\documentclass[11pt]{beamer}
\usepackage{amsmath,amssymb}
\usepackage{graphicx}
\usepackage{url}
\usepackage{textcomp}
\usetheme{iqss}

\title[Beamer Demo]{Beamer Class Demonstration\thanks{for beamer}}

\author{Me Myself and I }
\date{\today}
\institute{IQSS}
%\titlegraphic{\includegraphics[height=3em]{iqlogo.png}}

\begin{document}

\frame[plain]{\titlepage}

\iqsssectiontitle{Outline}

\section[Features]{Beamer Features}

\subsection{Some of Gary's Examples}
\begin{frame}\frametitle{What's this course about?}

  \begin{itemize}
  \item \alert{Specific statistical methods for many research problems}
    \begin{itemize}
    \item How to learn (or create) new methods
    \item Inference: \underline{Using facts you know to learn about
        facts you don't know}
    \end{itemize}
  \item \alert{How to write a publishable scholarly paper}
  \item \alert{All the practical tools of research} --- theory,
    applications, simulation, programming, word processing, plumbing,
    whatever is useful
  \item $\leadsto$ \alert{Outline and class materials:}
    \begin{itemize}
    \item[]    \begin{center}

      \mbox{{\huge\parbox[b][.5in][t]{1in}{\alert{j.mp/G2001}}}
        $\qquad\qquad$\includegraphics[width=.95in]{figs/phbAr.png}}
    \end{center}
  \item The syllabus gives topics, not a weekly plan.
  \item We will go as fast as possible subject to everyone following
    along
  \item We cover different amounts of material each week
  \end{itemize}
  \end{itemize}
\end{frame}

\begin{frame}\frametitle{How much math will you scare us with?}
  \begin{itemize}
  \item All math requires two parts: \alertb{proof} and
    \alertb{concepts \& intuition}
  \item Different classes emphasize:
    \begin{itemize}
    \item \alert{Baby Stats}: dumbed down proofs, vague intuition
    \item \alert{Math Stats}: rigorous mathematical proofs
    \item \alert{\underline{Practical Stats}}: deep concepts and
      intuition, proofs when needed
      \begin{itemize}
      \item Goal: how to do empirical research, in depth
      \item Use rigorous statistical theory --- when needed
      \item Insure we understand the intuition --- always
      \item Always traverse from theoretical foundations to practical
        applications
      \item Includes ``how to'' computation
      \item $\leadsto$ Fewer proofs, more concepts, better practical
        knowledge
      \end{itemize}
    \end{itemize}
  \item Do you have the background for this class? \uncover<+->{\alert{A
      Test: What's this?
    \begin{align*}
      b=(X'X)^{-1}X'y
    \end{align*} }}
  \end{itemize}
\end{frame}

\begin{frame}\frametitle{Systematic Components: Examples}
  \begin{center}
    \includegraphics<+->[width=8cm]{figs/functionalForms}
  \end{center}
  \begin{itemize}
  \item \alertb{$E(Y_i) \equiv \mu_i = X_i\beta = \beta_0 +
      \beta_1X_{1i} +\dots+\beta_kX_{ki}$}
  \item \alertc{$\Pr(Y_i=1) \equiv \pi_i =
      \frac{1}{1+e^{-x_i\beta}}$}
  \item \alertd{$V(Y_i)\equiv \sigma_i^2 = e^{x_i\beta}$}
  \item Interpretation:
    \begin{itemize}
    \item Each is a \alert{class of functional forms}
    \item Set $\beta$ and it picks out one \alert{member of the class}
    \item \alert{$\beta$} in each is an ``effect parameter'' vector,
      with different meaning
    \end{itemize}
  \end{itemize}
\end{frame}

\begin{frame}\frametitle{Negative Binomial Derivation} \uncover<+->{Recall:}
  \begin{equation*}
    \uncover<+->{\Pr(A|B)=\frac{\Pr(AB)}{\Pr(B)} \implies \alertb{\Pr(AB)}=\alerte{\Pr(A|B)}\alertd{\Pr(B)}}
  \end{equation*}
  \alertb<1-1>{one}
  \alertc<2-2>{two}
  \alertd<3-3>{three}
  \begin{align*}
    \uncover<+->{\text{NegBin}(y|\phi,\sigma^2) &= \int_0^\infty
      \alerte{\text{Poisson}(y|\lambda)}
      \times\alertd{\text{gamma}(\lambda|\phi,\sigma^2)}d\lambda\\}
    \uncover<+->{&= \int_0^\infty
      \alertb{\P(y,\lambda|\phi,\sigma^2) }d\lambda\\}
    \uncover<+->{&=
      \frac{\Gamma\left(\frac{\phi}{\sigma^2-1}+y_i\right)}
      {y_i!\Gamma\left(\frac{\phi}{\sigma^2-1}\right)}
      \left(\frac{\sigma^2-1}{\sigma^2}\right)^{y_i}
      \left(\sigma^2\right)^{\frac{-\phi}{\sigma^2-1}}}
  \end{align*}
\end{frame}

\section[ ]{Other features} % optional [ ] omits section head.

\subsection{Structural Features}
\begin{frame}
  \frametitle{Structural Features}
  \begin{block}{Levels of Structure}
    \begin{itemize}
      \item usual \LaTeX\ \textbackslash{}section, \textbackslash{}subsection
      commands

      \item `frame' environments provide slides

      \item `block' environments divide slides into logical sections

      \item `columns' environments divide slides vertically (example later)

      \item overlays (\`a la prosper) change content of slides dynamically
    \end{itemize}
  \end{block}

  \begin{example}[Overlay Alerts]
    On the first overlay, \alert<1>{this text} is highlighted (or \emph{alerted}).\\ On the second, \alert<2>{this text} is.
  \end{example}
\end{frame}

\begin{frame}[fragile]
  \frametitle{Code blocks}

\begin{lstlisting}[language=R]
# Say hello in R
hello <- function(name) paste("hello", name)
\end{lstlisting}

\begin{uncoverenv}<2->
\begin{lstlisting}[language=Python]
# Say hello in Python
def hello(name):
return("Hello" + " " + name)
\end{lstlisting}
\end{uncoverenv}

\begin{uncoverenv}<3->
\begin{lstlisting}[language=Haskell]
-- Say hello in Haskell
hello name = "Hello" ++ " " ++ name
\end{lstlisting}
\end{uncoverenv}

\begin{uncoverenv}<4->
\begin{lstlisting}[language=C]
/* Say hello in C */
#include <stdio.h>
int main()
{
  char name[256];
  fgets(name, sizeof(name), stdin);
  printf("Hello %s", name);
  return(0);
}
\end{lstlisting}
\end{uncoverenv}

\end{frame}

\begin{frame}
  \frametitle{Alerts}
  \begin{itemize}
     \item First level \alert{alert}
     \item Second level \alertb{alert}
     \item Third level \alertc{alert}
     \item Fourth level \alertd{alert}
     \item Fifth level \alerte{alert}
  \end{itemize}

\end{frame}

\section{More Features}

\begin{frame}
  \frametitle{Other Features}

  \begin{block}{Levels of Structure}
    \begin{itemize}
    \item Clean, extensively customizable visual style
    \item Hyperlinks (\href{http://github.com/izahn/iqss-beamer-theme}{click here})
      \item No weird scaling prosper
      \begin{itemize}
        \item slides are 96~mm~$\times$~128~mm

        \item text is 10-12pt on slide

        \item slide itself magnified with Adobe Reader/xpdf/gv to fill screen
      \end{itemize}

      \item pgf graphics framework easy to use

      \item include external JPEG/PNG/PDF figures

      \item output directly to pdf: no PostScript hurdles

      \item detailed User's Manual (with good presentation advice, too)
    \end{itemize}
  \end{block}
\end{frame}

\subsection{Blocks}

\begin{frame}
\frametitle{Theorems and Proofs}
\framesubtitle{The proof uses \textit{reductio ad absurdum}.}
\begin{theorem}
There is no largest prime number.
\end{theorem}
\begin{proof}
\begin{enumerate}
\item<1-| alert@1> Suppose $p$ were the largest prime number.
\item<2-> Let $q$ be the product of the first $p$ numbers.
\item<3-> Then $q+1$ is not divisible by any of them.
\item<1-> But $q + 1$ is greater than $1$, thus divisible by some prime
number not in the first $p$ numbers.\qedhere
\end{enumerate}
\end{proof}
\end{frame}

\begin{frame}
  \frametitle{Blocks}

  \begin{block}{Normal block}
A \alert{set} consists of elements.
\end{block}

\begin{alertblock}{Alert block}
$2=2$.
\end{alertblock}

\begin{exampleblock}{Example block}
The set $\{1,2,3,5\}$ has four elements.
\end{exampleblock}

\end{frame}

\appendix

\section{\appendixname}

\subsection{More stuff}

\frame{Details}
\frame{Text omitted in main talk.}
\subsection{Even more additional material}
\frame{More details}
\frame{Even more details}

\end{document}

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\documentclass[12]{scrartcl}
\usepackage{amssymb,amsmath,gensymb,dsfont,calc,multicol,fullpage}
\makeatletter
\newcommand\Aboxed[1]{
   \@Aboxed#1\ENDDNE}
\def\@Aboxed#1&#2\ENDDNE{%
   &
   \settowidth\@tempdima{$\displaystyle#1{}$}
   \setlength\@tempdima{\@tempdima+\fboxsep+\fboxrule}
   \kern-\@tempdima
   \boxed{#1#2}
}
\makeatother

\begin{document}

\title{Homework 33, Section 6.4: 2, 5, 7}
\author{Alex Gordon}
\date{\today}
\maketitle
\section*{Homework}
\subsection*{2.}
So basically, this isn't that hard. All I'm doing is picking some random numbers for the average number of women on the Committee. \\
$1 \cdot \frac{C(10,1)C(15,2)}{C(25,3)} + 2 \cdot \frac{C(10,2)C(15,1)}{C(25,3)} + 3 \cdot \frac{C(10,3)}{C(25,3)} = 1.2$

\subsection*{5.}
$1 \cdot \frac{C(4,1)C(48,4)}{C(52,5)} + 2 \cdot \frac{C(4,2)C(48,3)}{C(52,5)} + 3 \cdot \frac{C(48,2)C(4,3)}{C(52,5)} + 4 \cdot \frac{C(48,2)}{C(52,5)} = .39$
\subsection*{7.}
$\frac{10}{3} + \frac{5}{4} + \frac{1}{5} = 4.78$, multiplied by 5 courses so \$23.90 per semester.
\end{document}


\documentclass[main.tex]{subfiles}
\begin{document}

\subsection{No-stress source}

\marginpar{Monday\\ 2021-4-26, \\ compiled \\ \today}

We considered a source in the form \(T_{\mu \nu } = \rho t_\mu t_\nu \), where \(t^{\mu } = (1, \vec{0})\).

Now we will consider a source in the form
%
\begin{align}
T_{\mu \nu } = - 2 \rho t_\mu t_\nu + 2 J_{(\mu } t_{\nu )}
\,,
\end{align}
%
where \(J^{\mu } = \rho u^{\mu } = \rho (\gamma , \gamma v^{i} / c)\).

