# Is there an unbiased estimator of an Hellinger distance?

In a setting where one observes $X_1,\ldots,X_n$ distributed from a distribution with density $f$, I wonder if there is an unbiased estimator (based on the $X_i$'s) of the Hellinger distance to another distribution with density $f_0$, namely $$\mathfrak{H}(f,f_0) = \left\{ 1 - \int_\mathcal{X} \sqrt{f(x)f_0(x)} \text{d}x \right\}^{1/2}\,.$$

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So f0 is known and fixed. But is f known or from a parametric family or are doing this in a nonparametric framework with all you know about f coming from your sample? I think it makes a difference when attempting an answer. –  Michael Chernick Jun 1 '12 at 10:16
@MichaelChernick: assume all you know about $f$ is the sample $X_1,\ldots,X_n$. –  Xi'an Jun 1 '12 at 10:57
I do not think it has been calculated (if there exists). If there exists, then AIC has a lost brother. –  user10525 Jun 1 '12 at 12:49
I think I am in agreement with Procrastinator. I haven't got any idea about how this could be done nonparametrically. You could use a kernel density estimate of f. But the density surely could not be unbiased for all x. So how could you possibly choose a kernel that would make that function an unbiased estimator of the distance. This is even worse of a problem if f and f0 have unbounded range because your data gives you no information on the very extreme tails which could still play a role in the calculation of the integral. I can't prove that it is impossible but I think it is! –  Michael Chernick Jun 1 '12 at 14:48
An attack on this problem looks feasible if you assume $f$ and $f_0$ are discrete. This leads to an obvious estimator (compute the Hellinger distance between the EDF and $f_0$). Bootstrapping (theoretically, not via simulation!) will give us a handle on the possible bias as well as a way to reduce (or even eliminate) the bias. I hold out some hope to succeed with the squared distance rather than the distance itself, because it is mathematically more tractable. The assumption of a discrete $f$ is no problem in applications; the space of discrete $f$ is a dense subset anyway. –  whuber Jun 1 '12 at 16:00
I don't know how to construct (if it exists) an unbiased estimator of the Hellinger distance. It seems possible to construct a consistent estimator. We have some fixed known density $f_0$, and a random sample $X_1,\dots,X_n$ from a density $f>0$. We want to estimate $$H(f,f_0) = \sqrt{1 - \int_\mathscr{X} \sqrt{f(x)f_0(x)}\,dx} = \sqrt{1 - \int_\mathscr{X} \sqrt{\frac{f_0(x)}{f(x)}}\;\;f(x)\,dx}$$ $$= \sqrt{1 - \mathbb{E}\left[\sqrt{\frac{f_0(X)}{f(X)}}\;\;\right] }\, ,$$ where $X\sim f$. By the SLLN, we know that $$\sqrt{1 - \frac{1}{n} \sum_{i=1}^n \sqrt{\frac{f_0(X_i)}{f(X_i)}}} \quad \rightarrow H(f,f_0) \, ,$$ almost surely, as $n\to\infty$. Hence, a resonable way to estimate $H(f,f_0)$ would be to take some density estimator $\hat{f_n}$ (such as a traditional kernel density estimator) of $f$, and compute $$\hat{H}=\sqrt{1 - \frac{1}{n} \sum_{i=1}^n \sqrt{\frac{f_0(X_i)}{\hat{f_n}(X_i)}}} \, .$$
@Zen: Good point! I consider this answer as the answer because it made me realise $H$ sounds very much like a standard deviation, for which there exists no unbiased estimator. As for the variance of $\hat H^2_n$, no worries: $\mathbb{E}[(\sqrt{f_0(X)/f(X)})^2]=1$ implies that this estimator has a finite variance. –  Xi'an Oct 19 '12 at 5:42