Formulas for precisely and efficiently estimating the bias and variance of the length measurements

Abstract

Error analysis in length measurements is an important problem in geographic information system and cartographic operations. The distance between two random points—i.e., the length of a random line segment—may be viewed as a nonlinear mapping of the coordinates of the two points. In real-world applications, an unbiased length statistic may be expected in high-precision contexts, but the variance of the unbiased statistic is of concern in assessing the quality. This paper suggesting the use of a k-order bias correction formula and a nonlinear error propagation approach to the distance equation provides a useful way to describe the length of a line. The study shows that the bias is determined by the relative precision of the random line segment, and that the use of the higher-order bias correction is only needed for short-distance applications.

Appendix: k-Order approximation about the expectation of the length statistic

1.1 De-correlation on the variance matrix of the positional error

To simplify the discussion, the correlation between $\varepsilon_{i}$ and $\varepsilon_{j}$ of ${\varvec{\upvarepsilon}}$ for $i \ne j$ is reduced by applying the spectral decomposition as follows (Seber 2008)

$$ {\varvec{\Lambda}} = {\mathbf{S}}^{\text{T}} {\mathbf{DS}}: = {\text{diag(}}\lambda_{ 1} ,\lambda_{ 2} ,\ldots ,\lambda_{m} ) $$

(37)

where S is the $(m,m)$ orthogonal matrix, and $\lambda_{ 1} \ge \lambda_{ 2} \ge ,\ldots ,\ge \lambda_{m}$ are the eigenvalues. That is, for the random variable ${\varvec{\varepsilon}}^{\prime}$ given by the coordinate transformation

$$ \left\{ \begin{aligned} {\varvec{\varepsilon}}^{\prime} & = {\mathbf{S}}^{\text{T}} {\varvec{\upvarepsilon}} \\ {{\tilde{\varvec{\Delta}^{\prime}}}} & = {\mathbf{S}}^{\text{T}} {\tilde{\varvec{\Delta }}} \\ \end{aligned} \right. $$

(38)

we have

$$ \left\{ \begin{aligned} & E({\varvec{\varepsilon}}^{\prime}) = {\mathbf{0}} \\ & E({\varvec{\varepsilon}}^{\prime}{\varvec{\varepsilon}}^{\prime\text{T}} ) = {\varvec{\Lambda}}. \\ \end{aligned} \right. $$

(39)

Particularly for the normal distribution (11), $\text{cov} (\varepsilon^{\prime}_{i} ,\varepsilon^{\prime}_{j} ) = 0$ can result in

$$ E\left( {\varepsilon_{i}^{{{\prime }k_{1} }} \varepsilon_{j}^{{{\prime }k_{2} }} } \right) = E\left( {\varepsilon_{i}^{{{\prime }k_{1} }} } \right)E\left( {\varepsilon_{j}^{{{\prime }k_{2} }} } \right) $$

(40)

for arbitrary k ₁, $k_{2} \in {\mathbb{Z}}$.

Because the length statistic given by Eq. (7) is invariant under the transformation given by Eq. (38), we have

$$ d({\varvec{\Delta}}) = d({\tilde{\varvec{\Delta }}}^{{\prime }} ) = \sqrt {\tilde{d}^{2} + 2{\tilde{\varvec{\Delta }}}^{{{\prime }T}} {\varvec{\upvarepsilon}}^{{\prime }} + {\varvec{\upvarepsilon}}^{{{\prime }T}} {\varvec{\upvarepsilon}}^{{\prime }} } . $$

(41)

As indicated by this equation, there is no loss of generality in assuming that the random vector components are expressed in the principal axis frame. That is, the covariance matrix is diagonal with nonzero diagonal elements.

