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Let $(X_i, Y_i)_{i=1}^{\infty}$ be iid continuous random vectors with continuous joint density, where $X_1$ have support $\mathcal{X}$. Let $B_n\subset \mathcal{X}\subset\mathbb{R}$ be decreasing subsets (open intervals) such that $\cap B_n= x_0\in\mathcal{X}$.

Let $S = \{i\leq n: X_i\in B_n\}$. I want to show that $$ \frac{1}{|S|}\sum_{i\in S}Y_i \overset{P}{\to} \mathbb{E}[Y_1\mid X_1=x_0], \,\,as\,\,n\to\infty. $$ I assume that the necessary condition for this convergence is $|S|\to\infty$ or that $nP(X_i\in B_n)\to\infty$. Is it sufficient? Is there some theory that describes this?

Intuitively, this condition should be sufficient since, for large $n$ also $B_n$ will be large and $\frac{1}{|S|}\sum_{i\in S}Y_i \approx \mathbb{E}[Y_1\mid X_1\in B_n]\approx \mathbb{E}[Y_1\mid X_1=x_0]$. However proving this two-step intuition is harder than it looks (at least for me).

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  • $\begingroup$ Source of the problem? $\endgroup$
    – Zhanxiong
    Apr 2 at 23:44
  • $\begingroup$ @Zhanxiong I made it up, I need it to prove consistency of an estimator of mine $\endgroup$ Apr 3 at 9:14

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It's not sufficient. Take ${\cal X}=[-1,1]$ and let $X$ be uniform on ${\cal X}$. Suppose the distribution of $Y$ given $X=x$ is a mixture of $N(0,1)$ with probability $1-x$ and $N(1/x,1)$ with probability $x$. For any non-zero $x$, $E[Y|X=x]=1$, but $E[Y|X=x]=0$

Now take $x_0=0$ and take $B_n = (-n^{-1/2},n^{-1/2})$. Your average will converge to 1, but the true value is zero.

I now need to show that this does actually have a continuous density. It's obviously continuous except at $x=0$. Consider any sequence of points $(x_n,y_n)$ converging to a limit $(0,y)$. For any fixed $y$ and all large enough $n$, the $N(1/x_n,1)$ component of the density will be essentially zero in a neighbourhood of $Y=y$, and it is exactly zero at $x=0$, so $f(x_n,y_n)\to f(0,y)$.

You will need some sort of uniform continuity, or some sort of bound on the tails of the conditional distributions of $Y|X=x$

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  • $\begingroup$ Thanks for the thought. But I still don't belive that such density is continuous. The argument '...component of the density will be essentially zero in a neighbourhood of $Y=y$...' is not valid, since this component will have less and less weight on it but will be larger and larger. I don't think that if joint density is continuous, then conditionals can have a jump. If you write your density down for $x=0.001$ and for $x=0$ and for $x=-0.001$ I think they will be shifted by a constant. $\endgroup$ Apr 3 at 9:19
  • $\begingroup$ It's not uniformly continuous, but it is continuous at every $(x,y)$ $\endgroup$ Apr 3 at 22:56

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