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Let $(\Omega,\mathcal F,P)$ be a probability space, and suppose that we are given, for each $\gamma \in[0,1]$, an iid sequence of real integrable random variables $\{X_n(\gamma)\}_{n=1}^\infty$. Let $Y$ be a random variable taking values in $[0,1]$ which is independent from $X_n(\gamma)$ for all $n,\gamma$.

How can I show that

$$\frac{1}{n}\sum_{k=1}^n X_k(Y)\to E[X_n(Y)| \sigma(Y)] \,\,\text{ as } \,\,n\to \infty$$

in probability?

EDIT: Assume the following additional condition: for all $\delta>0$ we have

$$\sup_{\gamma\in[0,1]} P\bigg(\bigg|\frac{1}{n}\sum_{k=1}^n \Big(X_k(\gamma)-E[X_k(\gamma)] \Big) \bigg|>\delta\bigg)\to 0 \,\,\text{ as } \,\,n\to \infty.$$

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    $\begingroup$ Can you represent $X_n(\gamma)$ as $f(Z_n,\gamma)$ for some function $f$ and a r.v. $Z_n$? $\endgroup$
    – d.k.o.
    May 14 at 14:01
  • $\begingroup$ @d.k.o. Yes! We have $X_n(\gamma)(\omega)=f(\gamma,Z_n(\omega))$ for some rv $Z_n$ and some $\mathcal B[0,1]\otimes \mathcal F$ measurable function $f$. $\endgroup$
    – Alphie
    May 14 at 14:11
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Because of the assumed independence of $\{X_n(\gamma): n\ge 1, \gamma\in[0,1]\}$ and $Y$, you can assume that $(\Omega,\mathcal F,P)$ is the product of $(\Omega_1,\mathcal F_1,P_1)$ and $(\Omega_2,\mathcal F_2,P_2)$, with the $X_n(\gamma)$ depending only on $\omega_1\in\Omega_1$ and $Y$ depending only on $\omega_2\in\Omega_2$. By the standard SLLN, for each $\gamma\in[0,1]$, $$ n^{-1}\lim_n\sum_{i=1}^n X_i(\omega_1,\gamma)=g(\gamma):=E[X_1(\gamma)], $$ for $P_1$-a.e. $\omega_1$. Consequently, by Fubini's theorem, $$ n^{-1}\lim_n\sum_{i=1}^n X_i(\omega_1,Y(\omega_2))=g(Y(\omega_2)), $$ for $P_1\otimes P_2$-a.e. $(\omega_1,\omega_2)$, provided $\gamma\mapsto E[X_1(\gamma)]$ is measurable. If $(\omega_1,\omega_2)\mapsto X_1(Y(\omega_2))$ is $P_1\otimes P_2$-integrable, then the conditional expectation $E[X_1(Y)\mid\sigma(Y)]$ exists and coincides with $g(Y)$.

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  • $\begingroup$ Thank you for your answer. If the uniform bound I gave in my edit holds, do you think this is sufficient also? $\endgroup$
    – Alphie
    May 14 at 18:21
  • $\begingroup$ You need the integrability only to identify the limit at a conditional expectation. $\endgroup$ May 14 at 21:29
  • $\begingroup$ Sorry I meant the bound $\sup_{\gamma\in[0,1]} P\bigg(\bigg|\frac{1}{n}\sum_{k=1}^n \Big(X_k(\gamma)-E[X_k(\gamma)] \Big) \bigg|>\delta\bigg)\to 0 \,\,\text{ as } \,\,n\to \infty$. Can I use this assumption instead of independence to show the result? $\endgroup$
    – Alphie
    May 14 at 21:55
  • $\begingroup$ May I ask why $(\omega_1,\omega_2)\mapsto X_1(Y(\omega_2))$ is measurable? $\endgroup$
    – Alphie
    May 17 at 13:51
  • $\begingroup$ There seems to be an implicit assumption that $(\omega,\gamma)\mapsto X_k(\gamma)(\omega)$ is jointly measurable. $\endgroup$ May 17 at 15:23
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If $\{Z_n\}$ are i.i.d. and $X_n=f(Z_n,Y)$ for some Borel function $f$ and a random variable $Y$ which is independent of $\{Z_n\}$, then $\{X_n\}$ are conditionally i.i.d. given $\sigma(Y)$, and
$$ \frac{1}{n}\sum_{i=1}^n X_n\to \varphi(Y) \quad\text{a.s.}, $$ where $\varphi(y):=\mathsf{E}[f(Z_1,y)]$ (assuming that the latter expectation exists). (See, e.g., Theorem 4.2 in this paper.)


As for the updated version of the question, since $\{Z_n\}$ is independent of $Y$ (any sequence $\{Z_n\}$ works here), for any nonnegative Borel function $g_n$, \begin{align} \mathsf{E}[g_n(Z_1,\ldots,Z_n,Y)\mid Y]=\varphi_n(Y), \end{align} where $$ \varphi_n(y)=\mathsf{E}[g_n(Z_1,\ldots,Z_n,y)]\le \sup_{y\in [0,1]}\mathsf{E}[g_n(Z_1,\ldots,Z_n,y)]. $$

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  • $\begingroup$ Ah but the $Z_n$ are not iid in my set-up. In fact $Z_n=S^n$ for some measure preserving and ergodic map $S:\Omega \to \Omega$. Maybe ergodic Theorem? $\endgroup$
    – Alphie
    May 14 at 14:26
  • $\begingroup$ How about this paper? Specifically, Theorem 2.5 on page 71. $\endgroup$
    – d.k.o.
    May 14 at 14:42
  • $\begingroup$ Am not sure we need such machinery. I added details in my post. Do you see how to continue from there? $\endgroup$
    – Alphie
    May 14 at 15:06
  • $\begingroup$ I think you need a uniform law of large number for ergodic sequences. Btw, the RHS of the last equation in the question depends on $n$, which doesn't make sense. $\endgroup$
    – d.k.o.
    May 14 at 15:08
  • $\begingroup$ Yes sorry I corrected the typo. $\endgroup$
    – Alphie
    May 14 at 15:10

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