On the Statistical Significance Test for the Procedure of Polarity Classification by Types of Acoustic Emission Sources

Abstract—Using a mathematical statistics approach, we review the procedure for type classification of acoustic emission (AE) events into shear, tension, and collapse, proposed by Zang et al. (1998). The procedure is based on counting the signs of first pulses of waves arriving at acoustic sensors and is widely used in rock physics experiments. Under the assumption that the determination errors of first-pulse signs at sensors have uniform and independent distribution , the statistical significance and power of the type separation test are evaluated for a given number of sensors used. We consider and compare three methods of the construction of a statistical test based on the P-value approach and symmetric and asymmetric statistical hypothesis tests. Considering the results of the statistical study, we propose some practical recommendations for selecting a threshold to classify AE event types in experimental studies.

APPENDIX

From formula (7),

$$\begin{gathered} ~\beta = P\left( {{{m}_{0}} < {{\eta }} < n - {{m}_{0}}{\text{|}}{{H}_{1}}} \right) \\ = \mathop \sum \limits_{k = {{m}_{0}} + 1}^{n - \left( {{{m}_{0}} + 1} \right)} C_{n}^{k}{{p}^{k}}{{\left( {1 - p} \right)}^{{n - k}}}. \\ \end{gathered} $$

For convenience, we denote $m = \left( {{{m}_{0}} + 1} \right)$, then

$$~\beta = \mathop \sum \limits_{k = m}^{n - m} C_{n}^{k}{{p}^{k}}{{\left( {1 - p} \right)}^{{n - k}}}.$$

Let us find the maximum as a function of p. Note that β has a meaning of probability, therefore, $0 \leqslant \beta \leqslant 1$. At $p = 0$ and $p = 1$ , we obtain $\beta = 0$. As β is not an identical zero at $0 < p < 1$, from the condition $\frac{{\partial \beta }}{{\partial p}} = 0$ we can determine the value corresponding to the maximum β as a function of$p$.

Let us find the respective derivative:

$$\begin{gathered} \frac{{\partial \beta }}{{\partial p}} = \frac{\partial }{{\partial p}}\left( {\mathop \sum \limits_{k = m}^{n - m} C_{n}^{k}{{p}^{k}}{{{\left( {1 - p} \right)}}^{{n - k}}}} \right) = \mathop \sum \limits_{k = m}^{n - m} \frac{{n!}}{{k!\left( {n - k} \right)!}} \\ \times \,\,\left( {k{{p}^{{k - 1}}}{{{\left( {1 - p} \right)}}^{{n - k}}} - \left( {n - k} \right){{p}^{k}}{{{\left( {1 - p} \right)}}^{{n - k - 1}}}} \right) \\ = n!\mathop \sum \limits_{k = m}^{n - m} \left( {\frac{{{{p}^{{k - 1}}}{{{\left( {1 - p} \right)}}^{{n - 1 - \left( {k - 1} \right)}}}}}{{\left( {k - 1} \right)!\left( {n - 1 - \left( {k - 1} \right)} \right)!}} - \frac{{{{p}^{k}}{{{\left( {1 - p} \right)}}^{{n - 1 - k}}}}}{{k!\left( {n - 1 - k} \right)!}}} \right). \\ \end{gathered} $$

We split the expression into two parts:

$${{\varphi }_{1}} = \mathop \sum \limits_{k = m}^{n - m} \frac{{{{p}^{{k - 1}}}{{{\left( {1 - p} \right)}}^{{n - 1 - \left( {k - 1} \right)}}}}}{{\left( {k - 1} \right)!\left( {n - 1 - \left( {k - 1} \right)} \right)!}}~,$$

$${{\varphi }_{2}} = \mathop \sum \limits_{k = m}^{n - m} \frac{{{{p}^{k}}{{{\left( {1 - p} \right)}}^{{n - 1 - k}}}}}{{k!\left( {n - 1 - k} \right)!}}~.$$

