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Given $Z_1,Z_2,\cdots$ i.i.d with $E|Z_i|<\infty$. $\theta$ is an independent r.v. with finite mean and $Y_i=Z_i+\theta$.

If we define $F_n=\sigma(Y_1,\cdots,Y_n), F_\infty=\sigma(\cup_n F_n).$ Do we have $\theta\in F_\infty$?

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up vote 4 down vote accepted

How about: If $E[Z_i] = m$, then $$ \lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^n (Z_i+\theta) = m + \theta $$ a.e. so $m+\theta$ is $F_\infty$-measurable. So is the constant $m$.

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Yes you're right. Thank you. – Iew Jun 16 '12 at 2:46
And what if you don't assume $Z_i$ is integrable? Must $\theta$ still be $F_\infty$-measurable? – GEdgar Jun 16 '12 at 2:54

Concerning GEdgar's followup question; the moments are not necessary. We replace the sample average and true mean $m$ by the empirical distribution function and the true distribution function $F$.

Without loss of generality assume that the probability space is $\Omega\times \Omega^\prime$, a product space equipped with a product measure $\mathbb{P}\times\mathbb{Q}$. For any fixed $c\in\mathbb{R}$ we have (almost surely) $${1\over n}\sum_{i=1}^n1_{(-\infty,\, c]}(Z_i(\omega)+\theta(\omega^\prime)) = {1\over n}\sum_{i=1}^n1_{(-\infty,\,c-\theta(\omega^\prime)]}(Z_i(\omega)) \to F(c-\theta(\omega^\prime)),\tag1$$ where $F$ is the cumulative distribution function of $Z$.

For each fixed $\omega^\prime$, the convergence in (1) holds for $\mathbb{P}$-almost every $\omega$ by the law of large numbers and hence, by Fubini, holds for $(\mathbb{P}\times\mathbb{Q})$-almost every $(\omega,\omega^\prime)$ on the product space. For $0<u<1$ define $q(u)=\inf(x: F(x)\geq u)$, the quantile function of $F$. It is easy to show that $F(c-\theta(\omega^\prime))\geq u$ if and only if $c-\theta(\omega^\prime)\geq q(u)$. In particular, this shows that $$\{(\omega,\omega^\prime) : c-q(1/2)\geq \theta(\omega^\prime)\}\in {F}_\infty\vee{\cal N}$$ where ${\cal N}$ is the $\sigma$-field of $(\mathbb{P}\times\mathbb{Q})$-trivial sets. Since $c$ is arbitrary, we conclude that $\theta\in {F}_\infty\vee{\cal N}.$

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