\todo[inline]{Probably the first 2 is not there.}

The static source from before can be recovered from this expression in the low-velocity limit \(v^{i} / c \to 0\).
In that case, \(T_{ij} = 0\): we can see that \(T_{ij}\) is of order \(v^2 / c^2\), so to first order they vanish.

In this situation, we get the system
%
\begin{align}
\begin{cases}
    \square \overline{h}_{0 \mu } &= - 16 \pi T_{0 \mu } \\
    \square \overline{h}_{ij} &= 0
\,.
\end{cases}
\end{align}

In order to simplify, let us assume that \(\partial_{t} \overline{h}_{ij} = 0\): then, the solution to the second of these becomes \(\nabla^2 \overline{h}_{ij} = 0\), with flat boundary conditions at large distance.
By linearity, this leads to \(\overline{h}_{ij} = 0\).

\begin{claim}
If we define \(A_{\mu } = - (1/4) \overline{h}_{0 \mu } = - (1/4) \overline{h}_{\mu \nu } t^{\nu }\), then the metric becomes
%
\begin{align}
g_{00} &= - 1 + 2 A_0  \\
g_{0i} &= 4 A_i  \\
g_{ij} &= (1 + 2 A_0) \delta_{ij}
\,.
\end{align}
\end{claim}

In terms of this \(A_{\mu }\), the Dalambertian equation from before reads
%
\begin{align}
\square A_{\mu } = - \frac{16}{4} \pi J_\mu = - 4 \pi J_\mu
\,,
\end{align}
%
which are formally identical to the Maxwell equations!
Therefore, we can employ known techniques from electromagnetism.

For example, if \(\partial_{t} A_{\mu } = 0\) then
%
\begin{align}
\begin{cases}
    A_0 &= - \phi  \\
    A_{i} &= \int \dd[3]{x^{i}} \frac{J_i}{\abs{x - x^{i}}}
\,,
\end{cases}
\end{align}
%
which is the reason why the phenomena which can be described through this formalism are known as gravito-electric and gravito-magnetic effects.

\begin{claim}
For example, geodesics in a weak-field stationary (no stress) spacetime are described by a Lagrangian
%
\begin{align}
\mathscr{L} &= - mc \qty(- g_{\mu \nu } \dot{x}^{\mu } \dot{x}^{\nu })^{1/2}  \\
&= - mc^2 \qty(- g_{00} - 2 g_{0i} \frac{v^{i}}{c} - g_{ij} \frac{v^{i} v^{j}}{c^2})^{1/2}  \\
&\approx -mc^2 + \frac{m}{2} v^2 + m \phi + 4 mc A_{i}v^{i}
\,.
\end{align}
\end{claim}

We have a mass term, a kinetic term, a gravitational term, and a contribution to the Lorentz force.

The corresponding equations of motion read
%
\begin{align}
\ddot{\vec{x}} = \vec{E} + 4 \vec{v} \times \vec{B}
\,,
\end{align}
%
where \(\vec{E}\) and \(\vec{B}\) are the gravitoelectric and gravitomagnetic fields derived from our \(A_{\mu }\).
The differences from EM are: the absence of charge, and the factor of 4 before the magnetic term.

An example of a gravito-electromagnetic effect is the Lense-Thirring effect: a magnetic moment \(\vec{s}\) in a magnetic field precesses, according to
%
\begin{align}
\dv{\vec{s}}{t} &= \vec{s} \times \vec{\Omega}
\qquad \text{where} \qquad
\vec{\Omega} = - \frac{q}{m} \vec{B}_{EM}
\,,
\end{align}
%
so in order to generalize to the precession of a gyroscope in an EM field we need to map \(q \to m\) and \(\vec{B}_{EM} \to 4 \vec{B}\).

This way, we see for example that \(\Omega _g = - 4 B\).
A mission called Gravity Probe B measured this effect: they found precession with \(\Omega _g \sim \SI{.22}{arcsec / yr} (R_{\oplus} / r)^3\).
This is a \SI{20}{\percent} accurate test of GR in the weak field.

\todo[inline]{What does that mean?}

Another example is \textbf{frame dragging}, which applies in full GR: if we put the gyroscope around a BH a similar effect emerges.
Around a Kerr BH we have
%
\begin{align}
g_{0i}^{\text{Kerr}} \sim \Omega_{BH}
\,,
\end{align}
%
and if the particle is close to the BH a particle is ``locked'' to the BH rotation.

\section{Gravitational Waves in linear GR}

GW are solutions of weak-field GR in a vacuum.
There, the wave equation reads \(0 = \square _\eta \overline{h}_{\mu \nu }\). What are the properties of the solutions of these equations?
The simplest thing we can do is look for plane wave solutions. We take a wave vector \(k^{\mu } = (\omega , k^{i})\) and an amplitude \(A_{\mu \nu }\); then
%
\begin{align}
\overline{h}_{\mu \nu } = A_{\mu \nu } e^{i k_\mu x^{\mu }} = A_{\mu \nu } e^{i (- \omega t + \vec{k} \cdot \vec{x}) }
\,,
\end{align}
%
so \(\partial_{\mu } \overline{h}_{\alpha \beta } =  (i k_\mu ) \overline{h}_{\alpha \beta }\).

Substituting the plane wave ansatz yields
%
\begin{align}
0 = \square \overline{h}_{\alpha \beta } = - \eta^{\mu \nu } k_\mu k_\nu \overline{h}_{\alpha \beta }
\,,
\end{align}
%
therefore \(k_{\mu }k^{\mu } = 0\). The wavevector is null.

This implies that the GW propagates at the speed of light: \(\omega s^2 = \abs{\vec{k}}^2\).

How do we completely specify a gauge?
Any infinitesimal transformation such that \(\square \xi^{\mu } = 0\) preserves the Hilbert gauge, so we can make a residual gauge transformation.

The harmonic gauge implies that
%
\begin{align}
0 = - \partial^{\alpha } \overline{h}_{\mu \alpha } = i k^{\alpha } \overline{h}_{\mu \alpha }
\,,
\end{align}
%
which yields \(k^{\alpha } A_{\alpha \mu } = 0\). This means that GWs are \textbf{transverse} to the propagation direction.

We know that \(\overline{h}_{\mu \nu }\) maps to \(\overline{h}_{\mu \nu } + 2 \partial_{(\mu } \xi_{\nu )} + \eta_{\mu \nu } \partial_{\alpha } \xi^{\alpha }\).

Let us use \(\xi^{\mu } = B^{\mu } e^{i k_{\alpha } x^{ \alpha }}\) as an ansatz for our residual gauge transformation, since it automatically harmonic: we get
%
\begin{align}
A_{\mu \nu } \to A_{\mu \nu } - 2 i k_{(\mu } B_{\nu )} + i \eta_{\mu \nu } k_{\alpha } B^{\alpha }
\,,
\end{align}
%
and since we can pick \(B^{\mu }\) arbitrarily we can impose \(\overline{h} = A^{\mu }_{\mu } = 0\), the \textbf{traceless condition}, as well as \(\overline{h}_{\mu 0} = 0\), the \textbf{transverse condition}.
The second is suggested by the previously found result \(k_\alpha A^{\alpha \beta } = 0\).

In terms of \(B\), this is a linear algebraic system, and it is invertible.

In summary, we start from 10 variables, we use 4 equations to impose the Hilbert gauge, and 4 more to impose the TT gauge.
The two degrees of freedom which are left are the true degrees of freedom of a GW.

More explicitly, if we have \(k^{\mu } = (\omega, 0, 0, k_z)\) this means
\begin{enumerate}
    \item \(k^2 =0 \) implies \(- \omega = k_z\);
    \item the phase reads \(k_\alpha x^{\alpha } = \omega (t - z)\);
    \item the Hilbert gauge \(k^{\mu } A_{\mu \nu } = 0\) tells us that \(A_{0 \nu } = A_{3 \nu }\);
    \item the transverse condition tells us that \(A_{0 \mu } = 0\) (so also \(A_{3 \mu } = 0\));
    \item the traceless condition tells us that \(A^{\mu }_{\mu } = 0\).
\end{enumerate}

This leads to the usual formulation
%
\begin{align}
A_{\mu \nu }^{TT} = \left[\begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & A_+ & A_\times & 0 \\
0 & A_\times  & - A_+ & 0 \\
0 & 0 & 0 & 0
\end{array}\right]
\,.
\end{align}

Therefore,
%
\begin{align}
h^{TT}_{\mu \nu } = A_{\mu \nu }^{TT} \exp(i \omega (t - z))
\,.
\end{align}

In TT gauge we have \(\overline{h}_{\mu \nu } = h_{\mu \nu }\) since the trace is zero.
Importantly, the TT gauge can only be defined in vacuo! This is because in that case \(\square \overline{h}_{\mu \nu } \neq 0\), so while we can still exploit gauge freedom we cannot set components to zero inside the source.

The metric in TT gauge reads
%
\begin{align}
g &= - \dd{t^2} + \dd{z^2} + (1 + h_+) \dd{x^2} (1 - h_\times ) \dd{y^2} + 2 h_\times \dd{x} \dd{y}
g &= - \dd{t^2} + (\delta_{ij} + h^{TT}_{ij} ) \dd{x^{i}} \dd{x^{j}}
\,.
\end{align}

How do we identify the GW degrees of freedom in general?
We can impose the TT gauge outside the source (far away from the \(T_{\mu \nu }\)).

In general,
%
\begin{align}
h^{TT}_{\mu \nu } = \Lambda_{\mu \nu }{}^{\alpha \beta } \overline{h}_{\alpha \beta }
\,,
\end{align}
%
where \(\Lambda \) is a projection operator, defined as
%
\begin{align}
\Lambda_{\mu \nu }{}^{\alpha \beta } &= P_{\mu }^{\alpha } P_\nu^{\beta } - \frac{1}{2} P_{\mu \nu } P^{\alpha \beta }  \\
P_{\mu \nu } &= \delta_{\mu \nu } - n_\mu n_\nu
\,,
\end{align}
%
where \(n^\mu \) is the propagation direction.

The projection tensor \(P_{\mu \nu }\) is symmetric, it is transverse (\(P_{\mu \nu } n^{\nu } = 0\)), it is idempotent (\(P_{\mu \alpha } P_{\alpha \nu } = P_{\mu \nu }\)), and its trace is equal to \(2\).

The tensor \(\Lambda_{\mu \nu \alpha \beta } \) is also idempotent, transverse in all indices, traceless in \(\mu \nu \) and \(\alpha \beta \) separately, and symmetric in the swap of \(\mu \nu \) and \(\alpha \beta \).

In summary, we have found GW solutions, they propagate with \(c\), they are transverse, they have two degrees of freedom.

Symmetric, Transverse, Trace-Free tensors play an important role in GW theory. They can be used to obtain the \textbf{Multipolar expansion}.

``Living review of relativity'' (see webpage) describes all the tests of GR.

