Czy istnieje obiektywny estymator odległości Hellingera między dwiema dystrybucjami?

W otoczeniu, w którym obserwujemy $X_1,\ldots,X_n$ rozproszone z rozkładu o gęstości $f$ , zastanawiam się, czy istnieje obiektywny estymator (oparty na $X_i$ ) odległości Hellingera do innego rozkładu o gęstości $f_0$ , mianowicie

$$\mathfrak{H}(f,f_0) = \left\{ 1 - \int_\mathcal{X} \sqrt{f(x)f_0(x)} \text{d}x \right\}^{1/2}\,.$$

Dla f nie istnieje żaden obiektywny estymator ani ani H$\mathfrak{H}$$\mathfrak{H}^2$$f$ z dowolnego stosunkowo szerokim nieparametrycznego klasy dystrybucji.

Możemy to pokazać za pomocą pięknie prostego argumentu

Bickel and Lehmann (1969). Bezstronna ocena w rodzinach wypukłych . The Annals of Mathematical Statistics, 40 (5) 1523–1535. ( euclid projektu )

Napraw niektóre rozkłady , F i G o odpowiednich gęstościach f 0 , f i g . Niech H ( F ) oznaczają H ( f , f 0 ) , niech H ( X ) jest nieco estymatorem H ( F ), na podstawie n próbek IID x ı ~ F .$F_0$$F$$G$$f_0$$f$$g$$H(F)$$\mathfrak{H}(f, f_0)$$\hat H(\mathbf X)$$H(F)$$n$$X_i \sim F$

Załóżmy, że H jest obiektywne dla próbek z każdego podziału formy M alfa : = α F + ( 1 - α ) G . Ale potem Q ( α )$\hat H$

$$M_\alpha := \alpha F + (1 - \alpha) G .$$ $$\begin{align} Q(\alpha) &= H(M_\alpha) \\&= \int_{x_1} \cdots \int_{x_n} \hat H(\mathbf X) \,\mathrm{d}M_\alpha(x_1) \cdots\mathrm{d}M_\alpha(x_n) \\&= \int_{x_1} \cdots \int_{x_n} \hat H(\mathbf X) \left[ \alpha \mathrm{d}F(x_1) + (1-\alpha) \mathrm{d}G(x_1) \right] \cdots \left[ \alpha \mathrm{d}F(x_n) + (1-\alpha) \mathrm{d}G(x_n) \right] \\&= \alpha^n \operatorname{\mathbb{E}}_{\mathbf X \sim F^n}[ \hat H(\mathbf X)] + \dots + (1 - \alpha)^n \operatorname{\mathbb{E}}_{\mathbf X \sim G^n}[ \hat H(\mathbf X)] ,\end{align}$$ so that $Q(\alpha)$ must be a polynomial in $\alpha$ of degree at most $n$.

Now, let's specialize to a reasonable case and show that the corresponding $Q$ is not polynomial.

Let $F_0$ be some distribution which has constant density on $[-1, 1]$: $f_0(x) = c$ for all $\lvert x \rvert \le 1$. (Its behavior outside that range doesn't matter.) Let $F$ be some distribution supported only on $[-1, 0]$, and $G$ some distribution supported only on $[0, 1]$.

Now

$$\begin{align} Q(\alpha) &= \mathfrak{H}(m_\alpha, f_0) \\&= \sqrt{1 - \int_{\mathbb R} \sqrt{m_\alpha(x) f_0(x)} \mathrm{d}x} \\&= \sqrt{1 - \int_{-1}^0 \sqrt{c \, \alpha f(x)} \mathrm{d}x - \int_{0}^1 \sqrt{c \, (1 - \alpha) g(x)} \mathrm{d}x} \\&= \sqrt{1 - \sqrt{\alpha} B_F - \sqrt{1 - \alpha} B_G} ,\end{align}$$ where $B_F := \int_{\mathbb R} \sqrt{f(x) f_0(x)} \mathrm{d}x$ and likewise for $B_G$. Note that $B_F > 0$, $B_G > 0$ for any distributions $F$, $G$ which have a density.

$\sqrt{1 - \sqrt{\alpha} B_F - \sqrt{1 - \alpha} B_G}$ is not a polynomial of any finite degree. Thus, no estimator $\hat H$ can be unbiased for $\mathfrak{H}$ on all of the distributions $M_\alpha$ with finitely many samples.

Likewise, because $1 - \sqrt{\alpha} B_F - \sqrt{1 - \alpha} B_G$ is also not a polynomial, there is no estimator for $\mathfrak{H}^2$ which is unbiased on all of the distributions $M_\alpha$ with finitely many samples.

This excludes pretty much all reasonable nonparametric classes of distributions, except for those with densities bounded below (an assumption nonparametric analyses sometimes make). You could probably kill those classes too with a similar argument by just making the densities constant or something.

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 $$\hat{H}=\sqrt{1 - \frac{1}{n} \sum_{i=1}^n \sqrt{\frac{f_0(X_i)}{\hat{f_n}(X_i)}}} \, .$$