1.2 k-Order Taylor appropriation of the length statistic

To simplify the notation in the following discussion, let

$$ \left\{ \begin{aligned} z_{1} & = 2{\tilde{\varvec{\Delta }}}^{{\prime {\text{T}}}} {\varvec{\upvarepsilon}}^{\prime } = 2\tilde{\Delta }_{1}^{{\prime }} \varepsilon_{1}^{{\prime }} + 2\tilde{\Delta }_{2}^{{\prime }} \varepsilon_{2}^{{\prime }} + \cdots + 2\tilde{\Delta }_{m}^{{\prime }} \varepsilon_{m}^{{\prime }} \\ z_{2} & = {\varvec{\upvarepsilon}}^{{\prime {\text{T}}}} {\varvec{\upvarepsilon}}^{{\prime }} = \varepsilon_{1}^{{{\prime }2}} + \varepsilon_{2}^{{{\prime }2}} + \cdots + \varepsilon_{m}^{{{\prime }2}} \\ \end{aligned} \right. $$

(42)

then Eq. (41) reads as

$$ d({\varvec{\Delta^{\prime}}}) = \sqrt {\tilde{d}^{2} + z_{1} + z_{2} } = \tilde{d}\sqrt {1 + \frac{{z_{1} + z_{2} }}{{\tilde{d}^{2} }}} . $$

(43)

Applying the binomial series

$$ \sqrt {1 + x} = 1 + \frac{1}{2}x - \frac{1}{2 \cdot 4}x^{2} + \frac{1 \cdot 3}{2 \cdot 4 \cdot 6}x^{3} - \cdots $$

(44)

[cf. Spiegel (1968, p. 110)] to Eq. (43) about $z_{1} = 0, z_{2} = 0$, we have (Spiegel 1968)

$$ d({\varvec{\Delta^{\prime}}}) = \tilde{d} + \sum\limits_{i = 1}^{k} {\frac{{( - 1)^{i - 1} }}{{\tilde{d}^{2i - 1} }}\frac{(2i - 3)!!}{( 2i)!!}\left( {z_{1} + z_{2} } \right)^{i} } + s^{k} (\theta {\mathbf{z}}) $$

(45)

where ${\mathbf{z}} = \left[ {z_{1} , z_{2} } \right]^{\text{T}}$, $(2i - 3)!! = (2i - 3)(2i - 5) \ldots 1$, $( 2i)!! = ( 2i)( 2i - 2) \ldots 2$, $- 1!! = 1$, and

$$ s^{k} (\theta {\mathbf{z}}) = \frac{{( - 1)^{k + 1} (2k - 3)!!}}{{2\left( {k + 1} \right)!!d^{2k + 1} (\theta {\mathbf{z}})}} $$

(46)

with $\theta \in \left( {0,1} \right)$.

Ignoring the Eq. (46), we can obtain the k-order Taylor expansion of Eq. (43) as follows:

$$ d({\varvec{\Delta^{\prime}}}) \approx \tilde{d} + \sum\limits_{i = 1}^{k} {\frac{{( - 1)^{i - 1} }}{{\tilde{d}^{2i - 1} }}\frac{(2i - 3)!!}{( 2i)!!}\left( {z_{1} + z_{2} } \right)^{i} } . $$

(47)

The approximation given by Eq. (47) is a good basis for estimating the bias in the length statistic of the random line segment.

1.3 k-Order approximation about the expectation of the length statistic

By the approximation (47), the expectation given by Eq. (10) can be approximated as follows:

$$ \begin{aligned} M & \approx M_{k} : = \tilde{d} + \sum\limits_{i = 1}^{k} {\frac{{( - 1)^{i - 1} }}{{\tilde{d}^{2i - 1} }}\frac{(2i - 3)!!}{( 2i)!!}E\left( {z_{1} + z_{2} } \right)^{i} } \\ & = \tilde{d} + \sum\limits_{i = 1}^{k} {\frac{{( - 1)^{i - 1} }}{{\tilde{d}^{2i - 1} }}\frac{(2i - 3)!!}{( 2i)!!}\left( {E(z_{1}^{i} ) + \sum\limits_{{1 \le k_{1} \le i - 1, k_{1} + k_{2} = i}} {C_{i}^{{k_{1} }} E(z_{1}^{{k_{1} }} z_{2}^{{k_{2} }} )} + E(z_{2}^{i} )} \right)} \\ \end{aligned} $$

(48)

where $E(z_{1}^{i} )$, $E(z_{1}^{{k_{1} }} z_{2}^{{k_{2} }} )$ and $E(z_{2}^{i} )$ are the expectations of $z_{1}^{i}$, $z_{1}^{{k_{1} }} z_{2}^{{k_{2} }}$ and $z_{2}^{i}$, respectively, $C_{i}^{{k_{1} }} = {{i\,!} \mathord{\left/ {\vphantom {{i!} {\left[ {k_{1} !(i - k_{1} )!} \right]}}} \right. \kern-0pt} {\left[ {k_{1} !(i - k_{1} )!} \right]}}$ is the number of all combinations.