\end{document}


\section{Transition Systems Properties}
\label{sec:ts-properties}

\subsection{Transition-system traces}
\label{sec:ts-traces}

This section introduces the notion of traces induced by the transition systems
described in \cite{small_step_semantics}.

\begin{definition}[Traces]
  Given a state transition system $L=(S,T,\Sigma, R, \Gamma)$, the set of
  traces of $L$, denoted as $\fun{traces}_L$ is defined as:
  $$
  \{ (e, s, \txs, s') \mid e \in \Gamma,~ s \in S,~ \txs \in \seqof{\Sigma},~ s' \in S\}
  $$

\end{definition}

\begin{definition}[Valid traces]
  Given a state transition system $L=(S,T,\Sigma, R, \Gamma)$, we define the
  notion of valid traces inductively:

  \begin{itemize}
  \item For all $e \in \Gamma$, $s \in S$, $(e, s, \epsilon, s)$ is a valid
    trace of $L$.

  \item If $(e, s, \txs, s')$ is a valid trace, and
    $e \vdash s' \trans{}{\var{t}} s''$ is a valid transition according to the
    rules of $L$, then $(e, s, \txs; t, s'')$ is also a valid trace.
  \end{itemize}

\end{definition}

We denote the set of valid traces of $L$ as $\transtar{L}{}$, and we write
$e \vdash s \transtar{L}{\txs} s'$ as a shorthand for
$(e, s, \txs, s') \in \transtar{L}{}$. Furthermore, when the transition system
name is clear from the context we will omit it from the transition arrow label.

\subsection{Additional notation}
\label{sec:additional-notation}

We describe next additional notation needed in the properties of the transition
systems specified in this document.

\subsubsection{Sequence indexing}
\label{sec:seq-indexing}

Given a sequence $\mathcal{A} \in \seqof{\type{A}}$, and a natural number $i$
such that $0 \leq i < \size{\mathcal{A}}$, $\mathcal{A}_i$ refers to the
$i^{\text{th}}$ element of $\mathcal{A}$.

Given a sequence $\mathcal{A}$, a quantification symbol $\bigoplus$, e.g.
$\forall$ or $\exists$, a range predicate $R$, and a quantification term $T$
(both of which depend on an element of $\mathcal{A}$) we write:
%
$$
\bigoplus \mathcal{A}_i \cdot R~\mathcal{A}_i \cdot T~\mathcal{A}_i
$$
%
as a shorthand notation for:
%
$$
\bigoplus i \cdot 0 \leq i < \size{\mathcal{A}} \wedge R~\mathcal{A}_i \cdot T~\mathcal{A}_i
$$

For instance:
%
$$
\forall \txs_i \cdot \txins{\txs_i} \neq \emptyset
$$
is a shorthand notation for:
$$
\forall i \cdot 0 \leq i < \size{\txs} \Rightarrow \txins{\txs_i} \neq \emptyset
$$
Remember that the range of a universal quantification can be expressed by an
implication, and the range of an existential qualification by a conjunction.

\subsubsection{Quantifying over set operations}
\label{sec:quantifying-over-set-operators}

Given a sequence $\mathcal{A} \in \seqof{A}$, a set $B$, a set operation
$\bigoplus$, e.g. $\cup$ or $\unionoverrideRight$, and a function
$\fun{f} \in \type{A} \to \type{B}$, the term:
%
$$
\underset{\fun{f}}{\bigoplus} \mathcal{A}
$$
is a shorthand notation for:
%
$$
\underset{0 \leq i < \size{\mathcal{A}}}{\bigoplus} \mathcal{A}_i
$$

For instance:
$$
\bigcup_{\txins{}} \txs
$$
denotes the sequence of unions:
$$
\bigcup_{0 \leq i < \size{\txs}} \txins{\txs_i}
$$

In a set operation quantification over a sequence $\mathcal{A}$, the operation
is applied to the elements the order in which they appear in $\mathcal{A}$.
This is crucial in the case of non-commutative operations, such as union
override ($\unionoverrideRight$).

\subsection{UTxO Properties}
\label{sec:utxo-properties}

Property~\ref{prop:no-double-spending} expresses the fact that transaction
inputs cannot be used more than once. This property requires that the starting
UTxO does not contain any future outputs, which is a reasonable constraint.

\begin{property}[No double spending]\label{prop:no-double-spending}
  For all

  $$
  \left(
    \begin{array}{l}
      \var{utxo_0}\\
      \var{reserves_0}
    \end{array}
  \right)
  \transtar{\hyperref[fig:rules:utxo]{utxo}}{\txs}
  \left(
    \begin{array}{l}
      \var{utxo}\\
      \var{reserves}
    \end{array}
  \right)
  $$

  such that

  $$
    \forall \txs_i \cdot \dom~(\txouts{\txs_i}) \cap \dom~(\var{utxo_0}) = \emptyset
  $$

  we have:

  $$
  \forall \txs_i,~\txs_j \cdot i < j \Rightarrow \txins{\txs_i} \cap \txins{\txs_j} = \emptyset
  $$
\end{property}

Property~\ref{prop:utxo-out-min-in} expresses the fact that all inputs and
outputs are accounted for, in such a way that we can reconstruct the final
(UTxO) state by adding all the outputs to the initial state, and removing the
spent outputs.

\begin{property}[UTxO is outputs minus inputs]\label{prop:utxo-out-min-in}
  For all

  $$
  \left(
    \begin{array}{l}
      \var{utxo_0}\\
      \var{reserves_0}
    \end{array}
  \right)
  \transtar{\hyperref[fig:rules:utxo]{utxo}}{\txs}
  \left(
    \begin{array}{l}
      \var{utxo}\\
      \var{reserves}
    \end{array}
  \right)
  $$

  such that

  $$
    \forall \txs_i \cdot \dom~(\txouts{\txs_i}) \cap \dom~(\var{utxo_0}) = \emptyset
  $$

  we have:

  $$
  \bigcup_{\txins{}} \txs \subtractdom (\var{utxo_0} \cup \bigcup_{\txouts{}} \txs) = \var{utxo}
  $$

\end{property}

Property~\ref{prop:utxo-money-supply-cnst} models the fact that the amount of
money in the system (counted as Entropics) remains constant.

\begin{property}[Money supply is constant in the system]\label{prop:utxo-money-supply-cnst}
  For all

  $$
  \left(
    \begin{array}{l}
      \var{utxo_0}\\
      \var{reserves_0}
    \end{array}
  \right)
  \transtar{\hyperref[fig:rules:utxo]{utxo}}{\txs}
  \left(
    \begin{array}{l}
      \var{utxo}\\
      \var{reserves}
    \end{array}
  \right)
  $$

  we have:

  $$ \var{reserves} + \balance{utxo} =  \var{reserves_0} + \balance{utxo_0} $$
\end{property}

\subsection{Delegation Properties}
\label{sec:delegation-props}

Property~\ref{prop:no-dcert-replay} states that delegation certificates cannot be replayed. Remember
that $\Delta_i$ is the $i^{\text{th}}$ element of $\Delta$, which is a sequence of sequences of
delegation certificates, so $\Delta_i \in \seqof{\DCert}$ and $\Delta_{i_j} \in \DCert$ (assuming
$j$ is a valid index of $\Delta_i$).

\begin{property}[No replay of delegation certificates]\label{prop:no-dcert-replay}
  For all
  %
  $$
  \left(
    \begin{array}{l}
      \var{delegSt}
    \end{array}
  \right)
  \transtar{\hyperref[fig:rules:delegation-interface]{deleg}}{\Delta}
  \left(
    \begin{array}{l}
      \var{delegSt'}
    \end{array}
  \right)
  $$
  %
  we have:
  %
  $$
  \forall i, j, k, l \cdot j < l \Rightarrow \Delta_{i_j} \neq \Delta_{k_l}
  $$
\end{property}


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\title[DS-GA 1003 / CSCI-GA 2567]{Subgradient Descent}
\author{Xintian Han \& David S. Rosenberg }
\date{Feb 13, 2019}
\institute{NYU CDS}

\makebeamertitle
\mode<article>{Just in article version}

\begin{frame}{Contents}
\tableofcontents{}
\end{frame}

\section{Motivation and Review: Lasso}
\begin{frame}{The Lasso Problem}
\begin{itemize}
\item Lasso problem can be parametrized as
\[
\min_{w\in\reals^{d}}J(w) = \frac{1}{n}\sum_{i=1}^{n}\left\{ w^{T}x_{i}-y_{i}\right\} ^{2}+\lambda\|w\|_{1}
\]
\pause{}
\item We could solve Lasso by Shooting Method and Projected SGD.
\pause{}
\item How about using SGD?
\pause{}
\item $\|w\|_{1}=\mid w_{1}\mid+\mid w_{2}\mid$ is not differentiable!
\end{itemize}
\end{frame}
%
\begin{frame}{Gradient Descent on Lasso Objective?}
\begin{itemize}
\item The partial gradient of the Lasso objective is
\[
\del_{w}J(w)=\frac{1}{n}\sum_{j=1}^n 2\{w^Tx_j-y_j\}x_j+\lambda \cdot\text{sign}(w)
\]
when $w_i\neq 0$ for all $i$, and \textbf{otherwise
is undefined}.
\end{itemize}

\pause{}

Potential arguments for why we shouldn't care about the points of
nondifferentiability:
\begin{itemize}
\item If we start with a random $w$, will we ever hit exactly $w_i=0$?

\pause{}
\item If we did, could we perturb the step size by $\eps$ to miss such
a point?

\pause{}
\item Does it even make sense to check $w_i=0$ with floating
point numbers?
\end{itemize}
\end{frame}
%
\begin{frame}{Gradient Descent on Lasso Objective?}
\begin{itemize}
\item If we blindly apply gradient descent from a random starting point
\begin{itemize}
\item seems unlikely that we'll hit a point where the gradient is undefined.
\item So it's clear how we could apply gradient descent or SGD to a function that's not differentiable everywhere
\end{itemize}
\end{itemize}

\pause{}
\begin{itemize}
\item Still, doesn't mean that gradient descent will work if objective not
differentiable!
\end{itemize}

\pause{}
\begin{itemize}
\item Theory of subgradients and subgradient descent will clear up any uncertainty.
\end{itemize}
\end{frame}
%

\section{Convexity and Sublevel Sets}

\mode<article>{Just in article version}

\begin{frame}{Convex Sets}

\begin{definition}
A set $C$ is \textbf{convex} if the line segment between any two
points in $C$ lies in $C$.
\end{definition}

\begin{center}
\includegraphics[height=0.55\textheight]{fig7\lyxdot 4a}\includegraphics[height=0.55\textheight]{fig7\lyxdot 4b}\let\thefootnote\relax\footnotetext{\tiny{KPM Fig. 7.4}}
\par\end{center}

\end{frame}

\begin{frame}{Convex and Concave Functions}
\begin{definition}
A function $f:\reals^{d}\to\reals$ is \textbf{convex} if the line
segment connecting any two points on the graph of $f$ lies above
the graph. $f$ is \textbf{concave }if $-f$ is convex.
\end{definition}

\begin{center}
\includegraphics[height=0.5\textheight]{fig7\lyxdot 5a}\includegraphics[height=0.5\textheight]{fig7\lyxdot 5b}\let\thefootnote\relax\footnotetext{\tiny{KPM Fig. 7.5}}
\par\end{center}

\end{frame}
%
\begin{frame}{Examples of Convex Functions on $\reals$}
\begin{examples}

\begin{itemize}
\item $x\mapsto ax+b$ is both convex and concave on $\reals$ for all $a,b\in\reals$.

