sigma(X_1)$-measurable. Also by definition we have $\mathbb E[X_1|X_1]=\mathbb E[X_1|\sigma(X_1)]$. And also by definition $\sigma(X_1)$ is the smallest $\sigma$-algebra such that $\mathbb E[X_1|X_1]=X_1$ holds (almost surely). And sure, if you want to, you can use the argument with $f$ being the identity. The most general case is $\mathbb E[Y|X]=Y$ a.s. for any $X$-measurable random variable $Y$, you use it with $Y=f(X)$. This rule is also on Wikipedia, the first under Stability, with $X=Y$ and $\mathcal H=\sigma(X)$. – Matija Oct 01 '22 at 19:44
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    @Matija I see!! This helped a lot – user1294729 Oct 01 '22 at 19:45
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In Galen Shorack's Probability for Statisticians, there is a very nice compendium of properties of conditional expectation. I will paste a picture of about half of them (the proofs you can find by finding the book; there are copies online): enter image description here

The main properties you want to use for your problem are (21) and (20). In particular, i.i.d. means sigma-algebras of $X_k$ for $k\neq 1$ are independent of sigma-algebra of $X_1$ so you can use (21); and then use (20) on $Y=X_1, \mathcal D = \sigma(X_1)$, and $X=1$.

D.R.
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  • But if $X_1$ and $X_2$ are independent , then does this mean that $\sigma(X_1)$ and $\sigma(X_2)$ are independent because I'm confused since in probability 1 we haven't defined independence of random variable as independence of the sigma fields – user1294729 Oct 01 '22 at 19:25
  • ah so there is an equivalence between the two – user1294729 Oct 01 '22 at 19:27
  • that's very strange. How else would you define independence of random variables? – D.R. Oct 01 '22 at 19:28
  • That $E(f(X_1)g(X_2))=E(f(X_1))E(g(X_2))$ for $f,g$ measurable and bounded – user1294729 Oct 01 '22 at 19:32
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    it's a pretty standard exercise; going from this to sigma-algebra definition take characteristic functions; and conversely use the "standard machine" approximating characteristic -> simple -> non-negative (use MCT) -> general. See a similar question here https://math.stackexchange.com/questions/2884580/two-random-variables-are-independent-if-all-continuous-and-bounded-transformatio?noredirect=1&lq=1 – D.R. Oct 01 '22 at 19:40