To calculate $E(z_{2}^{i} )$ in Eq. (48), by connecting this problem with the Chi-square distribution, following Xue et al. (2015)

$$ E(\varepsilon_{j}^{{{\prime }2i}} ) = (2i - 1)(2i - 3) \cdots 1 \cdot \lambda_{j}^{i} = (2i - 1)!!\lambda_{j}^{i} \quad j = 1,2, \ldots, m $$

(49)

we have

$$ \begin{aligned} E\left( {z_{2}^{i} } \right) & = E\left( {\varepsilon_{1}^{{{\prime }2}} + \varepsilon_{2}^{{{\prime }2}} + \cdots + \varepsilon_{m}^{{{\prime }2}} } \right)^{i} \\ & = \left\{ {E\left[ {(\varepsilon_{1}^{{{\prime }2}} + \varepsilon_{2}^{{{\prime }2}} + \cdots + \varepsilon_{m}^{{{\prime }2}} )^{i} } \right] + \sum\limits_{i = 1}^{m} {\left[ {E(\varepsilon_{i}^{{{\prime }2}} )} \right]^{i} } } \right\} - \sum\limits_{i = 1}^{m} {\left[ {E(\varepsilon_{i}^{{{\prime }2}} )} \right]^{i} } \\ & = \left\{ {\left[ {E(\varepsilon_{1}^{{{\prime }2}} ) + E(\varepsilon_{2}^{{{\prime }2}} ) + \cdots + E(\varepsilon_{n}^{{{\prime }2}} )} \right]^{i} + \sum\limits_{i = 1}^{m} {E(\varepsilon_{i}^{{{\prime }2i}} )} } \right\} - \sum\limits_{i = 1}^{m} {\left[ {E(\varepsilon_{i}^{{{\prime }2}} )} \right]^{i} } \\ & = \left[ {(2i - 1)!! - 1} \right]tr({\mathbf{D}}^{i} ) + \left[ {tr({\mathbf{D}})} \right]^{i} \\ \end{aligned} $$

(50)

where $tr( \cdot )$ is the trace of the matrix.

Before calculating $E(z_{1}^{i} )$ and $E(z_{1}^{{k_{1} }} z_{2}^{{k_{2} }} )$ in Eq. (48), the variance of z ₁ must be computed first as follows

$$ \sigma_{{z_{1} }}^{2} : = E(z_{1}^{2} ) = 4{{\tilde{\varvec{\Delta}^{\prime}}}}^{\text{T}} {\varvec{\Lambda}} \tilde{\varvec\Delta^{\prime}} = 4{\varvec{\Delta}}^{\text{T}} {\mathbf{S}}{\varvec{\Lambda}}{\mathbf{S}}^{\text{T}} {\varvec{\Delta}} = 4\tilde{d}^{2} {{\tilde{\bf e}{\bf D}\tilde{\bf e}}}^{\text{T}} = 4\tilde{d}^{2} \sigma_{{{\tilde{\mathbf{e}}}}}^{2} $$

(51)

where $\sigma_{{{\tilde{\mathbf{e}}}}}^{2}$ is the directional variance. With regard to Eqs. (50) and (51), we then get

$$ E\left( {z_{1}^{i} } \right) = \left\{ {\begin{array}{*{20}l} 0 \hfill & {{\text{for }}i = 2s + 1} \hfill \\ {(i - 1)!!2^{i} \tilde{d}^{i} \sigma_{{{\tilde{\mathbf{e}}}}}^{i} } \hfill & {{\text{for }}i = 2s} \hfill \\ \end{array} } \right. $$

