5
$\begingroup$

Suppose $X_1,X_2,\ldots$ are $m$-dependent random variables. Let $F_i$ be the cdf of $X_i$. Let $F_n(x, \omega)$ be the empirical cdf of $X_1,\ldots,X_n$. What will be the variance of $F_n(x, \omega)$?

  • 0
    Yes and thank you for the clarification.2012-04-30

1 Answers 1

9

Let $Y(\omega) = F_n(x, \omega) = \frac{1}{n} \sum_{k=1}^n I(X_k(\omega) \leqslant x)$. Then, using $\mathbb{E}(I(X_k \leqslant x)) = \mathbb{P}(X_k \leqslant x)$, we have $ \mathbb{E}\left(Y\right) = \frac{1}{n} \sum_{k=1}^n \mathbb{P}(X_k \leqslant x) = \frac{1}{n} \sum_{k=1}^n F_k(x) $ $ \begin{eqnarray} \mathbb{E}(Y^2) &=& \frac{1}{n^2} \sum_{k=1}^n \sum_{\ell=1}^n \mathbb{E}\left( I(X_k \leqslant x) I(X_\ell \leqslant x) \right) \\ &=& \frac{1}{n^2} \sum_{k=1}^n \sum_{\ell=1}^n \mathbb{P}(X_k \leqslant x, X_\ell \leqslant x) \end{eqnarray} $ Therefore: $\begin{eqnarray} \mathbb{Var}(Y) &=& \frac{1}{n^2} \sum_{k=1}^n \sum_{\ell=1}^n \left( \mathbb{P}(X_k \leqslant x, X_\ell \leqslant x) - F_{X_k}(x) F_{X_\ell}(x) \right) \\ &=& \frac{1}{n^2} \sum_{k=1}^n F_{X_k}(x) \left(1-F_{X_k}(x)\right) + \frac{2}{n^2} \sum_{1 \leqslant k < \ell \leqslant n} \left( F_{X_k,X_\ell}(x,x) - F_{X_k}(x) F_{X_\ell}(x) \right) \end{eqnarray}$

For the case of independent variables in the sample, we get $ \mathbb{Var}(Y_\text{indep}) = \frac{1}{n^2} \sum_{k=1}^n F_{X_k}(x) \left(1-F_{X_k}(x)\right) $ For the case of identically distributed: $ \mathbb{Var}(Y_\text{i.i.d.}) = \frac{1}{n} F_X(x) (1-F_X(x)) $

  • 0
    Thanks, pretty clear explanation.2017-05-24