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Let $X$ and $\varepsilon$ be independent random vectors, $\mathcal{X} = \text{supp}(X)$, and $Y = f(X) + \varepsilon$ for some function $f$. For any $x \in \mathcal{X}$, let $y^i = y^i(\omega)$, $i \in \{1,\cdots,n\}$, be independent samples of $Y \mid X = x$ defined on a measurable space $(\Omega,\mathcal{F})$ and equipped with a probability measure $\mathbb{P}_x$. Suppose for each $x \in \mathcal{X}$, there exists a $\mathcal{F}$-measurable set $S(x)$ such that $\mathbb{P}_x\{S(x)\} = 0$ and for any $\omega\in \Omega \backslash S(x)$, $$\lim_{n \to \infty} \frac{1}{n}\sum_{i=1}^{n} g(y^i(\omega)) = \mathbb{E}[g(Y) \mid X = x]$$ for some function $g$.

Question: Can we find a measurable set $T$ such that $\mathbb{P}_x\{T\} = 0$, $\forall x \in \mathcal{X}$, and for any $\omega\in \Omega \backslash T$, $$\lim_{n \to \infty} \frac{1}{n}\sum_{i=1}^{n} g(y^i(\omega)) = \mathbb{E}[g(Y) \mid X = x]? \tag{1}$$ If this is not true in general, are there mild assumptions on the functions $f$ and $g$ and/or the distribution of $\varepsilon$ under which it holds?

Note: $\mathcal{X} \subset \mathbb{R}^m$ is the support of $X$, $f: \mathcal{X} \to \mathbb{R}^d$, and $g: \mathcal{Y} \to \mathbb{R}$, where $\mathcal{Y}$ is the support of $Y$.

Context: This question is based on Assumption (A6) in page 11 of this paper. I am not well-versed in measure theory, so forgive me for incorrect use of notation.

Thoughts: My rough interpretation is that $S(x)$ denotes the sample paths of $Y \mid X = x$ of probability zero over which the LLN-type equality does not hold. Generally, this set can depend on $x \in \mathcal{X}$, and the question is whether there exists a set (independent of $x$) $T \supset S(x)$, for a.e. $x \in \mathcal{X}$ also of zero probability for which the equality holds.

Clearly, this holds when $f \equiv 0$ (i.e., $Y$ is independent of $X$) since the set $S$ does not depend on $x$ in this case. When $f$ is not trivially zero, it seems like there cannot be (uncountably) many values that the set $S(x)$ can take, because the conditional distributions $Y \mid X = x_1$ and $Y \mid X = x_2$ only differ by a translation when $x_1 \neq x_2$.

Plausible argument for the case when $f$ is continuous and $g$ is Lipschitz continuous: Let $\bar{\mathcal{X}} = \mathcal{X} \cap \mathbb{Q}^m$ be the intersection of the support of $X$ with $m$-dimensional rational vectors. Then $T = \cup_{x \in \bar{\mathcal{X}}} S(x)$ satisfies (1). I think this is true because if $\omega \in S(x)$ for some $x \in \mathcal{X} \backslash \bar{\mathcal{X}}$, then we can pick $\bar{x} \in \bar{\mathcal{X}}$ that is arbitrarily close to $x$ (since $\mathbb{Q}^m$ is dense in $\mathbb{R}^m$) such that $\omega \in S(\bar{x})$.

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  • $\begingroup$ I don't mean to be annoying, but can you get rid of a lot of your question to ask a standalone question? You may certainly provide context, but after you have asked your question. It would be nice if you can ask a general, self-contained, question at the beginning. Like, do you need, $f$, $\epsilon$, etc? I have a headache now, so maybe I'm just too sensitive now to much information. $\endgroup$ – mathworker21 Apr 2 '20 at 23:22
  • $\begingroup$ @mathworker21 No worries! I tried my best to be concise. I'm afraid to change it too much from what's in the paper because I fear I might be imprecise with the notation/formulation. $\endgroup$ – madnessweasley Apr 3 '20 at 3:23
  • $\begingroup$ Thanks! Why do you have to suppose the existence of $S(x)$? Doesn't its existence follow from LLN? $\endgroup$ – mathworker21 Apr 3 '20 at 5:39
  • $\begingroup$ @mathworker21 Yes, I'm essentially assuming that the strong LLN holds for the function $g$ in question $\endgroup$ – madnessweasley Apr 3 '20 at 5:41
  • $\begingroup$ @mathworker21 It isn't obvious to me. The sets $S(x_1)$ and $S(x_2)$ can be different for $x_1 \neq x_2$, but it isn't clear to me that we can't find a single set $T$ of measure zero for which the equality holds for all $x$. For example, the paper says that if $\mathcal{X}$ is a finite set, then the result holds. I presume this is true by considering $T := \cup_{x \in \mathcal{X}} S(x)$. If so, then it seems like this trick would work even if $\mathcal{X}$ is countable. $\endgroup$ – madnessweasley Apr 3 '20 at 5:49
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The answer is "yes", assuming $g$ is continuous.

There are two extreme kinds of $\epsilon$. One kind is discrete $\epsilon$ (only atoms), and another is smooth $\epsilon$ (no atoms). I will deal with each, and let you handle mixtures.

Let's first do discrete. The limit condition is $\lim_{n \to \infty} \frac{1}{n}\sum_{i=1}^n g(f(x)+\epsilon_i) = \sum_{\epsilon'} p(\epsilon')g(f(x)+\epsilon')$, where $(\epsilon_i)_i$ is the sampled $\epsilon$'s and $p(\epsilon')$ is the probability of $\epsilon$ being a specific $\epsilon'$. We may let $T$ be the set of all $\omega$, equivalently the set of all $(\epsilon_i)_i$, such that there is some $\epsilon'$ with $\lim_{n \to \infty} \frac{1}{n}\#\{1 \le i \le n : \epsilon_i = \epsilon'\} \not = p(\epsilon')$ for each $\epsilon'$. It's easy to see $P_x[T] = 0$ for each $x$ (there's no dependence on $x$; $\epsilon$ is an independent thing, so by LLN, we have $\lim_{n \to \infty} \frac{1}{n}\#\{1 \le i \le n : \epsilon_i = \epsilon'\} = p(\epsilon')$ for each $\epsilon'$ with probability $1$).

Now let's do continuous. Let's suppose $\epsilon$ is supported in $0,1$, and $\mu$ is the distribution of $\epsilon$. We want $\lim_{n \to \infty} \frac{1}{n}\sum_{i=1}^n g(f(x)+\epsilon_i) = \int_0^1 g(f(x)+\epsilon')d\mu(\epsilon')$. We let $T$ be the set of all $(\epsilon_i)_i$ for which there is some subinterval $(a,b) \subseteq (0,1)$ with $\lim_{n \to \infty} \frac{1}{n}\#\{1 \le i \le n : \epsilon_i \in (a,b)\} \not \to \mu((a,b))$. Once again, $P_x[T]$ is independent of $x$, and it is $0$ by LLN. To see that $\lim_{n \to \infty} \frac{1}{n}\sum_{i=1}^n g(f(x)+\epsilon_i) = \int_0^1 g(f(x)+\epsilon')d\mu(\epsilon')$ for each $(\epsilon_i)_i$ not in $T$, we use continuity of $g$ (this is an easy analysis argument; let me know if you want me to sketch it out).

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