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?


1 Answer 1


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.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.