(52)

and

$$ \begin{aligned} E\left( {z_{1}^{{k_{1} }} z_{2}^{{k_{2} }} } \right) & = E\left[ {\left( {\Delta_{1}^{{\prime }} \varepsilon_{1}^{{\prime }} + \Delta_{2}^{{\prime }} \varepsilon_{2}^{{\prime }} + \Delta_{3}^{{\prime }} \varepsilon_{3}^{{\prime }} } \right)^{{k_{1} }} \left( {\varepsilon_{1}^{{{\prime }2}} + \varepsilon_{2}^{{{\prime }2}} + \varepsilon_{3}^{{{\prime }2}} } \right)^{{k_{2} }} } \right] \\ & = E\left[ {\sum\limits_{\begin{subarray}{l} 1 \le r, s \le k_{1}, r + s \le k_{1} \\ 1 \le t, q \le k_{2}, t + q \le k_{2} \end{subarray} } {C_{{k_{1} }}^{r} \left( {\Delta_{1}^{{{\prime }r}} \varepsilon_{1}^{{{\prime }r}} } \right)C_{{k_{1} - r}}^{s} \left( {\Delta_{2}^{{{\prime }s}} \varepsilon_{2}^{{{\prime }s}} } \right)\left( {\Delta_{3}^{{{\prime }k_{1} - r - s}} \varepsilon_{3}^{{{\prime }k_{1} - r - s}} } \right)C_{{k_{2} }}^{t} \left( {\varepsilon_{1}^{{{\prime }2t}} } \right)C_{{k_{2} - t}}^{q} \left( {\varepsilon_{2}^{{{\prime }2q}} } \right)\left( {\varepsilon_{3}^{{{\prime }2(k_{2} - t - q)}} } \right)} } \right] \\ & = \sum\limits_{\begin{subarray}{l} 1 \le r, s \le k_{1}, r + s \le k_{1} \\ 1 \le t, q \le k_{2}, t + q \le k_{2} \end{subarray} } {C_{{k_{1} }}^{r} \left( {\Delta_{1}^{{{\prime }r}} E(\varepsilon_{1}^{{{\prime }r}} )} \right)C_{{k_{1} - r}}^{s} \left( {\Delta_{2}^{{{\prime }s}} E(\varepsilon_{2}^{{{\prime }s}} )} \right)\left( {\Delta_{3}^{{{\prime }k_{1} - r - s}} E(\varepsilon_{3}^{{{\prime }k_{1} - r - s}} )} \right)C_{{k_{2} }}^{t} \left( {E(\varepsilon_{1}^{{{\prime }2t}} )} \right)C_{{k_{2} - t}}^{q} \left( {E(\varepsilon_{2}^{{{\prime }2q}} )} \right)\left( {E(\varepsilon_{3}^{{{\prime }2(k_{2} - t - q)}} )} \right)} \\ & = E\left( {\Delta_{1}^{{\prime }} \varepsilon_{1}^{{\prime }} + \Delta_{2}^{{\prime }} \varepsilon_{2}^{{\prime }} + \Delta_{3}^{{\prime }} \varepsilon_{3}^{{\prime }} } \right)^{{k_{1} }} E\left( {\varepsilon_{1}^{{{\prime }2}} + \varepsilon_{2}^{{{\prime }2}} + \varepsilon_{3}^{{{\prime }2}} } \right)^{{k_{2} }} \\ & = E(z_{1}^{{k_{1} }} )E(z_{2}^{{k_{2} }} ) \\ \end{aligned} $$

(53)

which can be further simplified by applying Eqs. (50) and (52) to (53). That is,

$$ E\left( {z_{1}^{{k_{1} }} z_{2}^{{k_{2} }} } \right) = \left\{ {\begin{array}{*{20}l} 0 \hfill & {{\text{for }}k_{1} = 2s + 1} \hfill \\ {\left\{ {(k_{1} - 1)!!2^{{k_{1} }} \tilde{d}^{{k_{1} }} \sigma _{{{{\tilde{\bf e}}}}}^{{k_{1} }} } \right\}\left\{ {\left[ {(2k_{2} - 1)!! - 1} \right]tr\left( {{\mathbf{D}}^{{k_{2} }} } \right) + \left[ {tr\left( {\mathbf{D}} \right)} \right]^{{k_{2} }} } \right\}} \hfill & {{\text{for }}k_{1} = 2s.} \hfill \\ \end{array} } \right. $$

(54)