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Given $T_n = \sum_{i=1}^n c_i X_i$, and integer $m$ with $0\leq m\leq n$, where

  • $X_1, \dots, X_n$ are $\{0,1\}$-valued random variables, and have a joint probability mass function which takes ${n \choose m}^{-1}$ whenever $\sum_{i=1}^n X_i = m$, and $0$ otherwise;

  • $c_i = \sqrt{mn(n-m)} (n+1) F^{-1}(\frac{i}{n+1})$, where $F$ is the cdf of a continuous distribution.

How should one determine the asymptotic variance of $T_n$ in terms of moments of $F$?

By the way, generally, what are some ways to determine asymptotic variance of a statistic?

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Note: Based on the working below, I think you may have mis-specified the form of the constants $c_i$ that you want to use in your expression. For reasons that will be clear in the solution, I am going to proceed on the basis that:

$$c_i \equiv \sqrt{\frac{n}{m(n-m)(n+1)}} \cdot q_i \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } q_i \equiv F^{-1}(\tfrac{i}{n+1}).$$

If you still want to use your specified values for $c_i$ this is just a simple change of constant multiple, so the corresponding result is easily obtained.


Your problem involves simple-random-sampling without replacement (SRSWOR) of $m$ values from a set of $n$ values. The values $X_1, ..., X_n$ are sampling indicators for each of the $n$ values. Under SRSWOR it can be shown that:

$$\mathbb{E}(X_i) = \frac{m}{n} \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \mathbb{V}(X_i) = \frac{m (n-m)}{n^2} \text{ } \text{ } \text{ } \text{ } \text{ } \text{ } \mathbb{C}(X_i, X_j) = - \frac{1}{n-1} \cdot \frac{ m(n-m)}{n^2},$$

where $i \neq j$ in the latter expression. Hence, the variance of any linear combination of these indicators (with constant weights) is given by:

$$\begin{equation} \begin{aligned} \mathbb{V}(T_n) = \mathbb{V} \Bigg( \sum_{i=1}^n c_i X_i \Bigg) &= \sum_{i} c_i^2 \mathbb{V}(X_i) + \sum_{i \neq j} c_i c_j \mathbb{C}(X_i, X_j) \\[6pt] &= \frac{m(n-m)}{n} \Bigg[ \sum_{i} c_i^2 - \frac{1}{n-1} \sum_{i \neq j} c_i c_j \Bigg]. \\[6pt] \end{aligned} \end{equation}$$

Denoting $q_i \equiv F^{-1}(\tfrac{i}{n+1})$ and substituting the above weight values $c_i$ (which are different to the ones you specified) we obtain:

$$\begin{equation} \begin{aligned} \mathbb{V}(T_n) &= \frac{m(n-m)}{n} \Bigg[ \sum_{i} c_i^2 - \frac{1}{n-1} \sum_{i \neq j} c_i c_j \Bigg] \\[6pt] &= \frac{m(n-m)}{n} \frac{n}{m(n-m)(n+1)} \Bigg[ \sum_{i} q_i^2 - \frac{1}{n-1} \sum_{i \neq j} q_i q_j \Bigg] \\[6pt] &= \sum_{i} \frac{q_i^2}{n+1} - \frac{n+1}{n-1} \sum_{i \neq j} \frac{q_i q_j}{(n+1)^2}. \\[6pt] \end{aligned} \end{equation}$$

This gives a general expression for the variance of interest based on the quantiles $q_1, ...., q_n$. Letting $Q = F^{-1}$ be the quantile function of $F$ we have limit:

$$\begin{equation} \begin{aligned} \lim_{n \rightarrow \infty} \mathbb{V}(T_n) &= \lim_{n \rightarrow \infty} \Bigg[ \sum_{i} \frac{q_i^2}{n+1} - \frac{n+1}{n-1} \sum_{i \neq j} \frac{q_i q_j}{(n+1)^2} \Bigg] \\[6pt] &= \int \limits_0^1 Q(p)^2 dp - \int \limits_0^1 \int \limits_0^1 Q(p)Q(q) dp dq \\[6pt] &= \mathbb{V}(Q(P) | P \sim \text{U}(0,1)) \\[6pt] &= \mathbb{V}(Y | Y \sim F). \\[6pt] \end{aligned} \end{equation}$$

Hence, we see that the limit is the variance of the underlying distribution $F$. (Note that this result follows from my own alternative specification of $c_i$. If you use yours you will get a similar result that is scaled by $n^7 \cdot \theta^2 (1-\theta)^2$ where $\theta = \lim_{n \rightarrow \infty} m/n$, which is a rather odd result.)

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