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\subsection{Quantum Chromo Dynamics}

\marginpar{Wednesday\\ 2020-4-8, \\ compiled \\ \today}

Asymptotic freedom is not a political slogan from the sixties!

With this lecture and the next we should finish discussing the standard model of particle physics.

Today we are going to explore a property which is profoundly different between EM and the strong interaction.

We introduced the quantity \(\alpha_{EM} = e^2 / 4 \pi \): how do we measure this parameter?
We know that the interaction term looks like
%
\begin{align}
\mathscr{L} \sim e j^{\mu }A_{\mu }
\,,
\end{align}
%
so we can measure \(e\) using the cross sections of processes, such as \(e^{+} e^{-} \to \mu^{+} \mu^{-}\); the free parameters are just \(e\) and the COM energy, so we are done.
But how do we do it without an accelerator? We could do the Millikan drop experiment, for example.

But we are doing Quantum Field Theory: beyond physical particles we can also have virtual particles, which despite the name can influence physical processes: inside the ``black box'' within which the interactions occur we can have off-shell processes.

In fact, in the aforementioned decay the photon is off mass shell; but we can go beyond. The photon can create and then annihilate an \(e^{+} e^{-}\) pair.

The vacuum is a \emph{quantum} vacuum, which does correspond to a minimal energy but it is filled by the continuous creation and annihilation of these particle-antiparticle pairs, since there is an indetermination between time and energy.
It is important that after being created these pairs are indeed destroyed.
This is called \emph{vacuum polarization}.

The pairs are virtual but they have an effect. The coupling constant \(\alpha \) has a numerical value of around \(1 / 137\), and the loop diagrams of increasing number of loops depend on increasing powers of \(\alpha \).
The fact that \(\alpha \) is small allows us to work perturbatively.

This works, the predictions of QED are extremely precise and correspond to experiment.
For instance, the anomalous magnetic moment of the electron: in a QFT, when an electron interacts with a photon, there can be other particles.

The Feynman diagram which governs the interaction of an external electromagnetic field with a fermion is a one-vertex one; however we can draw diagrams with loops (the first correction is shown in figure \ref{fig:fermion-external-EM-field-feynman-diagram}) to calculate the perturbative corrections.

\begin{figure}[ht]
\centering
\feynmandiagram[horizontal = e1 to e2]{
e1 -- [fermion] x -- [fermion] e2,
gamma [crossed dot] -- [photon] x
};
\qquad
\feynmandiagram[horizontal = e1 to e2]{
e1 -- [fermion] a -- [fermion] x -- [fermion] b -- [fermion] e2,
gamma [crossed dot] -- [photon] x,
a -- [photon] b
};
\caption{Interaction of a fermion with an external electromagnetic field, to first and third order respectively.}
\label{fig:fermion-external-EM-field-feynman-diagram}
\end{figure}

The polarization of the vacuum creates a screening effect, as shown in an example scattering in figure \ref{fig:vacuum-polarization-feynman-diagram}.
This is hard to compute analytically, but the result can be readily described: the vacuum acts as a dielectric medium, screening part of the electromagnetic charge and making the interaction weaker.

\begin{figure}[ht]
\centering
\feynmandiagram[vertical = x to y]{
e1 -- [fermion] x -- [fermion]  e2,
x -- [photon] y,
m1 -- [anti fermion] y -- [anti fermion] m2
};
\qquad
\feynmandiagram[vertical = x to y]{
e1 -- [anti fermion] x -- [anti fermion] e2,
x -- [photon] z,
z -- [fermion, half right] w,
z -- [anti fermion, half left] w,
w -- [photon] y,
m1 -- [fermion] y -- [fermion] m2
};
\caption{Vacuum polarization: second and fourth order contributions to \(e^{-}e^{-} \to e^{-}e^{-}\) scattering.}
\label{fig:vacuum-polarization-feynman-diagram}
\end{figure}

So, we expect the effective \(\alpha_{EM}\) to be larger: if the vacuum is filled with a sea of screening pairs, at large distances (low energies) it is effectively uniform, while if we get at distances lower than \(1/m_e\) we start to be on smaller scales than the maximum size of these dipoles, so screening is harder and we can see the ``bare charge'', which formally diverges at a point called the \textbf{Landau pole}.

This is an illustrative description of a mathematical process called \textbf{renormalization}: we impose the fact that the loop contributions to the theory should not diverge, and this forces us to have a running coupling to the theory, which corresponds to increased pair production (since the EM field is coupled more strongly with the fermions!).

We compare the Millikan experiment (\(E \sim 0\)), where we find \(\alpha \sim 1/137\), with LEP: now \(E \sim \SI{100}{GeV}\), and we find \(\alpha \sim 1/128\)!
A figure depicting the complete relation can be found in Peskin \cite[fig.\ 11.1]{peskinConceptsElementaryParticle2019}.
At very high energies we must also account for pair production of higher-mass fermions, such as muons.

What is the theoretical relation between \(\alpha \) and \(Q\), the momentum transfer of the process at hand? It comes out to be (the calculation can be found in an older book by Peskin and Schroeder \cite[sec.\ 7.5]{peskinIntroductionQuantumField1995})
%
\begin{align}
\alpha (Q) = \frac{\alpha_0 }{1 - \frac{2 \alpha_0 }{3 \pi } \log(\frac{Q}{Q_0 })}
\,.
\end{align}

At energies \(Q \lesssim m_e\) the correction is negligible and \(\alpha \sim 1/136\), it becomes significantly different at higher energies.

We also define the function \(\beta \): it is a function of the coupling constant \(\alpha \), defined by
%
\begin{align}
\beta (g_e ) = \dv{g_e }{\log Q}
\,.
\end{align}

What happens if we compute this for the strong interaction? We define, analogously to the electromagnetic case:
%
\begin{align}
\alpha_s = \alpha _{\text{strong}} = \frac{g^2_{s}}{4 \pi }
\,.
\end{align}

The problem is that this is a large number: but our tools are perturbative! How do we compute the cross sections then?
The thing is, with increasing energy this \(\alpha_s \) becomes \emph{smaller}!
Why is this? The gluons are self interacting, because of the \(A \wedge A\) term in the field strength of the strong interaction, as opposed to photons.