\pause{}
\item $x\mapsto\left|x\right|^{p}$ for $p\ge1$ is convex on $\reals$

\pause{}
\item $x\mapsto e^{ax}$ is convex on $\reals$ for all $a\in\reals$

\pause{}
\item Every norm on $\reals^{n}$ is convex (e.g. $\|x\|_{1}$ and $\|x\|_{2}$)

\pause{}
\item Max: $\left(x_{1},\ldots,x_{n}\right)\mapsto\max\left\{ x_{1}\ldots,x_{n}\right\} $
is convex on $\reals^{n}$
\end{itemize}
\end{examples}

\end{frame}
%
\begin{frame}{Simple Composition Rules }
\begin{examples}

\begin{itemize}
\item If $g$ is convex, and $Ax+b$ is an affine mapping, then $g(Ax+b)$
is convex.
\item If $g$ is convex then $\exp g(x)$ is convex.
\item If $g$ is convex and nonnegative and $p\ge1$ then $g(x)^{p}$ is
convex.
\item If $g$ is concave and positive then $\log g(x)$ is concave
\item If $g$ is concave and positive then $1/g(x)$ is convex.
\end{itemize}
\end{examples}

\end{frame}
%
\begin{frame}{Main Reference for Convex Optimization}
\begin{itemize}
\item Boyd and Vandenberghe (2004)
\begin{itemize}
\item Very clearly written, but has a ton of detail for a first pass.
\item See the \href{https://davidrosenberg.github.io/mlcourse/Notes/convex-optimization.pdf}{Extreme Abridgement of Boyd and Vandenberghe}.
\end{itemize}
\end{itemize}
\includegraphics[height=0.6\textheight]{bv_cvxbook_cover}
\end{frame}
%
\begin{frame}{Level Sets and Sublevel Sets}

Let $f:\reals^{d}\to\reals$ be a function. Then we have the following
definitions:
\begin{definition}
A \textbf{level set} or \textbf{contour line} for the value $c$ is
the set of points $x\in\reals^{d}$ for which $f(x)=c$.
\end{definition}

\pause{}
\begin{definition}
A \textbf{sublevel} set for the value $c$ is the set of points $x\in\reals^{d}$
for which $f(x)\le c$.
\end{definition}

\pause{}
\begin{theorem}
If $f:\reals^{d}\to\reals$ is \textbf{convex}, then the \textbf{sublevel
sets are convex}.
\end{theorem}

\pause{}

(Proof straight from definitions.)
\end{frame}
%
\begin{frame}{Convex Function}
\begin{center}
\includegraphics[height=0.65\textheight]{3d-plot-convex-fn}
\par\end{center}

\begin{center}
\let\thefootnote\relax\footnotetext{\tiny{Plot courtesy of Brett Bernstein.}}
\par\end{center}

\end{frame}
%
\begin{frame}{Contour Plot Convex Function: Sublevel Set}
\begin{center}
\includegraphics[height=0.65\textheight]{contour-plot-convex-fn}
\par\end{center}

Is the sublevel set $\left\{ x\mid f(x)\le1\right\} $ convex?
\begin{center}
\let\thefootnote\relax\footnotetext{\tiny{Plot courtesy of Brett Bernstein.}}
\par\end{center}

\end{frame}
%
\begin{frame}{Nonconvex Function}
\begin{center}
\includegraphics[height=0.7\textheight]{3d-plot-nonconvex-fn}\let\thefootnote\relax\footnotetext{\tiny{Plot courtesy of Brett Bernstein.}}
\par\end{center}

\end{frame}
%
\begin{frame}{Contour Plot Nonconvex Function: Sublevel Set}
\begin{center}
\includegraphics[height=0.65\textheight]{contour-plot-nonconvex-fn}
\par\end{center}

Is the sublevel set $\left\{ x\mid f(x)\le1\right\} $ convex?
\begin{center}
\let\thefootnote\relax\footnotetext{\tiny{Plot courtesy of Brett Bernstein.}}
\par\end{center}

\end{frame}
%
\begin{frame}{Fact: Intersection of Convex Sets is Convex}
\begin{center}
\includegraphics[height=0.65\textheight]{convex-set-intersection}
\par\end{center}

\begin{center}
\let\thefootnote\relax\footnotetext{\tiny{Plot courtesy of Brett Bernstein.}}
\par\end{center}

\end{frame}
%
\begin{frame}{Level and Superlevel Sets}
\begin{center}
\includegraphics[height=0.65\textheight]{superlevel-sets}
\par\end{center}

Level sets and superlevel sets of convex functions are \textbf{not}
generally convex.
\begin{center}
\let\thefootnote\relax\footnotetext{\tiny{Plot courtesy of Brett Bernstein.}}
\par\end{center}

\end{frame}
%
\begin{frame}{Convex Optimization Problem: Standard Form}
\begin{block}{Convex Optimization Problem: Standard Form}
\begin{eqnarray*}
\textrm{minimize} &  & f_{0}(x)\\
\textrm{subject to} &  & f_{i}(x)\le0,\;\;i=1,\ldots,m
\end{eqnarray*}
where $f_{0},\ldots,f_{m}$ are convex functions.
\end{block}

\pause{}
\begin{itemize}
\item What can we say about each constraint set $\left\{ x\mid f_{i}(x)\le0\right\} ?\pause$
(convex)
\item What can we say about the feasible set $\left\{ x\mid f_{i}(x)\le0,\,i=1,\ldots,m\right\} ?\pause$
(convex)
\end{itemize}
\end{frame}
%
\begin{frame}{Convex Optimization Problem: Implicit Form}
\begin{block}{Convex Optimization Problem: Implicit Form}
\begin{eqnarray*}
\textrm{minimize} &  & f(x)\\
\textrm{subject to} &  & x\in C
\end{eqnarray*}
where $f$ is a convex function and $C$ is a convex set.

\pause{}

An alternative ``generic'' convex optimization problem.
\end{block}
\end{frame}
%

\section{Convex and Differentiable Functions}
\begin{frame}{First-Order Approximation}
\begin{itemize}
\item Suppose $f:\reals^{d}\to\reals$ is \textbf{differentiable.}
\item Predict $f(y)$ given $f(x)$ and $\del f(x)$?

\pause{}
\item Linear (i.e. ``\textbf{first order}'') approximation:
\[
f(y)\approx f(x)+\del f(x)^{T}(y-x)
\]
\end{itemize}
\begin{center}
\let\thefootnote\relax\footnotetext{\tiny{Boyd \& Vandenberghe Fig. 3.2}}
\par\end{center}

\begin{center}
\includegraphics[width=0.7\columnwidth]{BVFig3\lyxdot 2-convexTangent}
\par\end{center}

\end{frame}
%
\begin{frame}{First-Order Condition for Convex, Differentiable Function}

\begin{itemize}
\item Suppose $f:\reals^{d}\to\reals$ is \textbf{convex} and \textbf{differentiable.}
\item Then for any $x,y\in\reals^{d}$
\[
f(y)\ge f(x)+\del f(x)^{T}(y-x)
\]

\pause{}
\item The linear approximation to $f$ at $x$ is a \textbf{global underestimator
}of $f$:
\end{itemize}
\let\thefootnote\relax\footnotetext{\tiny{Figure from Boyd \& Vandenberghe Fig. 3.2; Proof in Section 3.1.3 }}
\begin{center}
\includegraphics[width=0.7\columnwidth]{BVFig3\lyxdot 2-convexTangent}
\par\end{center}

\end{frame}
%
\begin{frame}{First-Order Condition for Convex, Differentiable Function}
\begin{itemize}
\item Suppose $f:\reals^{d}\to\reals$ is \textbf{convex} and \textbf{differentiable}
\item Then for any $x,y\in\reals^{d}$
\[
f(y)\ge f(x)+\del f(x)^{T}(y-x)
\]

\pause{}

\end{itemize}
\begin{corollary}
If $\del f(x)=0$ then $x$ is a global minimizer of $f$.
\end{corollary}

\pause{}

For convex functions, \textbf{local information gives global information.}
\end{frame}

\section{Subgradients }
\begin{frame}{Subgradients}
\begin{definition}
A vector $g\in\reals^{d}$ is a \textbf{subgradient} of $f:\reals^{d}\to\reals$
at $x$ if for all $z$,
\[
f(z)\ge f(x)+g^{T}(z-x).
\]

\pause{}
\end{definition}

\begin{center}
\includegraphics[height=0.4\textheight]{Subderivative_illustration}
\par\end{center}

Blue is a graph of $f(x)$. \\
Each red line $x\mapsto f(x_{0})+g^{T}\left(x-x_{0}\right)$ is a
global lower bound on $f(x)$.
\end{frame}
%
\begin{frame}{Subdifferential}
\begin{definitions}
\begin{itemize}
\item $f$ is \textbf{subdifferentiable} at $x$ if $\exists$ at least
one subgradient at $x$.
\item The set of all subgradients at $x$ is called the \textbf{subdifferential:}
$\partial f(x)$

\pause{}

\end{itemize}
\end{definitions}

\begin{block}{Basic Facts}
\end{block}
\begin{itemize}
\item $f$ is convex and differentiable $\implies$ $\partial f(x)=\left\{ \del f(x)\right\} $.

\pause{}
\item Any point $x$, there can be $0$, $1$, or infinitely many subgradients.

\pause{}
\item $\partial f(x)=\emptyset$ $\implies$ $f$ is not convex.
\end{itemize}
\end{frame}

\begin{frame}{Globla Optimality Condition}
\begin{definition}
A vector $g\in\reals^{d}$ is a \textbf{subgradient} of $f:\reals^{d}\to\reals$
at $x$ if for all $z$,
\[
f(z)\ge f(x)+g^{T}(z-x).
\]

\pause{}

\end{definition}

\begin{corollary}
If $0\in\partial f(x)$, then $x$ is a \textbf{global minimizer}
of $f$.
\end{corollary}

\end{frame}

\begin{frame}{Subdifferential of Absolute Value}
\begin{itemize}
\item Consider $f(x)=\left|x\right|$

\pause{}

\end{itemize}
\let\thefootnote\relax\footnotetext{\tiny{Boyd EE364b: Subgradients Slides}}

\begin{center}
\includegraphics[width=0.8\columnwidth]{subgradient-absolute-value}
\par\end{center}
\begin{itemize}
\item Plot on right shows $\left\{ (x,g)\mid x\in\reals,\;g\in\partial f(x)\right\} $
\end{itemize}
\end{frame}
%
\begin{frame}{$f(x_{1},x_{2})=\left|x_{1}\right|+2\left|x_{2}\right|$}
\begin{center}
\includegraphics[height=0.75\textheight]{3d-plot-abs-x1-plus-2absx2}\let\thefootnote\relax\footnotetext{\tiny{Plot courtesy of Brett Bernstein.}}
\par\end{center}

\end{frame}
%
\begin{frame}{Subgradients of $f(x_{1},x_{2})=\left|x_{1}\right|+2\left|x_{2}\right|$}
\begin{itemize}
\item Let's find the subdifferential of $f(x_{1},x_{2})=\left|x_{1}\right|+2\left|x_{2}\right|$
at $\left(3,0\right)$.
\end{itemize}

