TITLE: Connected components of homogeneous space attached to GL_2 over number fields
QUESTION [1 upvotes]: Basic setup:
Let $F$ be a number field, $G = GL_{2,F}$ and $G' = SL_{2,F}$. I write $\mathbb{A}$ for the adeles over $F$ and $\mathbb{A}_f$ for the finite adeles. Similarly for ideles we write $\mathbb{I}$ and $\mathbb{I}_f$. For simplicity, let $A = \prod_{\mathfrak{p}} \mathcal{O}_{F_{\mathfrak{p}}}$, where $F_{\mathfrak{p}}$ is the $\mathfrak{p}$-adic completion of $F$ at $\mathfrak{p}$. Let $K_f = GL_2(A)$, and $K_{\infty}$ be the maximal compact connected subgroup mod center, i.e. $K_{\infty} = \prod_{\nu \mid \infty} K_{\nu}$, with $K_{\nu}$ being either $\mathbb{R}^* SO_2(\mathbb{R})$ or $\mathbb{C}^* U_2$. Similarly let $U_{\infty}$ be the groups without the centers, i.e. the product of $SO_2(\mathbb{R})$'s and $U_2$'s for real and complex places respectively, and $U_f = SL_2(A)$.
We have
$$ X^1 = G'(F) \backslash G'(\mathbb{A}) / U_f U_{\infty} \hookrightarrow G(F) \backslash G(\mathbb{A}) / K_f K_{\infty} = X $$ induced by the inclusion of $G'$ to $G$.
Further we have the determinant map
$$ X = G(F) \backslash G(\mathbb{A}) / K_f K_{\infty} \to \mathbb{I} / F^* A^* F_{\infty}^* = \mathbb{I}_f / F^* A^*. $$ This latter group is (isomorphic to) the ideal class group of $F$.
It is often mentioned that the connected components of $X$ are in one to one correspondence with the elements of $\mathbb{I} / F^* A^*$, i.e. with the class group. It is said that this follows from strong approximation for $SL_2$.
The question is now the following: How?
My idea is that $X^1$ is the fibre over $1 \in \mathbb{I}_f / F^* A^*$ and connected. As the determinant is surjective and the fibres are isomorphic, the connected components of $X$ and $\mathbb{I}_f/F^*A^*$ agree. The latter is a finite discrete group, so the statement.
I believe, I am using the strong approximation propertry of $G'$ when I say, that $X^1$ is connected, but I don't really see this.
Note: For me, strong approximation for an algebraic group $H$ over $F$ is that $H(F) \subseteq H(\mathbb{A})$ is dense.
Final remark:
As it might be unclear, I want to stress that I would like to get an opinion on my idea of the proof and hints or proves of the missing details.
Edit:
The question can be generalized for arbitrary compact open subgroups of $G(\mathbb{A})$ instead of $K_f$. I believe, the idea of the proof would be essentially the same. The class group then becomes a class group with respect to a conductor.

REPLY [2 votes]: Most of this should be in https://www.jmilne.org/math/xnotes/svi.pdf, but let me at least answer one of your questions here:
First of all note that $G'(\mathbb{R})$ is connected, so that also $G'(\mathbb{R})/K_{\infty}$ is connected, so it suffices to determine the connected components of $G'(F)\backslash G'(\mathbb{A}_f)/K_f$. Strong approximation for $G'(F)$ tells us that $G'(F)$ is dense in $G(\mathbb{A}_f)$, which implies that $G'(\mathbb{Q}) K_f = G(\mathbb{A}_f)$. Indeed, $G(\mathbb{A}_f)/K_f$ has the discrete topology and the image of $G'(F)$ in $G(\mathbb{A}_f)/K_f$ is dense, so it must be everything.


TITLE: Continuity and differentiability for $\sin(\sqrt x)$ & $\sinh(\sqrt {-x})$?
QUESTION [0 upvotes]: Let $f: \Bbb R \to \Bbb R$ with
$$f(x)= \begin{cases}  {\sin(\sqrt{x})\over\sqrt{x}},&   \text{for } x>0\\
                 1,&                           \text{for } x=0\\
                 {\sinh(\sqrt{-x})\over\sqrt{-x}},& \text{for } x<0.\end{cases}$$
(i) Is $f$ continuous on $\Bbb R$ ?
(ii) Is $f$ differentiable on $\Bbb R$ ?
Well, I don't know whether this works or not:
(i)
For $x≠0$ $f$ is a composition and an operation of continuous functions.
For $x=0$ both equations $\sin$ and $\sinh$ have the same limit $1$.
So $f$ is continuous on $\Bbb R$.
(ii)Here I'm not sure:
For $x≠0$ f is a composition and an operation of direffentiable functions
And for $x=0$ ?
Thanks in advance and sorry for Latex,I haven't worked much with it.