In the EM case, we had \(\gamma \to e^{+} e^{-} \to \gamma \) as a virtual particle process; in the chromodynamics case instead we have \(g \to g + g \to g \): it is the case (even though it is not easy to prove) that this loop contribution has the opposite sign! We define \(\beta \) in the same way as the one from QED, so we find \cite[eq.\ 11.65]{peskinConceptsElementaryParticle2019}:
%
\begin{align}
\dv{g_{s}}{\log Q} = \beta (g_s)
\qquad \text{where} \qquad
\beta = - \frac{11}{3} C(G) +  \frac{4}{3} n_f C(R)
\,,
\end{align}
%
where \(C(G)\) is the Casimir\footnote{This means that it is the constant depending on the dimension of the representation which appears in
%
\begin{align}
\Tr[t^{a}_{N} t^{b}_{N}] = C(N) \delta^{ab}
\,.
\end{align}} of the \textbf{adjoint} representation: \(C(G) = 3\), while \(C(R) = 1/2\) pertains to the \textbf{fundamental} representation, to which fermions belong: \(n_f\), the number of fermions, is 6. Then, the final result is negative!

This is called asymptotic freedom: \(\alpha \) decreases as \(Q\) increases.
If the energy is high enough, the quarks and gluons are effectively decoupled.
On the other hand, at low energy their coupling is very large.

So, at the low energies we look at in atomic nuclei the quarks are confined by an extremely high coupling: no one has ever observed a free quark, not even at LHC.
This is called hadronization, since the quarks are always bound into hadrons.
ALICE, at LHC, is looking for the phase transition between this ``infrared slavery'' of the quarks and their free state.

Since \(\alpha_{S}\) decreases, while \(\alpha_{EM}\) increases, there should be a point at which they cross and the electromagnetic interaction becomes stronger than the strong one.

There might be a point at which they merge, like the weak and electromagnetic interactions do.
Actually, it might be possible that all of the interactions are manifestations of the same kind of interaction.

The crucial point is that the strength of an interaction depends on the energy. If we define \(b_0 = 11 - 2 n_f / 3\) we find
%
\begin{align}
\alpha_s (Q) = \frac{ 2 \pi / b_0 }{\log(Q / \Lambda )}
&&
\Lambda = Q_0 \exp(\frac{-2 \pi }{b_0 \alpha_{s}(Q_0 )})
\,,
\end{align}
%
where we defined \(\Lambda \), the QCD scale: it is the energy at which the QCD coupling becomes strong.

In QED (and more generally in abelian gauge theories) the vertex only has one photon line, so there can be no photon loops.
In QCD, instead, we can have gluon loops.

\subsection{The weak interaction}

How do we formally describe whether an interaction is ``strong''?
We know that the time needed for hadronization  is of the order \(\tau _{\text{had}} \sim \SI{e-23}{s}\), while the time for a pion's decay is \(\tau (\pi^{+}) \sim \SI{3e-8}{s}\). These are strong interaction-mediated processes.

The time for beta decay --- a weak interaction mediated process --- is \(\tau (n \to p + e^{-} + \overline{\nu}_{e}) \sim \SI{880}{s}\).

This can give us a first signal for the fact that the weak interaction is \emph{weak}.

We know that the neutron is \(u d d \) in terms of quarks, while the proton is \(u u d\): so, in the beta decay one down quark must become an up quark.
There is no term in QED or QCD in which this can happen: QED and QCD conserve flavour. This can be seen in the Lagrangian: there are terms which mix colors, while the flavour terms are diagonal.

But we do see beta decay: so, we must introduce a new \emph{weak interaction}.

Beta decay is experimentally observed to be a three-body process, which is why observations of it were the first indication of the existence of neutrinos.
Beta-decay is not \(P\)-symmetric, whereas QED and QCD are.
This was seen in the \(\beta^{-}\) decay of \(\ce{^{60}Co}\) nuclei, they always emitted left-handed electrons.

In QED and QCD we see current-current interaction by means of the mediator (photon or gluon, respectively); will we see current-current weak interactions?

\todo[inline]{Will we?}

\end{document}


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\begin{document}

\title{ToFu geometric tools\\ Intersection of a LOS with a cone}
\author{Didier VEZINET \and Laura S. Mendoza}
\date{02.06.2017}
\maketitle

\tableofcontents

\chapter{Definitions}

\section{Geometry definition in ToFu}

The definition of a fusion device in ToFu is done by defining the edge of a poloidal plane as a set of segments in a 2D plane. The 3D volume is obtained by an extrusion for cylinders or a revolution for tori.
We consider an orthonormal direct cylindrical coordinate system $(O,\ud{e}_R,\ud{e}_{\theta},\ud{e}_Z)$ associated to the orthonormal direct cartesian coordinate system $(O,\ud{e}_X,\ud{e}_Y,\ud{e}_Z)$. We suppose that all poloidal planes live in $(R,Z)$ and can be obtained after a revolution around the $Z$ axis of the user-defined poloidal plane at $\theta=0$, $\mathcal{P}_0$. Thus, the torus is axisymmetric around the $(O,Z)$ axis (see Figure~\ref{fig:tok-ab}).

\begin{figure}[h]
\centering{
\def\svgwidth{0.75\linewidth}
\import{figures/}{tore_cones12.pdf_tex}
\caption{Two examples of a circular torus approximated by a revolved octagon. For each segment $\overline{AB}$ of the octagon there is a cone with origin on the $(O,Z)$ axis.}
\label{fig:tok-ab}
}
\end{figure}

\section{Notations}

In order to simplify the computations, let $A$ and $B$ be the end points of a segment $\mathcal{S}_i$ such that $A\neq B$ and $\mathcal{P}_0 = \cup_{i=1}^{n} \mathcal{S}_i = \cup_{i=1}^n \overline{{\rm A}_i{\rm B}_i}$ with $n$ the number of segments given by the user defining the plane $\mathcal{P}_0$. We define a right circular cone $\mathcal{C}$ of origin $P = ({\rm A},{\rm B}) \cap ({\rm O}, {\rm Z})$ of generatrix $(A,B)$ and of axis $(O,Z)$ (see Figure~\ref{fig:tok-ab}). Thus we can define the edge of the torus as the union of the edges of the frustums $\mathcal{F}_i$ defined by truncating the cones $\mathcal{C}_i$ to the segment $\overline{AB}_i$.