\pause{}
\begin{itemize}
\item First coordinate of subgradient must be $1$, from $\left|x_{1}\right|$
part (at $x_{1}=3$).
\end{itemize}

\pause{}
\begin{itemize}
\item Second coordinate of subgradient can be anything in $\left[-2,2\right]$.
\end{itemize}

\pause{}
\begin{itemize}
\item So graph of $h(x_{1},x_{2})=f(3,0)+g^{T}\left(x_{1}-3,x_{2}-0\right)$
is a global underestimate of $f(x_{1},x_{2})$, for any $g=\left(g_{1},g_{2}\right),$
where $g_{1}=1$ and $g_{2}\in[-2,2]$.
\end{itemize}
\end{frame}
%
\begin{frame}{Underestimating Hyperplane to $f(x_{1},x_{2})=\left|x_{1}\right|+2\left|x_{2}\right|$}
\begin{center}
\includegraphics[height=0.75\textheight]{underestimating-3d-plot-abs-x1-plus-2absx2}\let\thefootnote\relax\footnotetext{\tiny{Plot courtesy of Brett Bernstein.}}
\par\end{center}

\end{frame}

\begin{frame}{Important Properties of Subdifferential}
\begin{itemize}
\item If $f_1,\dots,f_m:\reals^d \rightarrow \reals$ are convex functions and $f = f_1+\cdots + f_m$, then $\partial f(x) = \partial f_1(x) + \cdots +\partial f_m(x)$.
\item For $\alpha \geq 0$, $\partial (\alpha f) (x) = \alpha \partial f(x)$.
\end{itemize}

\end{frame}

\begin{frame}{Subgradients of $f(x) = \|x\|_1$}
\begin{itemize}
\item Let's find the subdifferential of $f(x) = \|x\|_1 = \sum_{i=1}^d |x_i|$ at any given point $x^0 = (x^0_1,x^0_2, \dots, x^0_d)$.
\item By an important property of subdifferential: If $f = f_1+\cdots+f_m$, then $\partial f(x) = \partial f_1(x) + \cdots \partial f_m(x)$.
\item We could calculate the subgradient of $f^i(x) = |x_i|$ and sum them up.
\item The subgradient $g^i = (g^i_1,\dots, g^i_d)$ of $f^i(x) = |x_i|$ at $x^0 = (x^0_1,x^0_2, \dots, x^0_d)$ is:
\[
g_j^i = 0, \quad {j\neq i}; \quad g_j^i = s(x^0_j), \quad j = i,
\]
where $s(x) = \text{sign}(x)$ if $x\neq 0$ and $s(x) \in [-1,1]$ if $x = 0$
\item We sum all the $g^i$ up to get the subgradient $g = (g_1,\dots, g_d)$ of $f(x)$ at $x^0$:
\[
g_i = s(x^0_i) \quad \text{for all } i
\]
\end{itemize}

\end{frame}

%

\section{Subgradient Descent}
\begin{frame}{Subgradient Descent}
\begin{itemize}
\item Suppose $f$ is convex, and we start optimizing at $x_{0}$.
\item Repeat
\begin{itemize}
\item Step in a negative subgradient direction:
\[
x=x_{0}-tg,
\]
where $t>0$ is the step size and $g\in\partial f(x_{0})$.
\end{itemize}
\end{itemize}

\pause{}
\begin{itemize}
\item $-g$ not a descent direction -- can this work?
\end{itemize}
\end{frame}

\begin{frame}{Convergence Theorem for Fixed Step Size}

Assume $f:\reals^{n}\to\reals$ is convex and
\begin{itemize}
\item $f$ is Lipschitz continuous with constant $G>0$:
\[
\left|f(x)-f(y)\right|\le G\|x-y\|\mbox{ for all }x,y
\]
\end{itemize}
\begin{theorem}
For fixed step size $t$, subgradient method satisfies:
\[
\lim_{k\to\infty}f(x_{\text{best}}^{(k)})\le f(x^{*})+G^{2}t/2
\]
\end{theorem}

\begin{center}
\let\thefootnote\relax\footnotetext{\tiny{Based on \url{https://www.cs.cmu.edu/~ggordon/10725-F12/slides/06-sg-method.pdf}}}
\par\end{center}

\end{frame}

\begin{frame}{Convergence Theorems for Decreasing Step Sizes}

Assume $f:\reals^{n}\to\reals$ is convex and
\begin{itemize}
\item $f$ is Lipschitz continuous with constant $G>0$:
\[
\left|f(x)-f(y)\right|\le G\|x-y\|\mbox{ for all }x,y
\]
\end{itemize}
\begin{theorem}
For step size respecting Robbins-Monro conditions,
\[
\lim_{k\to\infty}f(x_{\text{best}}^{(k)})=f(x^{*})
\]
\end{theorem}

\begin{center}
\let\thefootnote\relax\footnotetext{\tiny{Based on \url{https://www.cs.cmu.edu/~ggordon/10725-F12/slides/06-sg-method.pdf}}}
\par\end{center}
\end{frame}

\begin{frame}{Subgradient Descent for Lasso Problem}
\begin{itemize}
\item Lasso problem can be parametrized as
\[
\min_{w\in\reals^{d}}J(w) = \frac{1}{n}\sum_{i=1}^{n}\left\{ w^{T}x_{i}-y_{i}\right\} ^{2}+\lambda\|w\|_{1}
\]
\pause{}
\item Subgradients of $J(w)$ are
\[
\frac{1}{n}\sum_{i=1}^n 2\{w^Tx_i-y_i\}x_i+\lambda s,
\]
where $s_i = \text{sign}(w_i)$ if $w_i\neq 0$ and $s_i \in [-1,1]$ if $w_i=0$.
\end{itemize}
\end{frame}

\begin{frame}{Subgradient Descent for Lasso Problem: Potential Issues}
\begin{itemize}
\item Subgradient descent will work for all convex and Lipschitz continuous objective functions.
\pause{}
\item BUT, convergence can be very \textbf{slow} for non-differentiable functions
\pause{}
\item One can often find better approaches by closer examination of the objective function. For example, shooting method or projected SGD.
\pause{}
\item Taking small steps in the direction of the (sub)gradient usually may \textbf{not} lead to zero coordinates.
\pause{}
\item BUT, in practice, we can threshold small values.
\end{itemize}

\end{frame}
\begin{frame}{Appendix: Subgradient Gets Us Closer To Minimizer}
\begin{theorem}
Suppose $f$ is convex.
\begin{itemize}
\item Let $x=x_{0}-tg$, for $g\in\partial f(x_{0})$.
\item Let $z$ be any point for which $f(z)<f(x_{0})$.
\item Then for small enough $t>0$,
\[
\|x-z\|_{2}<\|x_{0}-z\|_{2}.
\]
\end{itemize}

\pause{}
\end{theorem}

\begin{itemize}
\item Apply this with $z=x^{*}\in\argmin_{x}f(x)$.
\end{itemize}
$\implies$\textbf{Appendix: Negative subgradient step gets us closer to minimizer}.
\end{frame}

\begin{frame}{Subgradient Gets Us Closer To Minimizer (Proof)}
\begin{itemize}
\item Let $x=x_{0}-tg$, for $g\in\partial f(x_{0})$ and $t>0$.
\item Let $z$ be any point for which $f(z)<f(x_{0})$.
\item Then
\begin{eqnarray*}
\|x-z\|_{2}^{2} & = & \|x_{0}-tg-z\|_{2}^{2}\\
\pause & = & \|x_{0}-z\|_{2}^{2}-2tg^{T}\left(x_{0}-z\right)+t^{2}\|g\|_{2}^{2}\\
\pause & \le & \|x_{0}-z\|_{2}^{2}-2t\left[f(x_{0})-f(z)\right]+t^{2}\|g\|_{2}^{2}
\end{eqnarray*}

\pause{}
\item Consider $-2t\left[f(x_{0})-f(z)\right]+t^{2}\|g\|_{2}^{2}$.
\begin{itemize}
\item It's a convex quadratic (facing upwards).
\item Has zeros at $t=0$ and $t=2\left(f(x_{0})-f(z)\right)/\|g\|_{2}^{2}>0$.
\item Therefore, it's negative for any
\[
t\in\left(0,\frac{2\left(f(x_{0})-f(z)\right)}{\|g\|_{2}^{2}}\right).
\]
\end{itemize}

\let\thefootnote\relax\footnotetext{\tiny{Based on Boyd EE364b: Subgradients Slides}}
\end{itemize}
\end{frame}

\end{document}


\documentclass[a4paper,12pt]{article}

\usepackage[latin2]{inputenc}
\usepackage{t1enc}
\usepackage{graphicx}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amssymb}
\usepackage{natbib}
\usepackage{longtable}
\usepackage[table]{xcolor}
\usepackage[affil-it]{authblk}

  \newcommand{\biexample}[2]{
* \hspace{1cm}   $v_A$: \texttt{#1} \\
* \hspace{1cm}   $v_B$: \texttt{#2} \\
}

\newcommand{\triexample}[3]{ \\
* \hspace{1cm}   $v_A$: \texttt{#1} \\
* \hspace{1cm}   $v_B$: \texttt{#2} \\
* \hspace{1cm}   $v_C$: \texttt{#3}
}

\newtheorem{thm}{Theorem}[section]
\newtheorem{lemma}{Lemma}[section]
\newtheorem{cor}{Corollary}[section]
\newtheorem{fact}{Fact}[section]
\newtheorem{claim}{Claim}[section]
\newtheorem{example}{Example}[section]
\newtheorem{defn}{Definition}[section]

\begin{document}

\title{Aggregate-monotonicity and Population-monotonicity in Cake-cutting}

\author{Erel Segal-Halevi}
\affil{Bar-Ilan University}

\author{Bal{\'{a}}zs Sziklai}
\affil{Centre for Economic and Regional Studies, Hungarian Academy of Sciences, \iffalse Buda\"{o}rsi 45. H-1112 Budapest,  \fi E-mail: koczy@krtk.mta.hu, sziklai.balazs@krtk.mta.hu}

\date{\today}

\bibliographystyle{apalike}

\maketitle

\section{Introduction}

A village is built near the sea shore and its lands are divided fairly among the founders. In time, the village grows in area and the lands are re-divided. Can we guarantee that all villagers gain from the growth, or at least not lose?
Some time later, a few villagers leave the village and their lands become free for re-division. Can we guarantee that all remaining villagers gain from this re-division, or at least not lose?

Similar questions may arise in inheritance cases and divorce settlements. Although the monotonicity issues of division rules have been extensively studied with respect to cooperative game theory \citep{Calleja2012}, the apportionment problem \citep{Balinski1982} or classical economic models \citep{Thomson_2011}, the cake-cutting literature so far focused on other properties of fairness, like proportionality and envy-freeness.

Aggregate monotonicity\footnote{Also known as resource monotonicity.} requires that whenever the cake is enlarged the welfare of the participants should weakly increase. Population monotonicity is concerned with changes in the number of participants. No one should profit from the arrival of a new agent, when more people share the same resource, and everyone should be weakly better of if someone leaves\footnote{\cite{Thomson_2011} uses a more subtle phrasing. He requires that, whenever a change happens in the environment and no one is responsible for it, the welfare of all agents should be affected in the same direction.}. Note, that population monotonicity fundamentally differs from the so called consistency axiom, where the leaving agents take their fair shares with them (cf. \cite{Young1987} or \cite{Thomson_2012}).

