"math inline">\(\epsilon\)</span>, <span class="math display">\[\begin{eqnarray*}
\mathbb{P}(|K_+ - \mathbb{E}[K_+]| &gt; \epsilon) &amp; \leqslant &amp; 2\exp\left(-\frac{2\epsilon^2}{N}\right), \\
\mathbb{P}(|K_- - \mathbb{E}[K_-]| &gt; \epsilon) &amp; \leqslant &amp; 2\exp\left(-\frac{2\epsilon^2}{N}\right).
\end{eqnarray*}\]</span> Let‚Äôs use these concentration inequalities to show that for a finite sample drawn, <span class="math inline">\(K_+\)</span> and <span class="math inline">\(K_-\)</span>