Then, any point $M$ with coordinates $(X,Y,Z)$ or $(R,\theta,Z)$ belongs to the frustum $\mathcal{F}$ if and only if
$$
\exists q \in [0;1] /
\left\{ \begin{array}{ll}
R-R_A = q(R_B-R_A)\\
Z-Z_A = q(Z_B-Z_A)
\end{array}\right.
$$

Now let us consider a LOS $L$ (i.e.: a half-infinite line) defined by a point $D$ and a normalized directing vector $u$, of respective coordinates $(X_D,Y_D,Z_D)$ or $(R_D,\theta_D,Z_D)$ and $(u_X,u_Y,u_Z)$.
Then, point M belongs to $L$ if and only if:
$$
\exists k \in [0;\infty[ / \ud{DM} = k\ud{u}
$$

\chapter{Derivation}

Let us now consider all intersections between the edge of a frustum $\mathcal{F}$ and a semi-line $L$.

\begin{equation}
\begin{array}{lll}
\exists (q,k) \in [0;1]\times [0;\infty[ /& &
\left\{\begin{array}{ll}
R-R_A = q(R_B-R_A)\\
Z-Z_A = q(Z_B-Z_A)\\
X-X_D = ku_X\\
Y-Y_D = ku_Y\\
Z-Z_D = ku_Z
\end{array}\right.\\
\end{array}
\label{eqn:201}
\end{equation}

Which yields (by combining to keep only unknowns $q$ and $k$):

\begin{equation}
\begin{array}{cc}
q(Z_B-Z_A) = Z_D-Z_A + ku_Z\\
q^2(R_B-R_A)^2 + 2qR_A(R_B-R_A) = \left(k\ud{u}_{//} + \ud{D}_{//}\right)^2 - R_A^2
\end{array}
\end{equation}

Where we have introduced $R_D = \sqrt{X_D^2+Y_D^2}$, $\ud{u}_{//} = u_X\ud{e}_X+u_Y\ud{e}_Y$ and $\ud{D}_{//} = X_D\ud{e}_X + Y_D\ud{e}_Y$.
We can then derive a decision tree.

Given that the parallelization will take place on the LOS (i.e.: not on the cones which are parts of the vacuum vessel), we will discriminate case based prioritarily on the components of $\ud{u}$ and $D$.
We will detail only the cases which have solutions, in order to make it as clear as possible for implementation of an efficient algorithm.
We will also only consider non-tangential solution, as we are looking for entry/exit points.

\section{Horizontal LOS: $u_Z=0$}

Let us consider an horizontal LOS, such that $u_Z=0$, then \eqref{eqn:201} becomes

$$
\begin{array}{lll}
\exists (q,k) \in [0;1]\times [0;\infty[ /& &
\left\{\begin{array}{ll}
R-R_A = q(R_B-R_A)\\
Z_D-Z_A = q(Z_B-Z_A)\\
X-X_D = ku_X\\
Y-Y_D = ku_Y\\
Z=Z_D
\end{array}\right.\\
\end{array}
$$

From here we can differentiate two cases regarding the frustum $\mathcal{F}$.

\subsection{Plane Frustum: $Z_B = Z_A$}

Let us consider first the case where $Z_B=Z_A$, when the frustum becomes an annulus on the $(X,Y)$ plane, then we will have two different cases.

\begin{itemize}
\item $Z_D\neq Z_A \Rightarrow$ the cone and the LOS stand in different parallel planes $\Rightarrow$ no solution.
\item $Z_D=Z_A \Rightarrow$ the cone stands in the same plane as the LOS (see \ref{fig:hoz-frus-hoz-los}) $\Rightarrow$ infinity of solutions, we consider no solutions as this is a limit case with no clearly identified intersection.
\end{itemize}

\begin{figure}[h]
\centering{
\def\svgwidth{0.5\linewidth}
\import{figures/}{horizontal_cone_hoz_LOS.pdf_tex}
\caption{Plane frustum and horizontal Line of Sight on the same $Z$-plane.}
\label{fig:hoz-frus-hoz-los}
}
\end{figure}

Hence, the only derivable solutions suppose that $Z_B\neq Z_A$.

\subsection{Non-horizontal cone: $Z_B\neq Z_A$}

Then $q=\dfrac{Z_D-Z_A}{Z_B-Z_A}$.
There are acceptable solution only if $q\in[0;1]$.
By introducing $$C = q^2(R_B-R_A)^2 + 2qR_A(R_B-R_A) + R_A^2,$$ we have
$$
\left(k\ud{u}_{//} + \ud{D}_{//}\right)^2 - C = 0\\
\Leftrightarrow k^2\ud{u}_{//}^2 + 2k\ud{u}_{//}\cdot\ud{D}_{//} + \ud{D}_{//}^2-C = 0
$$

Then introducing $\Delta = 4\left(\ud{u}_{//}\cdot\ud{D}_{//}\right)^2 - 4\ud{u}_{//}^2\left(\ud{D}_{//}^2-C\right) = 4\delta$, there are non-tangential solutions only if $\left(\ud{u}_{//}\cdot\ud{D}_{//}\right)^2 >\ud{u}_{//}^2\left(\ud{D}_{//}^2-C\right)$.
It is necessary to compute the solutions k because we need to check if $k>=0$.

$$
k_{1,2} = \dfrac{-\ud{u}_{//}\cdot\ud{D}_{//} \pm \sqrt{\delta}}{\ud{u}_{//}^2}
$$

Hence, we have solutions if:
$$
\left\{
\begin{array}{lll}
u_Z = 0\\
Z_B\neq Z_A\\
\dfrac{Z_D-Z_A}{Z_B-Z_A} \in [0;1]\\
k_{1,2} = \dfrac{-\ud{u}_{//}\cdot\ud{D}_{//} \pm \sqrt{\delta}}{\ud{u}_{//}^2} \geq 0
\end{array}
\right.
$$

\section{Non-horizontal LOS: $u_Z\neq0$}

Then $k=q\dfrac{Z_B-Z_A}{u_Z} - \dfrac{Z_D-Z_A}{u_Z}$, which means:

$$
\begin{array}{lll}
q^2 & (R_B-R_A)^2 + 2qR_A(R_B-R_A) + R_A^2\\
& =\left(\left(q\dfrac{Z_B-Z_A}{u_Z} - \dfrac{Z_D-Z_A}{u_Z}\right)\ud{u}_{//} + \ud{D}_{//}\right)^2\\
& = \left(q\dfrac{Z_B-Z_A}{u_Z} - \dfrac{Z_D-Z_A}{u_Z}\right)^2\ud{u}_{//}^2 + 2\left(q\dfrac{Z_B-Z_A}{u_Z} - \dfrac{Z_D-Z_A}{u_Z}\right)\ud{u}_{//}\cdot\ud{D}_{//} + \ud{D}_{//}^2\\
& =q^2\left(\dfrac{Z_B-Z_A}{u_Z}\right)^2\ud{u}_{//}^2 - 2q\dfrac{Z_B-Z_A}{u_Z}\dfrac{Z_D-Z_A}{u_Z}\ud{u}_{//}^2 \\
& + \left(\dfrac{Z_D-Z_A}{u_Z}\right)^2\ud{u}_{//}^2
+ 2q\dfrac{Z_B-Z_A}{u_Z}\ud{u}_{//}\cdot\ud{D}_{//} - 2\dfrac{Z_D-Z_A}{u_Z}\ud{u}_{//}\cdot\ud{D}_{//} + \ud{D}_{//}^2\\
\end{array}
$$