Experimental studies show that people value certain fairness criteria more than others. \cite{Herreiner2009} demonstrated that people are willing to sacrifice Pareto-efficiency in order to reach an envy free allocation. To our knowledge no study was ever conducted to unfold the relationship between monotonicity and efficiency or envy-freeness. However some indirect evidence points toward that monotonicity of the solution is in some cases as important as proportionality.

In the apportionment problem there is a parliament with a fixed number of seats and administrative regions with different number of voters. The question is -- assuming that electoral districts are of equal size within each of the administrative regions -- how to distribute the seats among the administrative regions to minimize the differences between the voters influences. The problem is analogous to cake-cutting where the cake corresponds to the parliamentary seats that have to be distributed. During the 1880 US census C.W. Seaton, a Chief Clerk of the Census Office, noted that an enlargement of the House of Representatives from 299 to 300 would result in loss of seat for State Alabama. This anomaly together with the later discovered population and new state paradoxes pressed the legislators to adopt newer and newer apportionment rules. The currently used seat distribution method is free from such anomalies, however it does not satisfy the so called Hare-quota, a basic guarantee of proportionality \citep{Balinski1975}.

In this paper we analyze aggregate and population monotonicity with respect to cake-cutting. We show that none of the traditional cake cutting protocols satisfy aggregate monotonicity. In particular all the methods which are based on the cut and choose scheme violate it. We show an aggregate monotonic two player method and prove the existence of a proportional and Pareto-optimal protocol which satisfies both aggregate and population monotonicity. In addition we prove several impossibility results. Most importantly we show that aggregate monotonicity is not compatible with proportionality and Pareto-efficiency if we only allow contiguous pieces.

%The size of the parliament changes due to various reasons, as it did in the US House of Representatives in 1880.

%These two questions motivate the two axioms of \emph{aggregate monotonicity (AM)} and \emph{population monotonicity (PM)}. These axioms have been studied in the context of resource allocation; we study these axioms in the context of the classic cake-cutting problem \cite{Steinhaus1948Problem}.

\iffalse
\section{Model}
There is a cake $C$, which is assumed to be a 1-dimensional interval in $\mathbb{R}$. There are $n$ agents with different value measures over subsets of the cake. The value measure of agent $i$ is $V_i$. The value measures are assumed to be \emph{absolutely continuous with respect to length}, or just \emph{continuous} for short. This means that there are no valuable "atoms" - all singular points have zero value to all agents. The value measures of all agents are normalized such that $\forall i: V_i(C)=1$.
All these assumptions are common in the cake-cutting literature.

The cake should be partitioned to $n$ measurable pairwise-disjoint subsets, which we call "pieces". The piece allocated to agent $i$ is marked $C_i$.

A \emph{division rule} is a function that takes as input $n$ continuous value measures and returns a single partition, or a set of partitions.

A partition is called \emph{whole} (WH) if the entire cake is divided, i.e: $\cup_{i=1}^{n}{C_i}=C$. In this paper we consider only whole partitions.

A partition is called \emph{connected} (CON) if each piece is connected (i.e, an interval).

A partition is called \emph{proportional} (PROP) if each agent values its piece as at least $1/n$: $\forall i V_i(C_i)\geq 1/n$. This was the first fairness criterion studied in the context of cake-cutting \cite{Steinhaus1948Problem}.

A partition is called \emph{Pareto-optimal} (PO) if there is no partition which is weakly better for all agents and strictly better for at least one agent.

Often, we consider only connected partitions. In these cases, we call a partition \emph{Pareto-optimal for Connected Allocations} (POCA) if there is no \emph{connected} partition which is weakly better for all agents and strictly better for at least one agent.

A division rule is called WH/CON/PROP/PO/POCA if it returns only WH/CON/PROP/PO/POCA partitions, respectively.

For division rules that return a single partition, the following definitions hold:
\begin{itemize}
\item{A rule is AM if, whenever the cake grows [to one side], the value of the piece allocated to each agent is weakly larger.}
\item{A rule is PM if, whenever one or more agents leave, the value of the piece allocated to each remaining agent is weakly larger.}
\end{itemize}

For division rules that return a set of partitions, the following definitions hold:
\begin{itemize}
\item{A rule is AM if, for every x in the set returned for the smaller (original) cake, there exists y in the set returned for the larger (new) cake,
such that y is at least as good as x for every agent.}
\item{A rule is PM if, for every x in the set returned for the larger number of agents, there exists y in the set returned for the smaller (new) number of agents, such that y is at least as good as x for every remaining player.}
\end{itemize}

The following table summarizes the abbreviations we use for the axioms we study in this paper.

\begin{table}
    \begin{tabular}{|l|l|}
        CON  & All players receive connected pieces \\
        PROP & Proportional - every player receives at least $1/n$ of total  \\
        PO   & Pareto-Optimal (no weakly better allocation) \\
        POCA & Pareto-Optimal assuming connected pieces (no weakly better allocation with connected pieces)  \\
        WH   & Whole cake is divided (note that PO $\to$ POCA $\to$ WH) \\
        AM   & Aggregate-monotonic  (one-sided) \\
        PM   & Population-monotonic \\
    \end{tabular}
\end{table}

We say that two or more axioms are \emph{compatible} if there exists a division rule which satisfies all of them. Our goal in the rest of the paper is to check what axioms are compatible with the monotonicity axioms AM and PM.

\section{Related Work}
The AM and PM axioms were studied in the context of fair division of homogeneous resources \cite{Moulin_1988} \cite{Moulin_1990} \cite{Thomson_1994} \cite{Thomson_2011}.

Other axioms were studied in the context of fair cake-cutting. \cite{Chambers_2005} studied the axiom of \emph{division independence}, which means that, if a land is divided to sub-plots and each sub-plot is divided according to a rule, then the outcome is identical to dividing the original land using that rule. He proves that the only rule which satisfies PO and division-independence is the \emph{utilitarian-optimal} rule - the rule which maximizes the sum of the agents' utilities. Unfortunately, this rule is incompatible with CON - the piece given to each agent might be a collection of infinitely many tiny fractions of land. We believe that in practical land division setting, connectivity is crucial.

\section{Aggregate Monotonicity with Connected Pieces}
In this section, we assume that CON is a non-negotiable requirement. This makes sense, for example, when the "cake" which has to be divided is a land. People usually want a single, connected piece of land, rather than a large collection of small patches.

It is well known that CON+PO are incompatible. Consider the following cake:

\biexample{0101}{1010}

The only PO allocation is to give slices 1 and 3 to B and the other slices to A, but this allocation is not CON. Therefore, when CON is a requirement, we use either POCA or WH.

\begin{thm}
The global PROP solution (the solution that returns all proportional allocations) is not AM (not even OSAM).
\end{thm}
\begin{proof}
Consider the following cake:

\biexample{0011}{1111}

The global solution returns some allocations in which the value of A is 2.

However, when the cake is increased to the right like this:

\biexample{001100}{111111}

the only proportional allocation gives A a value of 1.
\end{proof}

The following result is positive:

\begin{thm}
For two agents, CON+PROP+WH+AM (one-sided) are compatible.
\end{thm}
\begin{proof}
The proof is constructive: we show an AM cake-cutting procedure, which works for the two person case and needs only one cut. Its like a Dubins-Spanier moving knife method only its reversed (instead of left to right, the knife moves from right to left).
\begin{itemize}
\item{Ask both players to make a mark which cuts the cake in half according their own valuation.}
\item{Cut the cake at the righmost mark and give the slice on the right to the player who made the mark.}
\item{The remaining part is given to the other player.}
\end{itemize}

We now prove that this algorithm is AM when the cake grows to the right. Suppose that the order of marks on the smaller cake was A:B. So Bob got the part from mark B to the right of the cake, which is exactly half for him, and Alice got the part from B to the left of the cake, which is more than half for her. When the cake becomes larger, Bob necessarily receives at least half of a bigger total. For Alice, there are two possible cases:
\begin{itemize}
\item{If the order of marks on the larger cake is still A:B, then, since the mark of B moved rightwards, Alice receives a piece containing his previous piece.}
\item{If the order of the marks on the larger cake is B:A, then Alice now receives the part from A to the right of the cake, but this part is equal for him to the part from A to the left of the cake, which contains the part from B to the left of the cake, which contains Alice's previous piece.}
\end{itemize}
\end{proof}

The above procedure divides the whole cake (WH), but the allocation it finds is not necessarily POCA. Indeed, we have:

\begin{thm}
CON+PROP+POCA+AM (one-sided) are incompatible even for two agents.
\end{thm}
\begin{proof}
Consider the following cake:

\biexample{401}{022}

Any protocol that is CON+PROP+POCA must cut after the first slice. No other allocation is possible. Hence, A gets 4 and B gets 4.

Now the cake grows in the following way:

\biexample{4014}{0220}

Now by proportionality we must give A at least 4.5. No matter how it is done, B gets less than 4. Hence the protocol is not AM.
\end{proof}

\textcolor{red}{
We have to check whether CON+PROP+WH+AM are compatible for 3 or more agents.
If not, whether CON+PROP+AM are compatible for 3 or more agents.
}

\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.7\columnwidth]{TwoPlayerAM.pdf}
\caption{Replace this text with your caption%
}
\end{center}
\end{figure}

\section{Population Monotonicity with Connected Pieces}

\begin{thm}
CON+PROP+POCA+PM are incompatible.
\end{thm}

\begin{proof}
Consider the following preferences:
\triexample{0202020200}{0000000030}{2000002003}
If C receives the left piece, then by PROP the piece must touch the two "2" slices.
But then, only a value of 2 is left for A, which contradicts PROP.
B also obviously cannot receive the left piece.
Hence, A must receive the left piece.

If C receives the next piece to the right, it must touch the "2" and the "3", which leaves B with no value. Hence, B must receives the next piece, whose value is 3 (and C receives the right piece, whose value is 3).

By POCA, this means that A must receive all the cake up to the "3" slice of B, which means that A's value is 8.

Now, suppose player B leaves. Now $n=2$, so player C must get a value of at least 7/2, so he must get at least the "3" and "2". But this means that A receives a value of at most 6.
\end{proof}

Note that this proof relies on POCA. Without POCA, we could not prove that the initial value of A must be 8.

\textcolor{red}{
We have to check whether CON+PROP+WH+PM are compatible.
If not, check whether    CON+PROP+PM are compatible.
}

\section{Disconnected pieces}
In this section we assume that the piece given to each agent need not be connected; it is only required to be measurable. We are going to describe a division rule that satisfies PROP+AM+PM+PO. For ease of presentation, we first describe two simpler rules: a rule that satisfies PROP+AM+PM but is not PO, and a rule that satisfies PO+AM+PM but is not PROP. Then

\subsection{A PROP+AM+PM division rule}
Given integers $n\geq 1$ and $k\geq 1$, an \textit{exact division} of a cake, marked $Exact(n,k)$, is a partition of the cake to $k$ pieces, such that each of $n$ agents believes that each piece is worth \emph{exactly} $1/k$. Note that an $Exact(n,n)$ division is PROP.