REPLY [1 votes]: From the series definition of both functions, $\sin$ and $\sinh$, you get that
$$
f(x)=\sum_{k=0}^\infty (-1)^k\frac{x^k}{(2k+1)!}
$$
which, analogous to the exponential series, has infinite radius of convergence.
One might also note that
$$
\sinh(iz)=\frac{e^{iz}-e^{-iz}}2=i\sin(z)
$$
so that one might also recognize from this fact that the parts of $f$ really are the same function.


TITLE: Why does the minimal polynomial of α divide all polynomials for which α is a root?
QUESTION [3 upvotes]: Suppose α ∈ E with E being a field extension of F. Let S be the set of all polynomials in F[x] for which α is a root. Prove that the minimal polynomial of α over F divides every polynomial in S.

REPLY [12 votes]: Both these answers are much more complicated than is necessary. Simply note that by the division algorithm, we can always write
$$f(\alpha) = q(\alpha)m(\alpha) + r(\alpha)$$
for some $q \in F[x]$ and $f \in S$, with $m \in S$ being the minimal polynomial. Since $\alpha$ is a root of $f$, we either have $r=0$, as desired, or $r(\alpha) = 0$. But, since the degree of $r$ is smaller than the degree of $m$, this would violate the minimality of $m$, so we must have $r=0$.


TITLE: Confusion in conservation of angular momentum
QUESTION [0 upvotes]: Problem statment: A rod hinged at one end is released from the horizontal position. When it becomes vertical its lower half separates without exerting any reaction at the breaking point. Then find the maximum angle $θ$ in degrees made by the hinged upper half with the vertical.

Approach:
Let $w$ be angular velocity
Use energy conservation to get $w_1=(3g/l)^{1/2}$
Now, as there is no external torque acting when the rod is vertical, angular momentum must be conserved (according to me).
So $w_2 = 8*w_1$    ($Iw=const$)
Use energy conservation again.
We get $cos(\theta) = -3$ which is definitely not right
However, if we assume angular velocity to not change will land to the answer which is $60°$
I just need a reason for the angular velocity to not change, I know how to proceed.
PS: Sorry for bad formatting, I am new here.

REPLY [1 votes]: Let us consider an analogous situation for linear motion. Suppose two trains were coupled and were going with same velocity, with no net force on them. Suddenly they are uncoupled in such a way that no force is excerted between them. Then the velocities of the trains will not change just after the decoupling as the net force on them were zero.
Similarly, consider the upper half of the rod. At vertical position there is no torque due to gravity. Since the halves of the rod don't apply any force on each other during separation, the net torque on the upper half of the rod will be zero. Therefore its angular momentum (hence angular speed) will not change.
In your solution you missed the fact that the lower end of the rod also has some angular momentum and it will retain that after separation. The angular momentum of the upper end will remain constant before and after separation.
Similarly, the linear momentum of each train will be conserved after decoupling (not that the total momentum of the two trains will be transferred to a single train)


TITLE: In a play, 10 actors (5 males and 5 females) are to be selected for 3 male roles and 3 female roles. In how many ways can this distribution happen?
QUESTION [2 upvotes]: To solve that, I'm thinking of the rule of product. For each role, 5 different actors could be selected, so I should consider (5 * 5 * 5) + (5 * 5 * 5) = 250 as the answer?

REPLY [0 votes]: You should consider the number of ways to select the male roles, given by the coefficient $\binom{5}{3}$, which is the same for the female. Now, by the rule of product, the number of ways to complete the task is $\binom{5}{3}\binom{5}{3}=100$.


TITLE: Probability of Finding an Empty Chair
QUESTION [0 upvotes]: A similar question here: Probability of finding empty seats
If I have a table with ten seats, seven being taken, what is the likelihood that when I arrive, there will be an available grouping of two seats?
How would I start this? I considered the ways to distribute two seat pairings, and I got 28 for that, then I considered all ways to distribute the 3 chairs in between seven, and got 168, yielding a pleasing 1/6, but I don't think this is right.

