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[von-neumann] Use note admonition rather than blockquote #152

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46 changes: 23 additions & 23 deletions source/rst/von_neumann_model.rst
Original file line number Diff line number Diff line change
Expand Up @@ -599,18 +599,18 @@ following arbitrage true
x_0^T\left(B-\gamma^{* } A\right)p_0 &= 0
\end{aligned}

..

*Proof (Sketch):* Assumption I and II imply that there exist :math:`(\alpha_0,
x_0)` and :math:`(\beta_0, p_0)` that solve the TEP and EEP, respectively. If
:math:`\gamma^*>\alpha_0`, then by definition of :math:`\alpha_0`, there cannot
exist a semi-positive :math:`x` that satisfies :math:`x^T B \geq \gamma^{* }
x^T A`. Similarly, if :math:`\gamma^*<\beta_0`, there is no semi-positive
:math:`p` for which :math:`Bp \leq \gamma^{* } Ap`. Let :math:`\gamma^{*
}\in[\beta_0, \alpha_0]`, then :math:`x_0^T B \geq \alpha_0 x_0^T A \geq
\gamma^{* } x_0^T A`. Moreover, :math:`Bp_0\leq \beta_0 A p_0\leq \gamma^* A
p_0`. These two inequalities imply :math:`x_0\left(B - \gamma^{* } A\right)p_0
= 0`.
.. note::

*Proof (Sketch):* Assumption I and II imply that there exist :math:`(\alpha_0,
x_0)` and :math:`(\beta_0, p_0)` that solve the TEP and EEP, respectively. If
:math:`\gamma^*>\alpha_0`, then by definition of :math:`\alpha_0`, there cannot
exist a semi-positive :math:`x` that satisfies :math:`x^T B \geq \gamma^{* }
x^T A`. Similarly, if :math:`\gamma^*<\beta_0`, there is no semi-positive
:math:`p` for which :math:`Bp \leq \gamma^{* } Ap`. Let :math:`\gamma^{*
}\in[\beta_0, \alpha_0]`, then :math:`x_0^T B \geq \alpha_0 x_0^T A \geq
\gamma^{* } x_0^T A`. Moreover, :math:`Bp_0\leq \beta_0 A p_0\leq \gamma^* A
p_0`. These two inequalities imply :math:`x_0\left(B - \gamma^{* } A\right)p_0
= 0`.

Here the constant :math:`\gamma^{*}` is both an expansion factor and an interest
factor (not necessarily optimal).
Expand Down Expand Up @@ -747,17 +747,17 @@ Using this interpretation, they restate Assumption I and II as follows

V(-A) < 0\quad\quad \text{and}\quad\quad V(B)>0

..

*Proof (Sketch)*: \* :math:`\Rightarrow` :math:`V(B)>0` implies
:math:`x_0^T B \gg \mathbf{0}`, where :math:`x_0` is a maximizing
vector. Since :math:`B` is non-negative, this requires that each
column of :math:`B` has at least one positive entry, which is
Assumption I. \* :math:`\Leftarrow` From Assumption I and the fact
that :math:`p>\mathbf{0}`, it follows that :math:`Bp > \mathbf{0}`.
This implies that the maximizing player can always choose :math:`x`
so that :math:`x^TBp>0` so that it must be the case
that :math:`V(B)>0`.
.. note::

*Proof (Sketch)*: \* :math:`\Rightarrow` :math:`V(B)>0` implies
:math:`x_0^T B \gg \mathbf{0}`, where :math:`x_0` is a maximizing
vector. Since :math:`B` is non-negative, this requires that each
column of :math:`B` has at least one positive entry, which is
Assumption I. \* :math:`\Leftarrow` From Assumption I and the fact
that :math:`p>\mathbf{0}`, it follows that :math:`Bp > \mathbf{0}`.
This implies that the maximizing player can always choose :math:`x`
so that :math:`x^TBp>0` so that it must be the case
that :math:`V(B)>0`.

In order to (re)state Theorem I in terms of a particular two-player
zero-sum game, we define a matrix for :math:`\gamma\in\mathbb{R}`
Expand Down