Hence:
$$
\begin{array}{lll}
0 &=& q^2\left( (R_B-R_A)^2 - \left(\frac{Z_B-Z_A}{u_Z}\right)^2\ud{u}_{//}^2 \right)\\
& + & 2q\left( R_A(R_B-R_A) + \frac{Z_B-Z_A}{u_Z}\frac{Z_D-Z_A}{u_Z}\ud{u}_{//}^2 - \frac{Z_B-Z_A}{u_Z}\ud{u}_{//}\cdot\ud{D}_{//}  \right)\\
& - & \left(\frac{Z_D-Z_A}{u_Z}\right)^2\ud{u}_{//}^2 + 2\frac{Z_D-Z_A}{u_Z}\ud{u}_{//}\cdot\ud{D}_{//} - \ud{D}_{//}^2 + R_A^2
\end{array}
$$

We can then introduce:
$$
\left\{
\begin{array}{ll}
A = (R_B-R_A)^2 - \left(\frac{Z_B-Z_A}{u_Z}\right)^2\ud{u}_{//}^2\\
B = R_A(R_B-R_A) + \frac{Z_B-Z_A}{u_Z}\frac{Z_D-Z_A}{u_Z}\ud{u}_{//}^2 - \frac{Z_B-Z_A}{u_Z}\ud{u}_{//}\cdot\ud{D}_{//}\\
C = -\left(\frac{Z_D-Z_A}{u_Z}\right)^2\ud{u}_{//}^2 + 2\frac{Z_D-Z_A}{u_Z}\ud{u}_{//}\cdot\ud{D}_{//} - \ud{D}_{//}^2 + R_A^2
\end{array}
\right.
$$

Because of the shape of potential solutions, we have to discriminate the case $A=0$.

\subsection{$A=0$: LOS parallel to one of the cone generatrices}

Then, because of the shape of the potential solution, we have to discriminate the case $B=0$.
But in this case we have $C=0$.
\begin{itemize}
\item if $C=0 \Rightarrow$ no condition on q and k, the LOS is included in the cone $\Rightarrow$ we consider no solution
\item if $C\neq 0 \Rightarrow$ Impossible, no solution
\end{itemize}

Only the case $B\neq 0$ is thus relevant.

\subsubsection{$B\neq0$: LOS not included in the cone}

Then, there is either one or no solution:
$$
\left\{
\begin{array}{ll}
q = -\frac{C}{2B} & \in [0,1]\\
k = q\frac{Z_B-Z_A}{u_Z} - \frac{Z_D-Z_A}{u_Z} & \geq 0
\end{array}
\right.
$$

\subsection{$A\neq 0$: LOS not parallel to a cone generatrix}

Then, we only consider cases with two distinct solutions (i.e.: no tangential case):
$$
\left\{
\begin{array}{ll}
B^2 > AC\\
q = \frac{-B \pm \sqrt{B^2-AC}}{A} & \in [0,1]\\
k = q\frac{Z_B-Z_A}{u_Z} - \frac{Z_D-Z_A}{u_Z} & \geq 0
\end{array}
\right.
$$

\appendix
\chapter{Acceleration radiation from a unique point-like charge}

\section{Retarded time and potential}
\subsection{Retarded time}

\subsubsection{Deriving the retarded time}
\label{Ap:RetardTime}

Hence $\frac{dR(t_r)}{c} + dt_r = dt$

\subsection{Retarded potentials}
\subsubsection{Deriving the potential propagation equations}
\label{Ap:PotentialPropagation}

\end{document}


\section{Stern-Gerlach again}
This is where we have some real fun. Stern-Gerlach was one of a host
of experiments being developed around the first third of the previous
century that gave rise to quantum mechanics. Modeling Stern-Gerlach is
seen as one of the tests of quantum mechanical theory.

Here's the Wikipedia link describing the experimental set up and
relevance to the development of quantum mechanics:

\href{http://en.wikipedia.org/wiki/Stern\%E2\%80\%93Gerlach_experiment}{\underline{Stern-Gerlach on Wikipedia}}

\subsection{Modeling the experiment}

TBD


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\newcommand{\fst}{$F_{\mathrm{ST}}$}

\begin{document}
%\thispagestyle{firstpage}

\subsection*{Fixation index \fst{} (\numpoints\ points)}

The fixation index $(F_\mathrm{ST})$ is useful for estimating the amount of genetic structuring and gene flow among populations. \fst{} can be calculated by several related equations. The equations you will use are
\begin{multicols}{2}
\noindent
\begin{equation}
F_\mathrm{ST} = \dfrac{2pq-\sum c_i2p_iq_i}{2pq}
\end{equation}

%\[F_\mathrm{ST} = \dfrac{2pq-\sum c_i2p_iq_i}{2pq}\]

	\columnbreak

\noindent
\begin{equation}
F_\mathrm{ST} = \dfrac{1}{1+4N_em},
\end{equation}%\[F_\mathrm{ST} = \dfrac{1}{1+4N_em}, \]
\end{multicols}

where $p$ is the frequency of the first allele, $q$ is the frequency of the second allele, $c_i$ is the contribution of each population to the total (the “weight” of each sample), $N_e$ is effective population size, and $m$ is the migration rate.

Be sure to review the lecture notes to understand how to calculate the sample weights and mean heterozygosity when sample sizes are unequal.

\textbf{Round your final answers to three digits after the decimal point except for the final question. Show your work for each question!}

\subsubsection*{Problem set}

\begin{questions}
	\question[4]

	Calculate \fst{} for populations \textsc{a} and \textsc{b}.

	\begin{multicols}{2}

		\begin{tabular}{rccc}
			\toprule
			& $N$ & $p$ & $q$ \tabularnewline
			\midrule
			A & 50 & 0.44 & 0.56 \tabularnewline
			B & 50 & 0.72 & 0.28 \tabularnewline
			\bottomrule
		\end{tabular}

		\columnbreak

		\ifprintanswers \fst{} $= 0.080$ \else \phantom{\fst{} $= 0.080$} \fi
	\end{multicols}

\question[4]\label{q:migration}

Calculate \fst{} for populations \textsc{a}, \textsc{b}, \textsc{c}, and \textsc{d}.

\begin{multicols}{2}
	\begin{tabular}{rccc}
	\toprule
  & $N$ & $p$  & $q$ \tabularnewline
  \midrule
A & 50  & 0.44 & 0.56 \tabularnewline
B & 50  & 0.72 & 0.28 \tabularnewline
C & 50  & 0.20 & 0.80 \tabularnewline
D & 50  & 0.16 & 0.84 \tabularnewline
\bottomrule
\end{tabular}

\columnbreak

\ifprintanswers \fst{} $= 0.212$ \else \phantom{\fst{} $= 0.212$} \fi

\end{multicols}

\question[4]\label{q:unequal}

Calculate \fst{} for populations \textsc{a}, \textsc{b}, and \textsc{c}.

\begin{multicols}{2}
	\begin{tabular}{rccc}
	\toprule
	& $N$ & $p$  & $q$ \tabularnewline
	\midrule
	A & 40  & 0.81 & 0.19 \tabularnewline
	B & 100 & 0.27 & 0.73 \tabularnewline
	C & 60  & 0.35 & 0.65 \tabularnewline
	\bottomrule
\end{tabular}

\columnbreak

\ifprintanswers \fst{} $= 0.178$ \else \phantom{\fst{} $= 0.178$} \fi

\end{multicols}

The last two questions require you to algebraically rearrange equation~(2) above to solve for $m$ or $N_e$.

