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Let $y_1, y_2, \ldots , y_N$ be arbitrary real numbers and suppose a process of simple random sampling without replacement that selects $n$ out of $N$ elements. Then suppose that these $N$ elements are embedded in an asymptotic regime, indexed by $h \in \mathbb{N}$, whereby (1) the original set of $N$ real elements is copied $h − 1$ times, (2) within each of the $h$ copies, exactly $n$ units are selected under simple random sampling without replacement and (3) the $h$ copies are then collected into a single population with $Nh$ total units and $nh$ sampled units.

Denote the sample mean under simple random sampling without replacement for each $h \in \mathbb{N}$ by \begin{align*} \hat{\bar{y}}_h = (1/nh) \sum \limits_{i = 1}^{Nh} S_{h,i} y_{h,i}, \end{align*} where $S_{h,i}$ is an indicator variable equal to $1$ if $y_{h,i}$ is included in the sample and $0$ if not. For each $h$, there are $\binom{Nh}{nh}$ possible samples, the collection of which is denoted by $\mathcal{S}_h$. The vector $\mathbf{s}_h = \begin{bmatrix} s_{h,1} & \ldots & s_{h,N} \end{bmatrix}^{\top}$ is an element from the set $\mathcal{S}_h$.

This asymptotic regime implies that

(1) The average outcome, $\bar{y}_h = (1/Nh) \sum \limits_{i = 1}^{Nh} y_{h,i}$ is equal to $\bar{y}$, which is fixed over all $h \in \mathbb{N}$;

(2) $\max \limits_{1 \leq i \leq Nh} \left(y_{h,i} - \bar{y}_{Nh}\right)^2$ is fixed over all $h \in \mathbb{N}$, which implies that $\lim \limits_{h \to \infty} \max \limits_{1 \leq i \leq Nh} \left(y_{h,i} - \bar{y}_{Nh}\right)^2 / Nh = 0$; and

(3) the proportion of sampled units, $p_h \in (0, 1)$, is fixed at $p \in (0, 1)$ over all $h \in \mathbb{N}$.

These three conditions suffice for the almost sure convergence of the sample mean to the population's mean, i.e., \begin{align*} \hat{\bar{y}}_h \xrightarrow{\textrm{a.s.}} \bar{y}. \end{align*} I interpret this convergence statement as meaning, in essence, that, with a large enough $Nh$, the sample mean, $\bar{y}_h$, will always lie (i.e., with probability $1$) in the interval $(\bar{y} - \varepsilon, \bar{y} + \varepsilon)$, where $\varepsilon$ is an arbitrarily small constant greater than $0$.

My question is whether either of the two conditions hold: \begin{align*} \lim \limits_{h \to \infty} \left(\dfrac{1}{\left\lvert \mathcal{S}_h \right \rvert}\right)\sum \limits_{\mathbf{s}_h \in \mathcal{S}_h} \mathbb{1}\left\{\hat{\bar{y}}_h = \bar{y}\right\} & > 0 \text{ or } \\ \lim \limits_{h \to \infty} \int_{\bar{y}^* \in \mathcal{Y}} \left(\dfrac{1}{\left\lvert \mathcal{S}_h \right \rvert}\right)\sum \limits_{\mathbf{s}_h \in \mathcal{S}_h} \mathbb{1}\left\{\hat{\bar{y}}_h = \bar{y}^*\right\} \, d \, \bar{y}^*& > 0, \end{align*} where $\mathcal{Y} = \left\{ \bar{y}^*: \left\lvert \bar{y} - \bar{y}^* \right\rvert < \varepsilon \right\}$.

In other words, although the sample mean will, as $h \to \infty$, lie within a narrow interval around the population mean, $\bar{y}$, could the probability that the sample mean is exactly equal to $\bar{y}$ tend to $0$ as $h \to \infty$? Or is this probability always bounded away from $0$ in the limit?

It seems like this probability would not always be bounded away from $0$, but I am having difficulty constructing a counterexample that satisfies the three conditions above. I would be greatly indebted to anyone who can give a simple example that satisfies the three conditions above, but where the probability that the sample mean is equal to the true limiting mean tends to $0$.