REPLY [0 votes]: There seem to some unspoken assumptions: (i) are the seats arranged in a line? If they are in a circle (so that seat 10 is next to seat 1) this is of course increases the probability. And (ii), do you need a strict two-seat, or is it ok if the third person joins your loveseat? :)
For lined-up chairs where both 2- and 3-seats are ok, the previous answers are correct
$$\mathrm{P\textrm{[2-seat|line]}} = \frac{\textrm{# 3 empty seat combinations with 2 adjacent}}{\textrm{# 3 empty seat combinations}} = \frac{{9\choose 1}\cdot{8\choose 1} }{10\choose 3} = \frac{3}{5},$$
where $9\choose 1$ is the number of combinations to place one 2-seat (9 positions) and $8\choose 1$ is the number of combinations to place the last seat. If however they are seated in a ring, there are not 9, but 10 positions for the 2-seat, and not 8 but 9 for the other seat
$$\mathrm{P\textrm{[2-seat|ring]}} = \frac{{10\choose 1}\cdot{9\choose 1}}{10\choose 3} = \frac{3}{4}$$
Finally if the last seat must be elsewhere it changes the combinations to choose for this seat, in the example of the ring, there are not 9 but 7 places to choose from, so that
$$\mathrm{P\textrm{[strict 2-seat|ring]}} = \frac{{10\choose 1}\cdot{7\choose 1} }{10\choose 3} = \frac{7}{12}$$


TITLE: Problem regarding Row and Column Space, when does two matrices have same Row or Column Space?
QUESTION [0 upvotes]: Below I've summarised my understanding of the topic till now. Please correct me wherever I am wrong.
Row operations preserve the row space of a matrix but do not preserve the column space in general (the only preserve the dimension of column space)
Column operations preserve the column space but not the rowspace in general(only preserve dimension of rowspace)
Two matrices $A$ and $B$ have
(a) The same column space iff it is possible to column reduce $A$ to $B$ OR iff it is possible to row reduce transpose($A$) to transpose($B$) [Since column space of A is same as rowspace of its transpose]
(b) They have the same row space iff it is possible to row reduce $A$ to $B$
Is this correct?

REPLY [1 votes]: Yes, it is correct. If you want to dig a bit further, the reason is that when reduce $A $ to $B$ (by rows, for example) in each step you replace a row by a linear combination of the current ones, in such a way that the subspace spanned by the rows does not change.


text:
\chapter{Multi-risk  Compartment Model}\label{multi}

In this chapter we develop and analyze our first and simplest model  that is a multi-risk compartmental   model that can be used to help understand the spread of Ct and to quantify the relative effectiveness of different mitigation efforts.
Our model is closely related to the deterministic population-based model developed by Clarke et al.   \cite{clarke2012exploring} to explore the short-term impacts of increasing screening and  partner notification. Also, it  is related to the STI models for the spread of the HIV/AIDS virus in a heterosexual network \cite{hyman1988using, hyman1989effect}.

The number of partners a person has (his/her  risk), and the number of partners that their partners have (his/her partner's risk) both affect the spread of Ct.  That is, different  assumptions about the distribution of risk behavior of the population will result in different disease forecasts.
We use  the selective sexual mixing STI model developed by Hyman et al. \cite{hyman1997disease} to capture the
heterogenous mixing among people with different numbers of partners.

Although age, ethnicity, economic statues,  and the spatial location of the individuals all influence the assortative mixing of sexual acts,  the risk of contracting Ct is primarily a function of the number of partners a person has, the number of acts per partner, the probability that a partner is infected, and the use of prophylactics (e.g. condoms).

In our ordinary differential equation model (ODE), we consider defining the risk categories based on  the number of partners a person has.
The relative importance of the number of partners and the number of acts per partner on the spread of an STI depends on the disease infectiousness.  Ct is a very infectious disease and the probability of transmission per sexual act  from an infected person to uninfected one is high;  one act with an infected person is enough to catch the infection. Therefore, the number of people a person infects  depends mostly  upon the number of partners he/she has.

