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The weighted empirical distribution function estimator is given by:

$\hat{F}(x)=\frac{1}{\sum_{i=1}^{n}w_{i}}\sum_{i=1}^{n}w_{i}I(X_{i}\leq x)$

see also here: http://www.okstate.edu/sas/v8/sashtml/insight/chap38/sect25.htm

I fail to find proofs/discussions of the properties of this estimator. Is this estimator an unbiased estimator of $F(x)$? Or at least asymptotically unbiased? Or maybe consistent? Why? What assumptions are needed in order for it to be? You can assume anything you want, I am looking for a discussion.

Thanks.

leo
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2 Answers2

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If $w_i \geq 0$ with atleast one $w_i >0$, $X_i \sim F$ are identically distributed but not necessarily independent then your estimator is unbiased

$$ \begin{align} \mathbb{E}[\hat{F}(x)] &=\sum_{i=1}^{n}\frac{w_i}{\sum_{i=1}^n w_i}\mathbb{E}[I(X_{i}\leq x)] \\ &= \sum_{i=1}^{n}\frac{w_i}{\sum_{i=1}^n w_i}F(x) \\ &= F(x) \end{align} $$

The first line comes form linearity of expectation operator, second as the random variables are identically distributed and last because the sum of re-weighted weights is 1.

Kumar
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Couple things to point out.

1) Unbiasedness is for the estimator to some parameter of the underlying distribution. I’m not sure what you mean by “unbiased” distribution function. However, the plain vanilla empirical CDF converges to “true” CDF under SLLN.

2) The formula you listed is not correct. You wanted to use different subscript for weights and the random variable.

williamdotyang
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