## The Annals of Mathematical Statistics

### The Order of the Minimum Variance in a Non-Regular Case

Thomas Polfeldt

#### Abstract

Let $f(y)$ be a probability density function on the real line, and $F(y)$ the corresponding distribution function. It is assumed that \begin{equation*}\tag{1} F(y) = 0 \text{for} y \leqq 0,\quad F(y) > 0 \text{for} y > 0.\end{equation*} Let $\theta$ be a location parameter, and let $X = (x_1, \cdots, x_n)$ denote a sample of $n$ independent observations, with each $x_i$ distributed according to $F(x - \theta)$. In this paper, we study the minimum variance of unbiased estimators $t = t(X)$ of $\theta$, with special reference to the order, in $n$, of that variance. For example, if $F(y) = 1 - e^{-y}(y > 0)$, the minimum variance is $n^{-2}$ rather than of order $n^{-1}$ as in regular cases. We shall state conditions on $f(y)$ which determine this order. One of these is that $f(y)$ varies regularly at zero with exponent $c - 1 (c > 0)$ (cf. , chapter 8, sect. 8-9). Under the conditions imposed, the smallest attainable variance order is $n^{-1}$ if $c > 2$, but $(F^{-1}(n^{-1}))^2$ if $0 < c < 2$. The case $c = 2$ has special features. Since $F(y)$ varies regularly with exponent $c$, the minimum variance order will be $n^{-2/c}L(n^{-1})$ with slowly varying $L (0 < c < 2$; also true for $c = 2)$. When $c > \frac{1}{2}$, the Chapman and Robbins inequality  is used to obtain a lower bound for the minimum variance. For $0 < c \leqq \frac{1}{2}$, a new inequality is used; we then restrict slightly the class of unbiased estimators. The results carry over, of course, to distributions with $F(y) < 1$ for $y < 0, F(y) = 1$ for $y \geqq 0$. A generalization to biased estimators (or to mean square error) is straightforward, but some conditions on the bias function will be necessary. Some questions recently raised by Blischke et al.  are answered by the theorems. The conditions imposed here may probably be relaxed to some extent. Notation. $K$ and $K'$ denote positive, finite constants. If there exist $K$ and $K'$ such that $K < a(x)/b(x) < K'(|x| < x_0)$, we shall write $a(x) = \Omega(b(x)) (x \rightarrow 0)$. The qualification $(x \rightarrow 0)$ will often be omitted.

#### Article information

Source
Ann. Math. Statist., Volume 41, Number 2 (1970), 667-672.

Dates
First available in Project Euclid: 27 April 2007

https://projecteuclid.org/euclid.aoms/1177697111

Digital Object Identifier
doi:10.1214/aoms/1177697111

Mathematical Reviews number (MathSciNet)
MR256499

Zentralblatt MATH identifier
0193.47302

JSTOR