Then $\frac{{\partial \beta }}{{\partial p}} = ~\,\,n!\left( {{{\varphi }_{1}} - {{\varphi }_{2}}} \right)$. In ${{\varphi }_{1}}$ we replace the summation variable $k{\kern 1pt} ' = k - 1$, so that the form of sum ${{{{\varphi }}}_{1}}$ is as similar as possible to ${{{{\varphi }}}_{2}}$:

$${{\varphi }_{1}} = \mathop \sum \limits_{k' = m - 1}^{n - 1 - m} \frac{{{{p}^{{k{\kern 1pt} '}}}{{{\left( {1 - p} \right)}}^{{n - 1 - k{\kern 1pt} '}}}}}{{k{\kern 1pt} '!\left( {n - 1 - k{\kern 1pt} '} \right)!}}~.$$

It can be seen that the sums in ${{\varphi }_{1}}$ and ${{\varphi }_{2}}$ differ only by the terms with $k{\kern 1pt} ' = m - 1$ and $k = n - m$. Since ${{\varphi }_{1}}$ and ${{\varphi }_{2}}$ are present in the formula for $\frac{{\partial \beta }}{{\partial p}}$ with opposite signs, therefore, in $\frac{{\partial \beta }}{{\partial p}}$, all the similar terms of sums ${{\varphi }_{1}}$ and ${{\varphi }_{2}}$ are cancelled. Thus, we can record

$$\begin{gathered} \frac{{\partial \beta }}{{\partial p}} = ~\,\,n!\left( {\frac{{{{p}^{{m - 1}}}{{{\left( {1 - p} \right)}}^{{n - 1 - \left( {m - 1} \right)}}}}}{{\left( {m - 1} \right)!\left( {n - 1 - \left( {m - 1} \right)} \right)!}}} \right. \\ \left. { - \,\,\frac{{{{p}^{{n - m}}}{{{\left( {1 - p} \right)}}^{{n - 1 - \left( {n - m} \right)}}}}}{{\left( {n - m} \right)!\left( {n - 1 - \left( {n - m} \right)} \right)!}}} \right) \\ = \frac{{n!{{p}^{{m - 1}}}{{{\left( {1 - p} \right)}}^{{m - 1}}}}}{{\left( {m - 1} \right)!\left( {n - m} \right)!}}\left( {{{{\left( {1 - p} \right)}}^{{n + 1 - 2m}}} - {{p}^{{n + 1 - 2m}}}} \right). \\ \end{gathered} $$

We use an abridged multiplication formula ${{a}^{n}} - {{b}^{n}} = \left( {a - b} \right)$ $\left( {{{a}^{{n - 1}}} + {{a}^{{n - 2}}}b + \cdots + a{{b}^{{n - 2}}} + {{b}^{{n - 1}}}} \right)$ to convert the last multiplier in the above formula. This yields

$$\begin{gathered} \frac{{\partial \beta }}{{\partial p}} = \frac{{n!{{p}^{{m - 1}}}{{{\left( {1 - p} \right)}}^{{m - 1}}}}}{{\left( {m - 1} \right)!\left( {n - m} \right)!}}\left( {\left( {1 - p} \right) - p} \right) \\ \times \,\,\left( {{{{\left( {1 - p} \right)}}^{{n - 2m}}} + p{{{\left( {1 - p} \right)}}^{{n - 1 - 2m}}}} \right. \\ + \left. { \ldots + {{p}^{{n - 1 - 2m}}}\left( {1 - p} \right) + {{p}^{{n - 2m}}}} \right). \\ \end{gathered} $$

In this formula, at $0 < p < 1$, only the multiplier $\left( {\left( {1 - p} \right) - p} \right) = 1 - 2p$ can be zero, which is true at $p = {1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-0em} 2}$. Thus, $\frac{{\partial \beta }}{{\partial p}} = 0$ at $p =$${1 \mathord{\left/ {\vphantom {1 2}} \right. \kern-0em} 2}$, therefore, the probability β as a function of $p$ has a maximum.