In this chapter we first formulate the mathematical model, then  we derive the basic reproduction number, $\mathcal{R}_0$, for two main risk groups (high-risk and low-risk) for men and women.
We then use sensitivity analysis of $\mathcal{R}_0$  and the equilibrium points with respect to the model parameters
to study how the heterogeneous mixing affects spread of Ct \cite{azizi2016multi}.

\section{Ct Transmission Model Overview}\label{Model}
In modeling the spread of Ct,  the population is divided into the susceptible sexually active population (S), the exposed infected, but not infectious population (E) and the infectious population (I). Once a person has recovered from Ct infection, they are again susceptible to infection.  Therefore, the models all have a S$\rightarrow$E$\rightarrow$I$\rightarrow$S (SEIS) structure, or a SIS structure if the exposed state is combined with the infectious state.

Because the exposed (infected, but not infectious) time period is short compared to time in the infectious stage, we
do not include a exposed stage in our model.
We divide men and women into $n$ risk groups based on the number of partners an individual has in a year.
This SIS model can be written as the system of $2n$ ordinary differential equations:
\begin{equation}\label{multi_model}
\begin{split}
\frac{d S_k}{d t}&=\mu (N_k-S_k)-\lambda_kS_k+\gamma_k I_k,\\
\frac{d I_k}{d t}&=\lambda_kS_k-\gamma_k I_k-\mu I_k,\\
\end{split}
\end{equation}
where $k=1, \cdots, n$ denotes men  with risk from $1$ to $n$, and $k=n+1, \cdots, 2n$ denotes women  with risk from $1$ to $n$.
The migration rate, $\mu$, determines the rate at which people enter and leave the population,  $N_k = S_k+I_k$ is the total population  of group $k$,
$\lambda_k$ is  the rate at which a susceptible person in risk group $k$ is being infected, and $\gamma_k$ is the rate that a person recovers  either through screening, or natural recovery.

We model a population of $15-25$ year-old individuals and assume that
the primary mechanism for migration is by aging into, and out of, the population, where migration rate $\mu = 0.003 = [(25-15)~ \mbox{years}]^{-1}$, with the assumption  that death is negligible compared to the rate that
people enter and leave the modeled population. We assume that, in the absence of infection,
equilibrium population $N^o_k$ for each risk group of men and women is given, and that
everyone aging into the model population enters as a susceptible person.

The rate  the infected population is treated, $\gamma_k$, depends upon the sex of the person and  their risk level.  The treatment can be initiated when infection is identified through screening, partner notification, or a medical check-up.
Most infected people  are asymptomatic and, when a significant fraction of a population is infected, then screening has been found to be a cost-effective approach to identify, and treat,  infected people.

Natural recovery rate, $\gamma^n_k$, is determined by assuming an exponential distribution for the average time to recovery $1/\gamma^n_k$, and screening recovery rate, $\gamma^s_k$, is determined by assuming a log normal distribution for the average time to recovery through screening $1/\gamma^s_k$.
We also define the probability that an infected individual is screened and treated each day, $\sigma_d^k$, in terms of the fraction of the population that will be screened at least once within a year as $\sigma_y^k$.  That is,
\begin{equation}
\sigma_d^k=1 - (1-\sigma_y^k)^{1/365}.
\end{equation}

\subsection{Transmission rate}

We will derive the disease transmission rate for the heterosexual case where
a susceptible person in group $k$ can be infected by someone of the opposite sex in any of the infected groups $j$.

The force of infection, $\lambda_k$,  is the rate that people in risk group $k$ are infected through sexual acts.
We define $\lambda_k$ as the sum of the rate of disease transmission from
each infected group, $I_{j}$,  to the susceptible group, $S_k$:
\begin{equation} \label{infrate}
\lambda_k = \sum_{j=1}^{n}  \lambda_{kj}.
\end{equation}
The rate of disease transmission from the
infected people $I_{j}$ in group $j$ to the susceptible individuals $S_{k}$ in group $k$, $\lambda_{kj}$,
 is defined as the product of three factors:
 \begin{align*}
\lambda_{kj} &= \left( \begin{array}{c}\mbox{Number of partners} \\
                                       \mbox{a susceptible in group $k$} \\
                                       \mbox{has with someone in} \\
                                        \mbox{group $j$ per unit time}
                       \end{array}
                \right)
                \left( \begin{array}{c}\mbox{ Probability of}\\
                                       \mbox{disease transmission} \\
                                       \mbox{ per partner}
                       \end{array}
                \right)
                \left( \begin{array}{c}\mbox{Probability that} \\
                                       \mbox{partner in  group $j$ } \\
                                       \mbox{ is infectious}
                       \end{array}
                \right) \\
                &= \hspace{2cm}p_{kj}\hspace{5cm}\beta_{k}\hspace{4cm}P_I(t,j)~,
\end{align*}