It seems that one way to do this would be to given an example in which the sample mean produces unique values for every $h \in \mathbb{N}$, which would imply that, at best, $\Pr\left(\hat{\bar{y}}_h = \bar{y}\right) = \left(\dfrac{1}{\left\lvert \mathcal{S}_h \right \rvert}\right)$. Since $\lim \limits_{h \to \infty} \dfrac{1}{\left\lvert \mathcal{S}_h \right \rvert} = 0$, then $\Pr\left(\hat{\bar{y}}_h = \bar{y}\right)$ would also tend to $0$. However, I am having difficulty constructing such an example that satisfies the three aforementioned conditions.

Finally, if indeed the probability is not always bounded away from 0 as $h \to \infty$, are there additional (ideally mild) conditions one could impose on the infinite triangular array to ensure that the probability is always bounded away from 0?

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It's possible set it up so that $\hat{\bar{y}}_N$ equals $\bar y_N$ at most finitely often.

Make all the $y$ binary 0/1 and choose $n$ coprime to $N$,eg by taking $n$ to be the smallest prime greater than $pN$ that does not divide $N$.

In that case, $\hat{\bar{y}}_N$ will always be a multiple of $1/n$ and $\bar y_N$ will only be a multiple of $1/n$ if it's zero, so they can't be equal for large $N$.

Alternatively, if the $Y$ are real numbers sampled from some continuous distribution then for almost every sequence of $Y$, $P(\hat{\bar y}=\bar y)=0$ for all $N$ and all $0<n<N$

The two can only be equal if there is some subset of size $n$ such that $$N\sum_{\text{sample}}Y=n\sum_{\text{all }N}Y.$$ For a fixed $N$ and fixed sample that event has measure zero, so over all $N$ and all samples it's a countable union of sets of measure zero.

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  • $\begingroup$ I don't think your example satisfies the third condition in which $\lim \limits_{N \to \infty} n/N = p$, where $p \in (0, 1)$. Or does it? If so, can you please explain to me how letting $n$ be the smallest prime greater than $pN$ that does not divide by $N$ implies that $n/N$ limits to a fixed $p \in (0, 1)$? $\endgroup$ Commented Oct 1, 2022 at 14:34
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    $\begingroup$ The typical gap between primes at $n$ is $O(n/logn)$, so $n=pN+O(pN/log(pN)$, and $n/N=p+O(p/log(pN))=p+o(1)$. That neglects the possibility that the prime might divide $N$, but you can't have several successive primes all dividing $N$ if $p$ is bounded away from zero. $\endgroup$ Commented Oct 1, 2022 at 23:50
  • $\begingroup$ thank you, that is very, very helpful. One further question: In practice, it seems that one thing a researcher could control is the sampling process and it's hard to imagine a researcher choosing an $n$ to be the smallest prime greater than $pN$ that does not divide by $N$. Is there an example in which the limiting probability that the sample mean is equal to the population mean is $0$ but with $p \in (0, 1)$ fixed over every $N \in \mathbb{N}$? $\endgroup$ Commented Oct 2, 2022 at 13:53
  • $\begingroup$ It's not possible to have $p$ fixed over every $N$: eg you can't have $p=1/2$ for odd $N$. With Poisson sampling rather than SRS you could: sample each observation independently with probability $p$ $\endgroup$ Commented Oct 2, 2022 at 20:33
  • $\begingroup$ thank you again for your helpful response. What I'm trying to ascertain is how much the counterexample you gave depends on the use of a "weird" sampling process. Is there an example in which the limiting probability that the sample mean is equal to the population mean is 0, but the sampling process is more "typical"? Say, for example, that p is fixed at 0.5 but now $N \in \{2, 4, 6, 8, 10, \ldots \}$. Is there an example in which outcomes satisfy (1) and (2) in the original post, but the limiting probability that the sample mean is equal to the population mean is 0? $\endgroup$ Commented Oct 3, 2022 at 3:06

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