For every $n$ and $k$, there exists an $Exact(n,k)$ division of a cake with at most $n(k-1)$ cuts, and this is the smallest number of cuts that can be guaranteed \cite{Alon_1987}. In particular, there exists an $Exact(n,n)$ division of a cake with at most $n(n-1)$ cuts, and for every $n\geq 2$, such a division is not necessarily CON (since there are more than $n-1$ cuts).

An exact division is necessarily PM, since when an agent leaves, the value of each of the remaining agents rises from $1/n$ to $1/(n-1)$. It is also necessarily AM, since when the cake increases (to either side), all agents receive exactly $1/n$ of a larger whole. Hence we get:

\begin{thm}
PROP+AM+PM  are compatible.
\end{thm}

Note that for $n=2$, there is a moving-knife procedure that actually finds an exact division \cite{Austin_1982}. Hence, for $n=2$ the above theorem is constructive.

\subsection{An AM+PM+PO division rule}
Our PO solution relies maximizing the sum of the agents' values (also termed: \emph{utilitarian social welfare}). The relation between social welfare optimization and Pareto optimality is well-studied, e.g \cite{Varian_1976}.

We need some definitions. The following definitions are about a partition of a cake $C$ to $n$ pairwise-disjoint measurable pieces: $C = C_1\cup\cdots\cup C_n$:

\begin{defn}
(a) The \emph{utilitarian social welfare} of a partition is the sum of agents' values: $U = \sum_{i=1}^n{V_i(C_i)}$.
(b) A partition is called \emph{utilitarian-optimal} if its utilitarian social welfare is maximal among all partitions.
\end{defn}

A utilitarian-optimal partition is obviously PO, since every Pareto-improvement strictly increases the utilitarian-social-welfare.

Moreover, the rule that selects a utilitarian-optimal partition is division-independent \cite{Chambers_2005}. This means that a utilitarian-optimal partition of a cake can be calculated by combining utilitarian-optimal partitions of two sub-cakes. This implies that the utilitarian-optimal rule is also AM+PM:

\begin{lemma}
\label{lemma:utilitarian-optimal-is-am-pm}
The rule that selects a utilitarian-optimal partition is AM and PM.
\end{lemma}
\begin{proof}
For AM, suppose that a cake $C$ is enlarged by a piece $E$ such that the new cake is $C\cup E$. By division independence, a utilitarian-optimal division of $C\cup E$ can be composed of a utilitarian-optimal division of $C$ and a utilitarian-optimal division of $E$. This means that in the utilitarian-optimal division of $C\cup E$, all agents are weakly better-off.

For PM, suppose that agent $n$ leaves and abandons his share $C_n$. Mark $E=C_n$ and $C=C\setminus C_n$. Apply the argument in the previous paragraph to prove that all agents are weakly better-off.
\end{proof}

\begin{cor}
AM+PM+PO are compatible.
\end{cor}

\subsection{A PROP+AM+PM+PO division rule}
In general, an exact division is proportional but not Pareto-optimal, and a utilitarian-optimal division is Pareto-optimal but not proportional. Hence, to satisfy both PROP and PO we must work harder. We need some definitions.

\begin{defn}
Let $k$ be an integer such that $0\leq k\leq n$. Let $W=(W_1,\cdots,W_k)$ a vector of $k$ weights.

(a) A partition is called \emph{$W$-exact} if the values of the first $k$ agents are exactly determined by the vector $W$, i.e: $\forall i\in 1,\cdots,k: V_i(C_i)=W_i$.

(b) A partition is called \emph{$W$-optimal} if its utilitarian social welfare is maximal among all \emph{$W$-exact} partitions.
\end{defn}

So when $k=0$, $W$ is an empty vector and the set of $W$-optimal partitions coincides with the set of utilitarian-optimal partitions. On the other extreme, when $k=n$ the set of $W$-optimal partitions coincides with the set of $W$-exact partitions.

A necessary condition for the existence of $W$-exact partitions is that the sum of weights in $W$ is at most 1. This condition is also sufficient: whenever the agents' value measures are continuous and $\sum_{i=1}^{k}{W_i}\leq 1$, there exist $W$-exact partitions. This is an immediate corollary of Corollary 1.1 of \cite{Dubins_1961}.

Whenever the agents' value measures are continuous, there exists a utilitarian-optimal partition. This is proved in Theorem 2 of \cite{Dubins_1961}. The following generalization of this theorem can be proved analogously:

\begin{lemma}
Let $k$ be an integer such that $0\leq k\leq n$. Let $W=(W_1,\cdots,W_k)$ a vector of $k$ weights. If there exist $W$-exact partitions, then there exists a $W$-optimal partition.
\end{lemma}

\begin{cor}
Let $W_{n-1}$ be the following vector of length $n-1$: $(\frac{1}{n},\cdots,\frac{1}{n})$. Then there exists a $W_{n-1}$-optimal partition.
\end{cor}

A $W_{n-1}$-optimal partition is a partition in which the first $n-1$ agents receive exactly their fair share of $1/n$, while the $n$-th agent receives the largest value that can be given to him under this constraint.

A $W_{n-1}$-optimal partition is Pareto-optimal in certain conditions which we now define. The following definition and fact are adapted from basic measure-theory:

\begin{defn}
(a) A measure on an interval is called \emph{strictly positive} if it assigns a positive value to any non-empty open sub-interval.

(b) The \emph{support} of a measure $V$, written $supp(V)$, is the set of all points $x$ for which every open neighborhood of $x$ has positive measure.
\end{defn}

\begin{fact}
Every measure $V$ is strictly positive on $supp(V)$.
\end{fact}

In the context of cake-cutting, the support of $V_i$ is the subset of the cake from which agent $i$ derives a positive value.

\begin{lemma}
\label{lemma:W-optimal-is-PO}
If the value measure of agent $n$ is strictly positive, then any $W_{n-1}$-optimal partition is Pareto-optimal.
\end{lemma}
\begin{proof}
Let $P$ be a $W_{n-1}$-optimal partition. Suppose by contradiction that there exists another partition $Q$ which Pareto-dominates $P$. There are two cases:

(a) The values of all the first $n-1$ agents in $Q$ are the same as in $P$, while the value of agent $n$ is strictly larger in $Q$. But this contradicts our assumption that $P$ is $W_{n-1}$-optimal.

(b) There is one or more agents $i$, where $i<n$, whose value in $Q$ is strictly larger than in $P$. In this case, we can cut a small piece of agent $i$'s share in $Q$, such that the value of agent $i$ reduces to exactly $1/n$. Then we can give the removed piece to agent $n$. Because agent $n$'s value measure is strictly positive, this transfer strictly increases the value of agent $n$. We can do the same to all other agents $i$ where $i<n$. The result is a $W_{n-1}$-exact partition in which agent $n$'s value is strictly more than in the partition $P$. But this again contradicts our assumption that $P$ is $W_{n-1}$-optimal.
\end{proof}

\begin{lemma}
If the value measure of agent $n$ is strictly positive, then the rule that selects a $W_{n-1}$-optimal partition is AM and PM.
\end{lemma}
\begin{proof}
Because agents $1,\cdots,n-1$ receive a value of exactly $1/n$, it is clear that they are weakly better-off when the cake is enlarged or when an agent leaves. It remains to prove that this is also true for agent $n$.

Let $P_C$ be a $W_{n-1}$-optimal partition of the original cake $C$.

For AM, suppose that a cake $C$ is enlarged by a piece $E$ such that the new cake is $C\cup E$. Suppose the enlargement $E$ is divided in an $W_{n-1}$-exact way; call this partition $P_E$. Then, the partition composed of $P_C$ and $P_E$ is also a $W_{n-1}$-exact division of $C\cup E$. In this partition, agent $n$ receives the same piece as in $P_C$, so he is weakly better off. By definition, in any $W_{n-1}$-optimal partition the value of agent $n$ is weakly better-off.

For PM, suppose that an agent $i\in 1,\cdots,n-1$ leaves and abandons his piece $C_i$. Mark $E=C_i$ and $C=C\setminus C_i$. Apply the argument in the previous paragraph to prove that agent $n$ is weakly better-off.
\end{proof}

We are now ready to describe our AM+PM+PROP+PO division rule.

For every subset $K$ of the agents (a subset of $\{1,\cdots,n\}$), define:

$$ C_K := \cap_{i\in K}{supp(V_i)} \setminus \cup_{i\notin K}{supp(V_i)} $$

$C_K$ is the subset of $C$ which is valuable only for the agents in $K$.
Thus the cake $C$ is partitioned to $2^n$ pairwise-interior-disjoint sub-cakes, one sub-cake for each subset of the agents.

We define the \emph{exact-optimal division rule} as follows:

For every subset $K$ of agents, let $k=|K|$:
\begin{itemize}
\item{If $k=0$ ($K=\emptyset$), then discard the sub-cake $C_K$.}
\item{If $k=1$, then give $C_K$ to the single agent in $k$.}
\item{Otherwise, divide $C_K$ according to a $W_{k-1}$-optimal partition. I.e, give every agent in $K$ a value of $1/k$ of its valuation of $C_K$, except the agent with the largest index in $K$, which receives the largest possible value subject to this constraint.}
\end{itemize}

\begin{thm}
The exact-optimal division rule presented above is PROP+PO+AM+PM
\end{thm}
\begin{proof}
PROP is obvious, since every agent receives a proportional share from every sub-cake whose value for it is positive.

PO is also obvious, since by Lemma \ref{lemma:W-optimal-is-PO}, every division of $C_K$ is Pareto-optimal relative to that cake and the agents in $K$. Because all agents not in $K$ value $C_K$ as 0, this means that the entire division is PO.

For AM, suppose that a cake $C$ is enlarged by an enlargement $E$. Then for every subsets $K$ of the agents, the sub-cake $C_K$ is enlarged by the corresponding subset $E_K$ (which may be empty). By an argument analogous to the proof of Lemma \ref{lemma:utilitarian-optimal-is-am-pm}, in the new division of $C_K\cup E_K$, all agents of $K$ are weakly better-off. This implies that in the new division of $C$, all agents are weakly better-off.

For PM, suppose that agent $i$ leaves and abandons its share $C_i$. This has an effect only on the divisions of the sub-cakes $C_K$ where $i\in K$. In each of these sub-cakes, all agents are weakly better-off by an argument analogous to the proof of Lemma \ref{lemma:utilitarian-optimal-is-am-pm}.
\end{proof}

\begin{cor}
PROP+AM+PM+PO are compatible.
\end{cor}

\textcolor{red}{NOTE: I have a feeling that the \emph{optimal proportional} rule - the rule that selects the proportional allocation with the largest utilitarian welfare - is AM+PM. Of course it is also PROP+PO. If my feeling is correct than we can remove most of this section. But currently the proof eludes me.}

\section{Future Work}
In this paper we studied monotonicity properties in combination with the basic fairness axiom of proportionality. An interesting future research topic is to study the same axioms in combination with the stronger fairness axion of envy-freeness.

\fi

\bibliography{CCM}

\end{document}


\documentclass[main.tex]{subfiles}
\begin{document}

\marginpar{Tuesday\\ 2021-6-15, \\ compiled \\ \today}

We continue discussing the initial data problem.