where
\begin{itemize}
\item $p_{kj}$ is the number of sexual partners per unit time that each individual in group $k$ has with
someone in group $j$, and
\item $\beta_{k}$ is the probability of disease transmission per partner for a susceptible person  in group $k$, and
\item $P_I(t,j)$ is the probability of that the person in group $j$ is infected.
\end{itemize}

For this last factor, we assume that the partners in group $j$ are all equally likely to be infected.  That is,
the probability  the person in group $j$ is infected  is the same as the fraction of the people in
group $j$ that are infected, $\frac{I_{j}}{N_j}$.

\subsection{Partnership formation}
The extent that Ct  spreads through a population is sensitive to
the  heterogenous mixing (partnership selection) among the different risk groups.
The model approximates the  mixing  through  mixing probabilities $p_{kj}$
that define how many partners a typical person in group $k$ has with someone in group $j$.
These mixing functions must  dynamically change to
account for variations in the size of the groups
\cite{hyman1997behavior, del2013mathematical, busenberg1991general}.

The force of infection, $\lambda_k$,  depends on how many partners
people in group $k$ have, the number
of acts they have per partner, and the probability that their partners are infected.
The mixing is biased since people who only have a few sexual partners (low-risk)
typically have partners who are also at low risk.

We define the model parameters so that someone in group $k$ has, on average,
 $p_{kj}$ partners who are in group $j$ per day.
 Therefore, the total number of partnerships per day  between people in group
 $k$ and group $j$ is  $p_{kj}N_k = p_{jk}N_j$.
Since each partnership
may have more than one act, we define $a_{k}$ as the average number of sexual acts per partner for people in group $k$.

To determine $p_{kj}$, we use a heterogeneous mixing algorithm developed
in   \cite{hyman1997behavior}.
This approach starts by defining
$\bar p_k$ as the \emph{desired} number of partnerships someone in group
$k$ wishes to have per unit time.   Because there may not be sufficient
available partners for everyone to have their desired number of partners,
the actual number of partners could be different.

We define the \emph{proportional} partnership (mixing) as the desired fraction $\rho_{kj}$
of these partnerships that a person in group $k$ wants to have
with someone in group $j$.  That is, a person in group $k$ wants to have
 an average of $\rho_{kj} \bar p_k$ partnerships per unit time with someone in group $j$.
Unfortunately, there is no guarantee that the total number of desired partnerships
that people in group $k$ want to have with people in group $j$ will be the same
as the total number of desired partnerships
that people in group $j$ want to have with people in group $k$.  That is, in general
$\rho_{kj} \bar p_k N_k \ne  \rho_{jk} \bar p_j N_j$, and this must be reconciled.

Since not everyone can have their desired number of partners distributed
exactly as they wish, the different heterogenous mixing algorithms represent different
compromises to resolve  these conflicts.
All of the heterogenous mixing algorithms maintain the  detailed balance for
mixing where the total number of partnerships for people in group $k$ with
people in group $j$ is the same as the total number of partnerships that
people in group $j$ have with people in group $k$.
In our model, we use the heterogenous mixing algorithm based on the algorithm
described in  \cite{hyman1997behavior,  del2013mathematical}
to determine $p_{kj}$.

