an unbiased estimator of the Hellinger distance?

Here is a question I posted on Stack Exchange a while ago:

In a setting where one observes X1,…,Xn distributed from a distribution with (unknown) density f, I wonder if there is an unbiased estimator (based on the Xi‘s) of the Hellinger distance to another distribution with known density f0, namely

\mathfrak{H}(f,f_0)=\left\{1-\int\sqrt{f_0(x)/(x)}\text{d}x\right\}^{1/2}
Now, Paulo has posted an answer that is rather interesting, if formally “off the point”. There exists a natural unbiased estimator of if not of H, based on the original sample and using the alternative representation
\mathfrak{H}^2(f,f_0)=1-\mathbb{E}_f[\sqrt{f_0(X)/f(X)}]

for the Hellinger distance. In addition, this estimator is guaranteed to enjoy a finite variance since

\mathbb{E}_f[\sqrt{f_0(X)/f(X)}^2]=1\,.

Considering this question again, I am now fairly convinced there cannot be an unbiased estimator of H, as it behaves like a standard deviation for which there usually is no unbiased estimator!

3 Responses to “an unbiased estimator of the Hellinger distance?”

  1. […] Acabei de descobrir que não existe estimador não-viesado para o desvio-padrão. Sempre aprendi que havia um estimador não-viesado para a variância. Que os livros (e os professores)  silenciassem sobre um estimador não-viesado para o desvio-padrão nunca me chamou a atenção. Afinal, parecia natural que haveria um estimador não-viesado para o desvio padrão: a raiz quadrada da variância amostral. Porém, isso não funciona. Felizmente sou Bayesiano e não me preocupo com o viés. […]

  2. You intuition is probably right. We should have a proof of that.

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