5.5: 1/N'th Highest Maxima

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Franz S. Hover & Michael S. Triantafyllou
Massachusetts Institute of Technology via MIT OpenCourseWare

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We next define a statistic that is the average lowest value of the 1/N’th highest peaks. The construction goes this way: From the record, list all of the highest peaks. Rank and then collect the highest 1/N fraction of these numbers. The average lowest value of this set is the 1/N’th highest maxima. For instance, let the peaks be [6 7 11 5 3 4 8 5 9 4 2 5]. There are twelve numbers here, and the one-third highest maxima \(a^{1/3}\) is around 6.5, because [7 11 8 9] exceed it. We use the superscript \(1/N\) to denote the 1/N'th highest maxima of a quantity.

Building on the previous section, we have that

\begin{align} p(a > a^{1/N}) \, = \, \dfrac{1}{N} \, &\approx \, \dfrac{2q}{1+q} \exp \left( -(a^{1/N})^2 / 2 M_0 \right), \text{ so that} \\[4pt] a^{1/N} \, &\approx \, \sqrt{2 M_0 \ln \left( \dfrac{2q}{1+q}N \right) }. \end{align}