\textsc{Hint:} use $F_\mathrm{ST} = 0.2$ and $N_e = 100$ or $m = 0.01$ as needed to test your rearranged equation. Using $F_\mathrm{ST} = 0.2$ and $N_e = 100$ should give you $m = 0.01$. Using $F_\mathrm{ST} = 0.2$ and $m = 0.01$ should give you $N_e = 100$. If so, then you are ready to proceed to the last two questions. If not, find and fix your mistake(s).

\question[4]
Use the \fst{} value you calculated for question~\ref{q:migration} and $N_e = 2,200$ individuals to estimate $m$.

\ifprintanswers
	\fst{} $= 0.000422384$ \quad $4.22 \times 10^{-4}$

	\vspace{2\baselineskip}
\else
	\vspace{2\baselineskip}
\fi

\question[4]
Use the \fst{} value you calculated for question~\ref{q:unequal} and $m = 0.000005\ (5\times 10^{}-6)$ to estimate $N_e$. Round your answer to the nearest whole individual.

\ifprintanswers
	$N_e = 230,898$\newline
	\vspace{\baselineskip}
\else
	\vspace{2\baselineskip}
\fi

\end{questions}

\end{document}


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\begin{document}

\begin{center}
\rule{\textwidth}{3pt}
Master MVA \\
Probabilistic Graphical Models \\
HM2 - 21/11/2016 \\
\textsc{ Achari Berrada Youssef} \\
\rule{\textwidth}{.3pt}
\end{center}

\section{Entropy and Mutual Information}
\begin{enumerate}
\item $X$ a discrete variable on a finite space $\mathcal{X}$ of cardinal $k$.
\begin{enumerate}
\item We know that the entropy is
\[
H(X) = - \sum_{x \in \mathcal{X}} p(x) \log(p(x))  \geq 0
\]
Because $0 \leq p(x) \leq 1 \Rightarrow \log(p(x)) \leq 0 \Rightarrow p(x) \log(p(x)) \leq 0 $, $\forall x \in \mathcal{X} $. \\
So the entropy is greater than or equal to zero. \\
If $H(X) = 0$, then $\forall x \in \mathcal{X} , \quad p(x) \log(p(x)) = 0 \Rightarrow \forall x \in \mathcal{X} ,  p(x) = 1 \, \text{or} \, p(x) = 0 $ \\
As we know that $\sum_{x \in \mathcal{X}} p(x)   = 1 $, so we conclude that $\exists \, x \in \mathcal{X} \text{ such that } \, p(x) =1$ and null elsewhere. So X is constant.

\item Let $q$ be the uniform distribution over $\mathcal{X}$.
We have

\begin{align*}
D(p||q)  & = \sum_{x \in \mathcal{X} } p(x) \log\left( \frac{p(x) }{q(x)} \right)  \\
		& = - H(X) + \log(k)
\end{align*}

\item So we deduce from a) and b) that :
\[ D(p||q) \leq \log(k) \]
\end{enumerate}
\item
\begin{enumerate}
\item
\begin{align*}
I(X_1,X_2)  & = \sum_{(x_1,x_2)\in \mathcal{X}_1 \times \mathcal{X}_2} p(x_1,x_2) \log\left( \frac{p(x_1,x_2)}{p(x_1) p(x_2)}\right) \\
			& = \mathbb{E}_{x_1,x_2} \left[  - \log\left( \frac{p(x_1) p(x_2)}{p(x_1,x_2)}\right) \right]  \\
-\log \text{ is convex} 		\qquad	& \geq   - \log\left(\mathbb{E}_{x_1,x_2} \left[  \frac{p(x_1) p(x_2)}{p(x_1,x_2)} \right]  \right) = -\log (1) = 0
\end{align*}

\item
\begin{align*}
I(X_1,X_2)  & = \sum_{(x_1,x_2)\in \mathcal{X}_1 \times \mathcal{X}_2} p(x_1,x_2) \log\left( \frac{p(x_1,x_2)}{p(x_1) p(x_2)}\right) \\
				& = \sum_{(x_1,x_2)\in \mathcal{X}_1 \times \mathcal{X}_2} p(x_1,x_2) \log\left( p(x_1,x_2) \right) - \sum_{(x_1,x_2)\in \mathcal{X}_1 \times \mathcal{X}_2} p(x_1,x_2) \log\left( p(x_1)  \right)  \\
				& \qquad  - \sum_{(x_1,x_2)\in \mathcal{X}_1 \times \mathcal{X}_2} p(x_1,x_2) \log\left( p(x_2)  \right)   \\
				& = \sum_{(x_1,x_2)\in \mathcal{X}_1 \times \mathcal{X}_2} p(x_1,x_2) \log\left( p(x_1,x_2) \right) - \sum_{x_1\in \mathcal{X}_1 } p(x_1) \log\left( p(x_1)  \right) \\
				& \qquad - \sum_{x_2 \in \mathcal{X}_2} p(x_2) \log\left( p(x_2)  \right)   \\
	I(X_1,X_2)			& = H(X_1) + H(X_2) - H(X_1,X_2)
\end{align*}
\item Given $X_1,X_2$, we have that
\[
H(X_1,X_2) \leq H(X_1) + H(X_2)
\]
With equality when $p_1$ and $p_2$ are independent random variables. \\
So the joint distribution of the maximal entropy is $p_{1,2} = p_1 p_2 $.
\end{enumerate}
\end{enumerate}

\section{ Conditional independence and factorizations }

\begin{enumerate}
\item The direct implication is straight forward from the conditional independence axiom : \\

If $X \amalg Y | Z$, then $p(x |y,z) = p(x|z)$ for all pairs $(y,z)$ such that $p(y,z) >0 $. \\

For the converse, and consider a pair $(y,z)$ such that $p(y,z) >0$ and assume $p(x|y,z) = p(x|z)$, we have :
\begin{align*}
p(x,y|z) & = p(x|y,z) p(y|z)   = p(x|z) p(y|z)
\end{align*}
And this implies that $X \amalg Y | Z$.
\item Let's $p \in \mathcal{L}(G)$, we have :
\[
p(x,y,z,t) = p(x) p(y) p(z|x,y) p(t|z)
\]

It is not true that $X \amalg Y | T$. Following a Bayes Ball algorithm, when we shade the node T, a ball can pass from X to through Z then bounce back at Z and finally pass through Y.