\subsection{Conformal Thin Sandwich}

We can give a different decomposition of \(\widetilde{A}_{ij}\):
%
\begin{align}
\mathscr{L}_m \widetilde{\gamma}_{ij} &= \partial_{t} \widetilde{\gamma}_{ij} - \mathscr{L}_\beta \widetilde{\gamma}_{ij} = \partial_{t} \widetilde{\gamma}_{ij} + \qty( \widetilde{L} \beta )^{ij} + \frac{2}{3} \widetilde{D}_{k} \beta^{k} \widetilde{\gamma}_{ij}  \\
&= 2 \alpha \hat{A}_{ij} + \frac{2}{3} \widetilde{D}_{k} \beta^{k} \hat{\gamma}_{ij}
\,.
\end{align}

The idea is to combine these two equations in order to express \(\hat{A}_{ij}\) as follows:
%
\begin{align}
\hat{A}_{ij} = (2\alpha )^{-1} \qty[ \widetilde{\dot{\gamma}}_{ij} + \qty(\widetilde{L} \beta )_{ij}]
\,,
\end{align}
%
and similarly for \(\overline{A}_{ij}\) if we wanted to use the conformal rescaling.

This equation replaces the \(L + TT\) decomposition of \(\hat{A}_{ij}\) in CTS.
We can plug this into the momentum constraint \(C_i = 0\), and we find
%
\begin{align}
\widetilde{D}_{j} \qty[ \widetilde{a}^{-1} \qty(\widetilde{L} \beta )^{ij}] + \widetilde{D}_{j} \qty[\widetilde{\alpha}^{-1} \widetilde{\dot{\gamma}}^{ij}]
- \frac{4}{3} \psi^{6} \widetilde{D}^{i} K
- 16 \pi \widetilde{P}^{i} = 0
\,.
\end{align}

This must be paired with the Lichnerowicz equation:
%
\begin{align}
\widetilde{D}^{i} \widetilde{D}_{i} \psi + \dots = 0
\,.
\end{align}

The free data will be \(\widetilde{\gamma}_{ij}\), \(K\), \(\widetilde{\dot{\gamma}}_{ij}\), \(E\) and \(P^{i}\).
The constrained data can be \(\psi \) and \(\beta^{i}\).

The difference from before is that we have substituted this TT part with the time derivative of the conformal metric; in some sense specifying \(\widetilde{\dot{\gamma}}\) can help in specifying free data, for example stationary data.

The term \(\widetilde{\alpha} = \psi^{-6} \alpha \) is a conformally rescaled lapse.
The maximal slicing condition \(K=0\) means that the CTS equations decouple.
The momentum constraint for \(P^{i}\) turns out to be \emph{linear} in this case.

York and Pfeiffer in 2003 proposed a formalism called XCTS, extended CTS, where the idea is to specify an equation for the conformal lapse \(\widetilde{\alpha}\) in order not to have to specify it.

Let us consider
%
\begin{align}
\mathscr{L}_m K &= \dot{K} - \beta^{i} \widetilde{D}_{i} K   \\
&= - \psi^{-4} \qty(\widetilde{D}_{i} \widetilde{D}^{i} K + 2 \widetilde{D}_{i} \log \psi \widetilde{D}^{i} \alpha )
+ \alpha \qty[ \dots]
\,:
\end{align}
%
the piece multiplied by \(\psi^{-4}\) can be written as
%
\begin{align}
\widetilde{D}_{i} \widetilde{D}^{i} K + 2 \widetilde{D}_{i} \log \psi \widetilde{D}^{i} \alpha =
\psi^{-1} \qty[\widetilde{D}_{i} \widetilde{D}^{i} (\alpha \psi ) + \alpha \widetilde{D}^{i} \widetilde{D}_{i} \psi ]
\,,
\end{align}
%
but using the Hamiltonian constraint \(C_0 = 0\) we can write this as
%
\begin{align}
\widetilde{D}_{i} \widetilde{D}^{i} \qty(\hat{a} \psi^{7})
- \qty(\hat{a} \psi^{7}) \qty[ \frac{\hat{K}}{8}
+ \frac{5}{12} K^2 \psi^{4} + \frac{7}{8} \hat{A}_{ij} \hat{A}^{ij}
\psi^{-8} + 2 \pi \qty( E + 25 \psi^{8}) \psi^{-4}]
+ \qty(\dot{K} - \beta^{i} \widetilde{D}_{i} K ) \psi^{5} = 0
\,.
\end{align}

The XCTS scheme uses this equation, the Lichnerowicz equation and the one previously dubbed equation 1.

Under \(K = 0\) these do not decouple; however there is a nice feature: free data includes \(\widetilde{\gamma}_{ij} \), \(\widetilde{\dot{\gamma}}\), \(K\) and \(\dot{K}\). Being able to specify these is very convenient.

There exist examples for which XCTS yields non-unique solutions.
Let us give a simple example: the solution of these for conformally, asymptotically flat \& zero-derivative vacuum initial data.

The equation reads
%
\begin{align}
\triangle \psi + \frac{1}{8} \hat{A}_{ij} \hat{A}^{ij} \psi^{-7} &= 0  \\
D_j \qty[\widetilde{\alpha }^{-1} (L \beta )^{ij} ] &= 0 iln
\triangle \qty(\widetilde{a} \psi^{7}) - \frac{7}{8} \hat{A}_{ij} \hat{A}^{ij} \widetilde{\alpha} \psi^{-7} &= 0
\,.
\end{align}

Asymptotically flat means \(\psi = 1\), \(\beta^{i} = 0\), \(\alpha = 1\) as \(r \to \infty \).
Then, \(\beta^{i} = 0\) is a solution we can take, together with \(\widetilde{\dot{\gamma}} = 0\), which means \(\hat{A}^{ij} = 0\).

The equations then read
%
\begin{align}
\triangle \psi &= 0  \\
\triangle (\hat{\alpha} \psi^{7}) &= 0
\,.
\end{align}

The inner boundary condition as before is a punctured space: \(\Sigma_0 = \mathbb{R}^3 \setminus 0\).
This yields \(\psi = 1 + M / 2r\).

The solution of the second equation is in the same form: \(\phi = \widetilde{\alpha} \psi^{7} = 1 + a / r\) for some constant \(a\).
Now, \(\alpha = \psi^{6} \widetilde{\alpha} = \phi \psi^{-1}\); therefore the lapse reads
%
\begin{align}
\alpha = \qty(1 + \frac{a}{r}) \qty(1 + \frac{M}{2r})^{-1}
= \frac{r + a}{r + M/2}
\,.
\end{align}

What is the constant \(a\), though? Remember that we are solving this equation on the punctured \(\mathbb{R}^3\): \(a\) is fixed by the choice of the value of the lapse \(\alpha \) at a certain radius, say \(r \to 0\).
The simplest choice is \(\alpha_0 = +1\): this yields \(a = M/2\), which also means that the lapse is identically equal to 1.

Therefore, we have re-found Schwarzschild in isotropic coordinates and geodesic gauge,
%
\begin{align}
g = - \dd{t^2} + \psi^{4} \qty(\dd{r^2} + r^2 \dd{\Omega^2})
\,.
\end{align}

What if we set \(\alpha_0 = -1\)? We know that \(\alpha > 0\), but let us explore this crazy condition anyways.
This yields
%
\begin{align}
a = - \frac{M}{2}
\,,
\end{align}
%
which implies
%
\begin{align}
\alpha = \qty(1 - \frac{M}{r}) \qty(1 + \frac{M}{2r})
\,.
\end{align}

This is the lapse in isotropic coordinates.

We have a moment of time symmetry, but the time development of \(\Sigma_0 \) are different!
If we take \(\alpha_0 = +1\), \(\Sigma_0 \) evolves in a nontrivial way!
If, instead, \(\alpha_0 = -1\) then \(\Sigma_0 \) does not evolve, and \(\partial_{t}\) is a Killing vector.

What do we do with this negative lapse? In some points the \(\vec{n}\) vector will remain future-pointing, but we will be in a weird coordinate system such that the time starts to run backwards at small radii.
We have to learn to live with it.

Where is XCTS useful?
Imagine a binary system: here there is no \(\partial_{t}\) timelike Killing vector, and despite the rotation there is no rotational Killing vector \(\partial_{\varphi }\).

Suppose we are considering a circular orbit with frequency \(\Omega \):
we can define
%
\begin{align}
n^a = \qty(\partial_{t})^{a} + \Omega \qty(\partial_{\varphi })^{a}
\,,
\end{align}
%
which is conserved! This is because the orbit looks like a helix in \(t, r, \varphi \) space: the direction of \(n\) is always along this helix.

This is an idealized, approximate situation, in which there is no radiation.

In a comoving frame with \(\dot{\gamma}_{ij} = 0 = \dot{K}\) we can use a CTT + Bowman-Yosk puncture method.

The other part of the discussion for today is about gauge conditions.

\subsection{Gauge condition}

We need to make a choice of the foliation and a choice of the spatial coordinates.
We want our choice to
\begin{enumerate}
    \item avoid singularities;
    \item enhance symmetries;
    \item minimize grid distortions.
\end{enumerate}

Something useful is to have our slicing be as close as possible to some Kerr coordinates.
If some Killing vector is present we want to stay close to it.

Let us start with \textbf{slicing} (\(\alpha \)).
We have already discussed \emph{geodesic slicing} in which \(\alpha = 1\) and \(\beta^{i} = 0\): the coordinates follow freely-falling observers.

Here, \(n^a = (\partial_{t})^{a}\), and \(\alpha^{a} = D_a \log \alpha \). This also means that \(t = \tau \).

This is certainly not singularity-avoiding: free-falling observers fall into the singularity. Also, it does not seek symmetries, and it distorts the grid a lot.

In this gauge, the ADM equation reads
%
\begin{align}
\partial_{t} K &= K_{ij} K^{ij} + 4 \pi (E - S)  \\
\partial_{t} \log \sqrt{ \gamma } &= - K
\,.
\end{align}

The behaviour of this gauge can be seen from these equations if we think about gravitational collapse:
we can draw a \(r\) versus \(t\) diagram and see that matter eventually falls into the horizon.

\(E-S\) gets large, \(\partial_{t} K\) gets large, and \(\partial_{t} \log \sqrt{\gamma }\) gets small. The curvature increases, and the coordinate volume element decreases. These are all characteristics of the \emph{bad gauges}.

Another gauge we have seen is \textbf{maximal slicing}: \(0 = K\), and we already discussed its geometric meaning.

If we look at \(0 = K = - \nabla_a n^a\), the picture we can draw suggests the motion of an incompressible fluid: the coordinate volume element cannot be squeezed.

The Maximal Slicing equation can be also written as
%
\begin{align}
D_i D^{i} \alpha = - \alpha \qty[ 4 \pi (E-S) + K_{ij} K^{ij} ] = 0
\,,
\end{align}
%
to be compared with the previous geodesic slicing ones.
Here the acceleration of an Eulerian observer will be different from 0, and the lapse \(\alpha \) will evolve in a certain way which counteracts the effect of this volume element compression, as well as counteracting the ``singularity seeking'' property of geodesic slicing.

\end{document}