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Make prop2 proof more consistent

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Jonathan Berrisch
2025-06-25 16:01:49 +02:00
parent 1c58c4bf89
commit 99e7b2fd7b
2 changed files with 4 additions and 4 deletions
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4
index.html
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@@ -26438,8 +26438,8 @@ w_{t,k}^{\text{Naive}} = \frac{1}{K}\label{eq:naive_combination}
<p><span class="math display">\[\begin{align} <p><span class="math display">\[\begin{align}
\rho_p(z) &amp;= \rho_p(z) &amp;=
\begin{cases} \begin{cases}
z(1 - p) = z - zp, &amp; \text{if } z &gt; 0 \\ z(1 - p) = z - zp, &amp; \text{if } 0 &lt; z \\
z(0 - p) = -zp, &amp; \text{if } z \leq 0 z(0 - p) = -zp, &amp; \text{if } 0 \geq z
\end{cases} \\ \end{cases} \\
\rho_p&#39;(z) &amp;= \rho_p&#39;(z) &amp;=
\begin{cases} \begin{cases}

4
index.qmd
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@@ -1464,8 +1464,8 @@ To find $\mathcal{Q}''_p(x)$ we rewrite \eqref{eq:check}:
\begin{align} \begin{align}
\rho_p(z) &= \rho_p(z) &=
\begin{cases} \begin{cases}
z(1 - p) = z - zp, & \text{if } z > 0 \\ z(1 - p) = z - zp, & \text{if } 0 < z \\
z(0 - p) = -zp, & \text{if } z \leq 0 z(0 - p) = -zp, & \text{if } 0 \geq z
\end{cases} \\ \end{cases} \\
\rho_p'(z) &= \rho_p'(z) &=
\begin{cases} \begin{cases}