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We have $X_1, \dots, X_n$ iid random variables with finite expectations. Defining $S_n = \sum_{i=1}^{n} X_i$, we want to show that $$E[X_i|S_n ] = \frac{S_n}{n}. \quad \forall 1 \leq i \leq n$$

Presently, I'm going with: $E[X_i|S_n ] = E[X_j|S_n]$ for all $1 \leq i,j \leq n$ because of identical distributions. Thus$$S_n = \sum_{i=1}^{n} E[X_i|S_n] = n E[X_i|S_n] \implies E[X_i|S_n] = \frac{S_n}{n}.$$

However, I feel I should probably be more formal/rigorous in the $E[X_i|S_n ] = E[X_j|S_n]$ step but am unable to find a way of expressing so. Would appreciate some help.

Also, in the same setup, in the notes I'm reading, I came across this statement I haven't been able to work out: $\sigma(S_m) \subset \sigma(X_1, \dots X_m)$. Shouldn't there be an equality?

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    $\begingroup$ The $\sigma$ field generated by a few functionals will be finer than the $\sigma$ field generated by a single functional. $\endgroup$ – copper.hat Feb 8 '18 at 4:55
  • $\begingroup$ But since $S_m = X_1 + \dots + X_m$ isn't it effectively being generated by the same functions? $\endgroup$ – anktsdmcknsy Feb 8 '18 at 4:58
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    $\begingroup$ if $\sigma (S_m)= \sigma(X_1,... X_m)\supset \sigma(X_i) $ then we should have had $E(X_i|S_n)= X_i$ $\endgroup$ – clark Feb 8 '18 at 5:03
  • $\begingroup$ Okay, I feel I am getting some sense. So, $\sigma(S_1, \dots, S_m) = \sigma(X_1, \dots, X_m)$ would be true though, right? $\endgroup$ – anktsdmcknsy Feb 8 '18 at 5:04
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    $\begingroup$ The field generated by $\sin,\cos$ are finer than that generated by $\sin+\cos$. $\endgroup$ – copper.hat Feb 8 '18 at 5:04
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However, I feel I should probably be more formal/rigorous in the $E[X_i|S_n ] = E[X_j|S_n]$ step but am unable to find a way of expressing so. Would appreciate some help.

The symmetry argument is okay.   The terms in sequence $(X_i)_{i=1}^n$ are iid and their coefficents in the series $S_n$ are identically $1$, so they are interchangeable. $\mathsf E(X_i\mid S_n)=\mathsf E(X_j\mid S_n)$ for all integer $i,j: (1\leq i{,}j\leq n)$

$$\mathsf E(X_j\mid S_n) {= \tfrac 1n\sum_{i=1}^n\mathsf E(X_i\mid S_n) \\ = \tfrac 1n \mathsf E(S_n\mid S_n) \\ = \tfrac 1n S_n }$$

Also, in the same setup, in the notes I'm reading, I came across this statement I haven't been able to work out: $\sigma(S_m) \subset \sigma(X_1, \dots X_m)$. Shouldn't there be an equality?

Take, for example, Benoulli random variables $X_1, X_2$ occuring in a sample space represented by outcomes that are the pair of their results. Ie: $\Omega=\{0,1\}^2$   Then consider the sigma-algebra generated by the atoms.

Then $~~~\sigma(X_1,X_2)= \sigma\big\{\{(0,0)\},\{(0,1)\},\{(1,0)\},\{(1,1)\} \big\} = \mathcal P(\Omega)\\ \sigma(X_1+X_2)=\sigma\big\{\{(0,0)\},\{(0,1),(1,0)\},\{(1,1)\}\big\} \subsetneq \mathcal P(\Omega)$

So we have at least one example where there is not equality.  

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