Limiting distribution of order statistics with binomial weights

Question

Let $G$ be a CDF whose support is $[0,1]$ and $x\in(0,1)$ is a constant.

Define a CDF $H^n$ by $$H^n(t)=\sum^n_{k=1}{n-1\choose k-1}x^{k-1}(1-x)^{n-k}G_{k;n}(t).$$ where $G_{k;n}$ is a CDF of k-th order statistics when number of samples is $n$. More precisely, it is $G_{k;n}(t)=\sum^n_{i=k}{n\choose i}G^i(t)(1-G(t))^{n-i}$. Thus $H^n$ is associated with a random variable whose value is $k$-th lowest one with probability ${n-1\choose k-1}x^{k-1}(1-x)^{n-k}$. $H^n$ certainly is a distribution function because $\sum^n_{k=1}{n-1\choose k-1}x^{k-1}(1-x)^{n-k}=\sum^{n-1}_{k=0}{n-1\choose k}x^{k}(1-x)^{n-1-k}=(x+1-x)^{n-1}=1$.

My question is What would be the limiting distribution of $H^n$?

Answer 1

You are examining the properties of "random central order statistics" and the asymptotic distribution of these statistics. This topic has been examined at length in Puri and Ralescu (1968), where the authors derive a number of useful asymptotic convergence results and a central limit theorem for a scaled version of these statistics. Rather than starting with the equation relating the distributions $H^n$ and $G$, it is simpler and more instructive to frame your problem in terms of random variables. In the analysis below I will define the "random central order statistics" of the distribution $G$ and show that this leads to the equation of interest in your question. I will then present some relevant limiting results from the literature.

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Defining the "random central order statistics" of $G$: In the following analysis, to use standard notation, I am going to use the parameter $\theta$ in place of your value $x$, and I am going to use the notation $G_{n,k}$ in place of your $G_{k;n}$. Define the independent sequences of random variables:

$$X_1,X_2,X_3,... \ \sim \text{IID Bern} (\theta) \quad \quad \quad \quad \quad T_1,T_2,T_3,... \ \sim \text{IID } G.$$

We will let $T_{n,k}$ be the $k$th order statistic in the sample of the first $n$ values, and denote its CDF by $G_{n,k}$. We will also define the random variable $K_n = 1 + \sum_{i=1}^{k-1} X_i \sim 1+\text{Bin}(n-1,\theta)$. Using these values we define the random variables $H_n = T_{n,K_n}$ and this gives us the following two sequences:

$$\begin{matrix} \text{Random central order statistics} & & & \{ H_n | n \in \mathbb{N} \}, \\[6pt] \text{Random central rank sequence} \ \ & & & \{ K_n | n \in \mathbb{N} \}. \\[6pt] \end{matrix}$$

From the strong law of large numbers, we know that $K_n/n \rightarrow \theta$ as $n \rightarrow \infty$ (with probability one) and so the random central rank sequence meets the requirements set out in Puri and Ralescu (1968). You are looking for the CDF of the random central order statistic, which is:

$$\begin{equation} \begin{aligned} H^n(t) &\equiv \mathbb{P}(H_n \leqslant t) \\[6pt] &= \mathbb{P}(T_{K_n,n} \leqslant t) \\[6pt] &= \sum_{k=1}^n \mathbb{P}(K_n=k) \ \mathbb{P}(T_{K_n,n} \leqslant t | K_n=k) \\[6pt] &= \sum_{k=1}^n \mathbb{P}(K_n=k) \ \mathbb{P}(T_{k,n} \leqslant t) \\[6pt] &= \sum_{k=1}^n {n-1 \choose k-1} \theta^{k-1} (1-\theta)^{n-k} \cdot G_{k;n}(t). \\[6pt] \end{aligned} \end{equation}$$

Denote the quantiles of $G$ by $Q(\theta) \equiv \inf \{ t \in \mathbb{R} | G(t) \geqslant \theta \}$. The various theorems in Puri and Ralescu (1968) are used to show that (under certain conditions) we have $H_n \rightarrow Q(\theta)$, and a normalised version of $H_n$ obeys a central limit theorem.

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CLT for random central order statistics: Suppose that the distribution $G$ is differentiable and has density $g$. Theorem 3.1 in Puri and Ralescu (1968) (p. 452) gives a central limit theorem for the random central order statistics. The theorem requires two conditions; the first is that the quantile $Q(\theta)$ is in the support of the distribution (i.e., $g(Q(\theta))>0$) and the second is that the distribution of $K_n$ must meet the following condition:

$$\frac{K_n - n \theta}{\sqrt{n}} \overset{\mathbb{P}}{\longrightarrow} \text{const}.$$

If these conditions are met then you have the asymptotic distribution:

$$\sqrt{n} \cdot g(Q(\theta)) \cdot \frac{H_n - Q(\theta)}{\sqrt{\theta(1-\theta)}} \overset{\text{Dist}}{\longrightarrow} \text{N}(0,1).$$

Unfortunately, in the present case, the limiting result on $K_n$ is not met, so the conditions of the theorem are not satisfied. You will therefore need to look into CLT results that have weaker conditions on the random central rank sequence.

Answer 2

I thought about this question a lot these days, and came up with an answer. If someone can verify my answer, I'd really appreciate it.

The answer is $H^n\rightarrow H$, where $$H(t)=0~\textrm{ if }t<G^{-1}(x)~\textrm{and}$$ $$H(t)=1~\textrm{ if }t\geq G^{-1}(x).$$ It is because

My answer is as follows:

Let $Y_1,Y_2,\cdots,Y_n$ be iid RV with CDF $G$. Let's denote $k$-th order statistics of it by $Y_{k;n}$ so that $Y_{1;n}\leq Y_{2;n}\leq\cdots\leq Y_{n;n}$. Then, for $\frac{k}{n} \rightarrow z$, we have $$Y_{k;n}\overset{p}{\to}G^{-1}(z).$$ I referred this result from here. It's on page 15.

On the other hand, if we look at the binomial part, define $Q^n$, an RV, whose value is $k/n$ with probability ${n−1\choose k−1}x^{k−1}(1−x)^{n−k}$. Then $Q^n\overset{p}{\to} x$ because the mean of $Q^n$ is $\frac{n-1}{n}x$ and variance goes to zero as $n$ goes to infinity. More precisely, $Var(Q^n)=\frac{n-1}{n^2}x(1-x)$. I used properties of binomial distribution.

Thus, at the summation, for $k$'s such that $\frac{k}{n}\neq x$, ${n-1\choose k-1} x^{k-1}(1-x)^{n-k}$ all converges to zero, while $k'$ such that $\frac{k'}{n}\rightarrow x$, the value ${n-1\choose k'-1} x^{k'-1}(1-x)^{n-k'}$ converges to 1. This implies $H^n$ converges to $H$.

This is my answer, anyone can confirm or elaborate more? I'd really appreciate it.