\item Statement "$ X \amalg Y $ and $ X \amalg Y | Z$ then $ ( X \amalg Z$ or $Y\amalg Z)$"
\begin{enumerate}

\item $Z$ is a binary variable. Let's note $p(x|z=0) = p_0(x) $ , $p(x|z=1) = p_1(x) $ , $p(z=1) = q$ and $p(z=0) = 1-q$ with $ q \in [0,1]$.
We have :
\begin{align}
p(x,y) & = \sum_{z}  p(x,y,z) = \sum_z p(x,y|z) p(z)  = \sum_z p(x|z) p(y|z) p(z)
\end{align}

On the other side,
\begin{align}
p(x,y) &= p(x)p(y) = \sum_{z} p(x|z)p(z) \sum_{z'} p(y|z') p(z')
\end{align}

By equaling the equation (1) and (2), we obtain:
\[
\left( (1-q) p_0(x) + q p_1(x) \right) . \left( (1-q) p_0(y) + q p_1(y) \right) =  (1-q) p_0(x) p_0(y) + q p_1(x) p_1(y)
\]

\[
\Leftrightarrow  \qquad p_0(x) p_0(y)  + p_0(x) p_1(y) + p_1(x) p_0(y)  - p_1(x) p_1(y) = 0
\]

\[
\Leftrightarrow  \qquad  (p_0(x) - p_1(x) ) . ( p_0(y)  - p_1(y))  = 0
\]
\[
\Leftrightarrow  \qquad  p_0(x) =  p_1(x) \quad \text{or} \quad  p_0(y)  = p_1(y)
\]

\[
\Leftrightarrow  \qquad  X \amalg Z \quad \text{or} \quad Y \amalg Z
\]
\item In general, the statement is not true.
\end{enumerate}

\end{enumerate}

\section{Distributions factorizing in a graph}
\begin{enumerate}
\item We know that in a DAG, G = (V,E) , we have that:
\[
p(x) = \prod_{k \in V} p(x_k |x_{\pi_k})
\]

We know that $\pi_j = \pi_i \cup \{ j\}$, then we compute:
\begin{align*}
p(x)  & = p(x_i|x_{\pi_i}) \quad p(x_j|x_{\pi_j})  & \prod_{k \in V \setminus \{i,j\}} p(x_k |x_{\pi_k}) \\
		& = p(x_i|x_{\pi_i}) \quad  p(x_j|x_i, x_{\pi_i}) &  \prod_{k \in V \setminus \{i,j\}} p(x_k |x_{\pi_k}) \\
		& =  p(x_j , x_i | x_{\pi_i}) &  \prod_{k \in V \setminus \{i,j\}} p(x_k |x_{\pi_k}) \\
		& =   p(x_j|x_{\pi_i}) \quad p(x_i|x_j, x_{\pi_i}) &  \prod_{k \in V \setminus \{i,j\}} p(x_k |x_{\pi_k}) \\
\end{align*}
Which prove that $\mathcal{L}(G) = \mathcal{L}(G')$.

\item G is a directed tree and does not contain any v-structure, hence \\$\exists ! y \in V, \text{s.t.} |\pi_y| = 0$ and $\forall x \in V \setminus \{ y\}, |\pi_x|  = 1$. \\$y$ is the root of the directed tree.\\
G' is an undirected tree hence it has only cliques of size 2, because the chromatic number of an undirected tree is 2.

We know that
\[
p(x_V) = p(y) \prod_{x \in v_y} p(x|y)  \prod_{(x_1,x_2) \in E , \neq y } p(x_2|x_1)
\]

So for all cliques, we define the potential funtion:
\begin{itemize}
\item If $C = \{y,x\} $ then $\psi_C(x_C) = p(y) p(x|y)$
\item if $C = \{x_1,x_2\}$ such that $x_1 \neq y$ and $x_2 \neq y$ and $(x_1,x_2) \in E $, we define $\psi_C(x_C) =p(x_2 | x_1) $.
\end{itemize}
Then we have proven that $\mathcal{L}(G) = \mathcal{L}(G')$.

\end{enumerate}

\newpage
\section{Implementation - Gaussian mixtures}

\begin{enumerate}[label=(\alph*)]
\item kMeans is sensible to the randomized centroids initialization, see figure\ref{Fig=kMeans}.

\begin{figure*}[h!]
    \centering
    \begin{subfigure}[t]{0.5\textwidth}
        \centering
       \includegraphics[width=\linewidth]{classification_data_HWK2/goodKMeans.eps}
        \caption{Good kMeans}
    \end{subfigure}%
    ~
    \begin{subfigure}[t]{0.5\textwidth}
        \centering
        \includegraphics[width=\linewidth]{classification_data_HWK2/badKMeans.eps}
        \caption{Bad kMeans}
    \end{subfigure}
    \caption{kMeans' sensibility on the centroids initialization}
    \label{Fig=kMeans}
\end{figure*}

\item The parameters of the EM algorithm are:
\[
 \theta   = ( \pi_k , ( \mu_k , \Sigma_k ) )_{k=1,..,K}
\]
With $ p (z) = \prod_{k=1}^K \pi_k^{z_k} \qquad$ and $\qquad p(x|z;( \mu_k , \Sigma_k )_k) = \sum_k \pi_k \mathcal{N}(x ;  \mu_k , \Sigma_k )  $ \\
\newline
We define
\[
q(z) = p(z | x, \theta) \quad \, \quad \tilde{l}(\theta) = \mathbb{E}_q [\log p(x,z,\theta) ]
\]
The goal is to maximize $  \mathbb{E}_q [\log p(x,z,\theta) ]  $. \\
First we initialize $ \mathbf{ \theta} = \mathbf{ \theta}_0$, and to do so, we use the k-Means result to estimate the parameters of the gaussians.
At iteration $t$, we have :
\[
q_{ik}^{(t)} = \mathbb{P}( z^{(i)}_k =1 ) = \mathbb{E}_{q_i^{(t)}} \left[z_k^{(i)} \right]
\]

\begin{enumerate}[label=\textbf{\arabic*-} ]
\item \textbf{E}xpectation step :
\[
q_{i,k}^{(t)} \leftarrow \dfrac{\pi_k^{t-1} \mathcal{N}(x^{(i)},\mu_k^{(t-1)},\Sigma_k^{(t)} )  }{ \sum_{j=1}^K \pi_j^{t-1} \mathcal{N}(x^{(i)},\mu_j^{(t-1)},\Sigma_j^{(t)} ) }
\]
\item \textbf{M}aximization step :
\[
\mu_k^{(t)} = \dfrac{\sum_i x^{(i)} q_{i,k}^{(t)} }{\sum_i q_{i,k}^{(t)} } \qquad , \qquad \Sigma_k^{(t)} = \dfrac{\sum_i (x^{(i)} - \mu_k )  (x^{(i)} - \mu_k )^T q_{i,k}^{(t)}  }{\sum_i q_{i,k}^{(t)}}
\]
and
\[
\pi_k^{(t)} = \dfrac{\sum_{i} x^{(i)} q_{i,k}^{(t)}}{\sum_{i,k'}  x^{(i)} q_{i,k'}^{(t)}}
\]
\end{enumerate}

