M-estimation using unbiased median variance estimate

Abstract

This paper first proves that the traditional median variance estimate is biased when the sample number is small and then proposes an unbiased median variance estimate to calibrate for the bias of the variance estimate. The scaled median variance estimate is firstly derived, and the unbiased median variance estimate is formed with independent residuals in an adjustment model no matter whether the measurements are contaminated by outliers or not. Using the unbiased median variance estimate, the M-estimate is constructed to mitigate for the biases caused by the variance estimate. The IGGIII reduction factor is used to verify the proposed algorithms by a levelling network example. Numerical analysis confirms that the proposed median variance estimate can achieve better unbiasedness for contaminated measurement set, but the dispersion of our estimate is unfortunately larger than that for the least-squares estimate.

Appendix

The detailed proof for Eq. (12) is as follows.

Letting $x = \frac{{\left| { \varepsilon_{1} } \right|}}{2}$ and $y = \frac{{\left| { \varepsilon_{2} } \right|}}{2}$, from Eqs. (1) and (9); it is deduced that the corresponding PDF is

$$ f\left( x \right) = \frac{4}{{\sqrt {2\pi } \sigma_{0} }}\exp \left\{ { - \frac{{2x^{2} }}{{\sigma_{0}^{2} }}} \right\}, \quad x \ge 0 $$

(71)

and

$$ f\left( y \right) = \frac{4}{{\sqrt {2\pi } \sigma_{0} }}\exp \left\{ { - \frac{{2y^{2} }}{{\sigma_{0}^{2} }}} \right\}, \quad y \ge 0 $$

(72)

Letting $z = x + y$, since variable $x$ and $y$ are independent, there is

$$ \begin{aligned} & f\left( z \right) = \mathop \smallint \limits_{ - \infty }^{ + \infty } f\left( x \right)f\left( {z - x} \right){\text{d}}x = \mathop \smallint \limits_{0}^{ + \infty } \frac{8}{{\pi \sigma_{0}^{2} }}\exp \left\{ { - \frac{{2x^{2} + 2\left( {z - x} \right)^{2} }}{{\sigma_{0}^{2} }}} \right\}{\text{d}}x \\ & \quad = \frac{8}{{\pi \sigma_{0}^{2} }}\exp \left\{ { - \frac{{z^{2} }}{{\sigma_{0}^{2} }}} \right\}\mathop \smallint \limits_{0}^{ + \infty } \exp \left\{ { - \frac{{4\left( {x - \frac{z}{2}} \right)^{2} }}{{\sigma_{0}^{2} }}} \right\}{\text{d}}x \\ & \quad = \frac{4\sqrt 2 }{{\sqrt \pi \sigma_{0} }}\exp \left\{ { - \frac{{z^{2} }}{{\sigma_{0}^{2} }}} \right\}\frac{1}{{\sqrt {2\pi } \frac{{\sigma_{0} }}{2}}}\mathop \smallint \limits_{0}^{ + \infty } \exp \left\{ { - \frac{{\left( {x - \frac{z}{2}} \right)^{2} }}{{\left( {\frac{{\sigma_{0} }}{2}} \right)^{2} }}} \right\}{\text{d}}x \\ & \quad = \frac{4\sqrt 2 }{{\sqrt \pi \sigma_{0} }}\exp \left\{ { - \frac{{z^{2} }}{{\sigma_{0}^{2} }}} \right\}\frac{1}{{\sqrt {2\pi } }}\mathop \smallint \limits_{0}^{ + \infty } e^{{ - t^{2} }} {\text{d}}t \\ \end{aligned} $$

(73)

Since

$$ \frac{1}{{\sqrt {2\pi } }}\mathop \smallint \limits_{0}^{ + \infty } e^{{ - t^{2} }} {\text{d}}t = \frac{1}{2} $$

(74)

Substituting Eq. (74) into Eq. (73), it is deduced

$$ f\left( z \right) = \frac{4}{{\sqrt {2\pi } \sigma_{0} }}\exp \left\{ { - \frac{{z^{2} }}{{\sigma_{0}^{2} }}} \right\} $$

(75)

Therefore, the expectation of $z$ is

$$ \begin{aligned} & E\left( z \right) = \mathop \smallint \limits_{ - \infty }^{ + \infty } zf\left( z \right){\text{d}}z = \frac{4}{{\sqrt {2\pi } }}\mathop \smallint \limits_{0}^{ + \infty } \frac{z}{{\sigma_{0} }}\exp \left\{ { - \frac{{z^{2} }}{{\sigma_{0}^{2} }}} \right\}{\text{d}}z \\ & \quad = \frac{{4\sigma_{0} }}{{\sqrt {2\pi } }}\mathop \smallint \limits_{0}^{ + \infty } \frac{z}{{\sigma_{0} }}\exp \left\{ { - \frac{{z^{2} }}{{\sigma_{0}^{2} }}} \right\}d\frac{z}{{\sigma_{0} }} \\ & \quad = \frac{{4\sigma_{0} }}{{\sqrt {2\pi } }}\mathop \smallint \limits_{0}^{ + \infty } te^{{ - t^{2} }} {\text{d}}t = \frac{{2\sigma_{0} }}{{\sqrt {2\pi } }}\varGamma \left( 1 \right) = \sqrt {\frac{2}{\pi }} \sigma_{0} \\ \end{aligned} $$

(76)