See figure \ref{Fig:CirGM} for the implementation of this methods in a particular case\footnote{I used to plot the Gaussion contour the function PLOT\_GAUSSIAN\_ELLIPSOID given by Gautam Vallabah in MathWorks.com website. }.

\begin{figure}[!h]
\includegraphics[width=\linewidth]{classification_data_HWK2/circleGM.eps}
\caption{Gaussian Mixture for gaussian with covariance matrices proportional to the identity.}
\label{Fig:CirGM}
\end{figure}

\item For the general case, see figure \ref{Fig:GenGM}.

\begin{figure}[!h]
\includegraphics[width=\linewidth]{classification_data_HWK2/goodGenEM.eps}
\caption{Gaussian Mixture for the general case.}
\label{Fig:GenGM}
\end{figure}

\item We see clearly from figures \ref{Fig:CirGM} and \ref{Fig:GenGM}, that the gaussian mixture scales better for the general case than for the limited case of identity proportional covariance matrices. This also can be seen very well in the log-likelihood estimations.

\begin{table}[!h]
\centering
\caption{Log-Likelihood Estimation.}
\label{my-label}
\begin{tabular}{lccll}
\multicolumn{1}{l|}{\begin{tabular}[c]{@{}l@{}} \end{tabular} } & \multicolumn{1}{c|}{Particular Case} & General Case         &  &  \\ \cline{1-3}
\multicolumn{1}{l|}{Training Data}                                                        & \multicolumn{1}{c|}{$3.7190$}                &           $15.6802$          &  &  \\ \cline{1-3}
\multicolumn{1}{l|}{Test Data}                                                            & \multicolumn{1}{c|}{$7.6726$}                &           $16.3715$           &  &  \\
                                                                                & \multicolumn{1}{l}{}                 & \multicolumn{1}{l}{} &  &
\end{tabular}
\end{table}

\end{enumerate}

\end{document}


\chapterquote{%
To see what is general in what is particular and\\
what is permanent in what is transitory is\\[.4ex]
the aim of scientific thought.}
{-- Alfred North Whitehead, \textit{An Introduction to Mathematics} (1911)}

Recent advances on wireless communications systems are pointing towards heterogeneous networks solutions with dense deployment of nodes \cite{hossain2014evolution}.
%
These wireless networks are expected to provide reliable connectivity to high-mobility devices, to users with millisecond delay constraints, and applications with gigabits per second of data rate requirements \cite{peng2015recent}.
%
This unceasing demand for higher throughput and lower latency leads to an ever-increasing traffic of packets.
%
Thus, the performance of wireless networks tends to be, each day more, limited by interference. % which depends on the spatial distribution of the interferers, the arrival process of packets, the queueing dynamics.

In this thesis, the studied telecommunication problem is the analysis of spatio-temporal traffic on wireless networks limited by interference.
%
In this first chapter, we provide a simple and introductory system model along with the approach for the analysis and the studied scenarios.
%
Note that a rigorous, general, and detailed model is provided in Chapter~\ref{cap:P2_00}.

\newpage
%-%-%-%-%-%-%-%-%-%-%-%-%-%-%-%-%-%
\section{The Problem}

In a wireless network, we have a set of transmitters and a set of receivers located across the space.
%
As time passes, for each transmitter is given packets that must be transmitted to a determined receiver as soon as possible.
%
Each packet is stored in a buffer and discarded upon successful reception by the corresponding receiver.

When a transmitter sends a packet, it interferes with other transmissions, and vice-versa. This aggregate interference is determined according to the transmit power of each transmitter, the transmission distance, and some random effects (e.g. fading).
%
Then, we determine whether a transmission was successful depending on the signal and the aggregate interference received by the receiver.

\textbf{What are the \textit{givens} of the problem?}%
\setlist{nolistsep}%
\begin{enumerate}[itemsep=1pt]
    \item The spatial position distribution of the transmitters and receivers.

    \item The arrival process of packets for each transmitter.

    \item The propagation model that affects the transmitted signals, e.g. path loss, fading.

    \item The network operation model or queueing system model, that describes the rules of packet buffering and transmitting, e.g. Aloha, CSMA.

    \item The transmission success probability model that decides whether a packet is successfully received based on the received signal and the received interference.
\end{enumerate}

\textbf{What do we want to \textit{know}?}%
\setlist{nolistsep}%
\begin{enumerate}[itemsep=1pt]
    \item Reliability metrics, i.e., metrics related to the transmission and packet success probabilities.

    \item Throughput metrics, i.e., metrics related to the number of successfully delivered packets, such as mean throughput, stability.

    \item Delay metrics, i.e., metrics related to the amount of time to deliver packets, such as mean delay, queue load, age-of-information.
\end{enumerate}

Regarding the \textit{givens}, we can relax them in some aspects.
%
For example, instead of working with a specific path loss model, we let the path loss model be an arbitrary function with some constraints.
%
When the problem is analytically tractable, usually there is little cost to let some well-chosen parameters (or even functions) be arbitrary.
%
Then, once we let a set of \textit{givens} arbitrary, we attain a better comprehension of the relationship between these \textit{givens} and the \textit{knowns} we are interested in.
%
We also obtain a generalization of the problem.

% In this sense, our approach is to study the tractable models used by the scientific community and let some of the ``givens'' be arbitrary.
% %
% More specifically, we choose the ``givens'' (e.g. arrival rate, access probability) for which we are interested in better understanding their relationship with the ``knowns'' (e.g. throughput, delay).

More specifically, the standard \textit{given} scenario of this thesis assumes a single class of users sharing the same frequency bandwidth and the following \textit{givens}:
\begin{enumerate}
    \item High-mobility Poisson network, i.e., the nodes' positions follow a Poisson point process.
    \item Packets arrive according to a Poisson (or Bernoulli) process.
    \item The signals are subjected to Rayleigh fading and power law path loss model.
    \item Packets are queued in infinite buffers, and transmitters access the channel according to a Poisson (or Bernoulli) process.
    \item Transmission success is decided by the capture model.
\end{enumerate}

% a high-mobility Poisson network under random-access, limited by interference, subject to Rayleigh fading and power law path loss model, packets are queued in infinite buffers, transmission success is decided by the capture model, and there is a single class of users sharing the same frequency bandwidth.
%
This standard model from the literature is easily (analytically) tractable, as it is shown in Section~\ref{sec:P1_03_02}.
%
For each scenario studied in this thesis, we list in Table~\ref{tab:givens} the \textit{givens} that we modified, flexibilized and generalized from the standard model, and the \textit{knowns} for which we established a connection.

\begin{table}[htb]
    \centering
    \begin{tabular}{c|l|l}
    \hline
    Location & Change on the \textit{givens}  & The \textit{knowns} \\
        \hline\hline
    Ch.~\ref{cap:P2_01} & transmission success models, no buffer
        & throughput, reliability \begin{tabular}[c]{@{}l@{}}~\\ ~ \end{tabular}\\\hline
    Ch.~\ref{cap:P2_02} & \begin{tabular}[c]{@{}l@{}}maximum number of retransmissions,\\ general transmission success model \end{tabular}
        & \begin{tabular}[c]{@{}l@{}}throughput, reliability,\\ delay, age-of-information\end{tabular}\\\hline
    Sec.~\ref{sec:high-mobility} & $N$ classes of users
        & \begin{tabular}[c]{@{}l@{}}stability, throughput, reliability,\\ delay, queue load\end{tabular} \\\hline
    Sec.~\ref{sec:bandwidth} & $2$ classes of users, bandwidth partition
        & \begin{tabular}[c]{@{}l@{}}stability, throughput, reliability,\\ delay, queue load\end{tabular}\\\hline
    Ch.~\ref{cap:P2_04} & \begin{tabular}[c]{@{}l@{}}static elements, general path loss model,\\ general distribution of link distances\end{tabular}
        & \begin{tabular}[c]{@{}l@{}}stability, queue load,\\
            stationary distribution\end{tabular}\\
    \hline
    \end{tabular}
    \caption{Variations on the \textit{givens} of the base model.}
    \label{tab:givens}
\end{table}

A thorough review of the literature models used in this thesis and their applications can be found in the tutorials \cite{haenggi2021stochastic, cardieri2010modeling}.

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\section{Outline of the Thesis}

We assume the reader is familiar with probability theory, calculus, and optimization.
%
However, we do not assume familiarity with measure theory nor stochastic geometry. That is why we provide a succinct review on these subjects.
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This knowledge is necessary for stating and understanding definitions and theorems from this thesis and from the well-celebrated books of stochastic geometry and wireless networks \cite{baccelli2010stochastic} and \cite{haenggi2012stochastic}.
%
Nevertheless, the reader that is not interested in mathematical details and proofs can skip some parts as indicated throughout the manuscript in notes highlighted with light-blue boxes.
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Furthermore, for ease of reading, on the left side of every proof, there is a vertical line. Therefore, the proofs that do not interest the reader can be easily skipped.

In this manuscript, we differentiate \textit{lemmas}, \textit{theorems}, and \textit{propositions} in the following manner.
%
We let the short and mathematical results that are used to prove other statements be the \textit{lemmas}.
%
For the results which are related to the physical problem, yet they have a mathematical basis in the sense that their existence is independent of the physical problem, those we call \textit{theorems}.
%
Finally, the results which are strongly related to the physical problem and alone do not have significance, those we call \textit{propositions}.

Throughout the thesis, we also have the \textit{remarks} and the \textit{notes}, which are highlighted in gray and light-blue boxes, respectively.
%
Their purpose is to comment on aforementioned results.
%
We reserve the more contextualized and formal comments to the \textit{remarks}, and the more digressive and informal comments to the \textit{notes}.

% We decided to include several definitions, because when we say that the system model is over a ``stationary and ergodic Poisson network'' the reader is able to check what exactly that means.

% \begin{itemize}
%     \item Mathematical optimization.
%     %
%     This knowledge is necessary for stating and understanding definitions and theorems from stochastic geometry presented on the well celebrated books \cite{baccelli2010stochastic} and \cite{haenggi2012stochastic}.

%     \item (alternatively) In this thesis we need measure theory (applied to integration and probability) to present and prove some results. Since this is not a common subject in the engineering curriculum, we provide a succinct introduction of this topic in Chapter~\ref{cap:P1_02} focusing on the results we need.

%     \item The thesis is readable without measure theory, except for Chapter~\ref{cap:P1_03} and some proofs.

%     \item We tried to be mathematically rigorous because we have general results and it is not possible to simulate every representative choice of model to guarantee them;

%     \item General results in the sense that they are valid for arbitrary functions that represent, e.g, the path loss, the success transmission model, the link distance distribution.

%     \item We are following the formalism adopted in the well celebrated books in the area of stochastic geometry applied to wireless networks \cite{baccelli2010stochastic}, \cite{haenggi2012stochastic}.
% \end{itemize}

This thesis is organized in nine chapters and two appendices.
%
This first chapter presents, in an intuitive and general manner, the main problem and the approach adopted to study it.
%
The other chapters are summarized below.

\begin{description}
    \item[Chapter~2:] We present the notations and definitions used throughout the manuscript. We also provide a succinct review of measure theory applied to integration and probability. They are necessary to understand stochastic geometry and several proofs of this thesis.

    \item[Chapter~3:] In this chapter, we present the theory of point processes and random measures. Then, we present some known and extensively used results of stochastic geometry applied to wireless networks.

    \item[Chapter~4:] This is the first chapter with contributions.
    We present a general system model for a wireless network and some definitions, which are used until the end of the thesis.

    \item[Chapter~5:] We consider a Poisson process to model packet transmissions and characterize the distribution of the interference in this model.

    \item[Chapter~6:] This chapter includes retransmission and analyzes the impact of this strategy in the throughput and delay.

    \item[Chapter~7:] Here we include the spatial positions of the nodes in the analysis. We also model the buffers of the transmitters as queues and study their stability.

    \item[Chapter~8:] In this chapter we introduce static elements. More precisely, the link distances between transmitters and receivers are fixed across time. This consists of a challenging class of problems because static elements hinder tractability.

    \item[Chapter~9:] We conclude the thesis listing the main contributions and prospects for future research.

    \item[Appendix~A:] Some definitions and results of real analysis used in the thesis.

    \item[Appendix~B:] A brief review of queueing theory (delay, stability).
\end{description}

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\section{Author's Publications}

This thesis is based on three published international journal papers \cite{dester2018,dester2021unique, dester2021retrans}, and one submitted paper \cite{dester2021part}.
%
Besides, the author has three published international journal papers \cite{dester2017stationary, francisco2020, bergamo2021} and two conference papers \cite{dester2018mtc, helder2019lora} that are not covered by this thesis.