The population in group $k$ desires $\rho_{kj} \bar p_k N_k$ partners from group $j$,
and the population in group $j$ desires $\rho_{jk} \bar p_j N_j$ partners from group $k$.
As a compromise, we  set the total number of partners the people in group
$k$  have with people in group $j$, and vice versa, to be the harmonic mean
\begin{equation}
p_{kj} N_k= p_{jk}N_j= \frac{2(\rho_{kj} \bar p_k N_k) (\rho_{jk} \bar p_j N_j)}{(\rho_{kj} \bar p_kN_k)+ (\rho_{jk} \bar p_j N_j)}.
\end{equation}
Other possibilities include the geometric mean or minimum of
$(\rho_{kj} \bar p_kN_k)$ and  $(\rho_{jk} \bar p_j N_j)$.
All of these averages satisfy the balance condition
to have the property
that if $\rho_{jk} =0$ then $p_{kj} =p_{jk}=0$, where if one group refuses to have a
partnership with another group, then this partnership does not happen.
In our model, we use the  harmonic mean and define
\begin{equation}
p_{kj} = \frac{1 }{ N_k} \frac{2(\rho_{kj} \bar p_kN_j) (\rho_{jk} \bar p_j N_k)}{(\rho_{kj} \bar p_kN_k)+ (\rho_{jk} \bar p_j N_j)}~.\end{equation}
Hence, $p_k= \sum_{j}p_{kj}$ is the actual average number of partners someone in group $k$ has per day.

Note that this approach is only appropriate if the desired number of partners between any two groups is in close agreement,
that is,  $\rho_{kj} \bar p_k N_k \approx \rho_{jk} \bar p_j N_j$.
This is because, the approach assumes that if the partners are not available from the desired group, then the individuals
will not change their preferences to seek partners in other risk groups.  The model can be extended to
handle these situations where the people adjust their desires to be in closer alignment with the availability of partners
through a simple iterative algorithm.  However,
we avoid this complication in our simulations and initialize the populations so the groups desires are close to the
availability of partnerships.

\subsection{Probability of transmission per partner}
The probability of a susceptible person catches infection from their infected partner depends upon the number of sexual acts between the people.
We allow the number of acts per partner for a person in group $k$, $a_k$, to depend upon the number of his/her actual partners  and  his/her total number of acts per unit time,
$A_k=a_k p_k$, where $A_k$ is total number of acts per unit time.

 The probability of transmission per act, $ \beta$, can be used to define  the probability that a susceptible person will not be infected by a single act with an infected person, $1- \beta$.
Therefore, the probability of someone in group $k$ not being infected after $a_{k}$ acts with an
infected person  is  $(1-\beta)^{a_{k}}$.
Hence, the  probability of being infected per partner is   \cite{hyman2003modeling}
\begin{equation} \label{betaij}
\beta_{k} = 1-(1-\beta)^{a_{k}}.
\end{equation}

\begin{figure}[htp] \label{ fig:cir}
\begin{center}
  \includegraphics[width=4in]{Figures/conpart.pdf}\\
  \caption[\textbf{The number of acts per partnership per unit time vs number of partners}]{The number of acts per partnership per unit time (year), $a_k$,  is a decreasing function of $p_{k}$, that
  is, people with more partners have fewer acts per partner per unit time than people with fewer partners. If $A_k$ is total number of acts
  for a person in group $k$, $a_k$ is defined as $a_k=\frac{A_k}{p_k}=\frac{A_k}{\sum_j p_{kj}}.$}
  \end{center}
\end{figure}

\begin{table}[htbp]
\centering
\resizebox{\columnwidth}{!}{
\begin{tabular}{ lp{9cm}lll }
\toprule[1.5pt]
 \normal{\head{Parameter}} & \normal{\head{Description}}
  & \normal{\head{Unit}} & \head{Baseline}\\
  \cmidrule(lr){1-3}\cmidrule(l){3-4}
  N& Total population. & people & $20000$\\
 $N_1$ & High-risk men population. & people &  $2400$\\
   $N_2$ & Low-risk men population. & people &  $4600$\\
 $N_3$ & High-risk women population. & people &  $2500$\\
 $N_4$ & Low-risk women population. & people &  $10,500$\\
$\mu$ & Migration rate. & 1/days& $0.00$ \\
  $\beta$ & Probability of transmission per act. & 1/act & $0.11$\\
        $1/\gamma_i^n~ (i=1,2,3,4)$& Average time to recover without treatment.   & days&  $365$   \\
        $1/\gamma_i^s~ (i=1,2,3,4)$& Average time to recover with treatment.   & days&  $7$   \\
        $A_1$& Total number of acts per time for a high-risk man.  & 1/days&  $0.14$   \\
        $A_2$& Total number of acts per time for a low-risk man.   & 1/days&  $0.07$   \\
        $A_3$& Total number of acts per time for a high-risk woman.   & 1/days&  $0.07$   \\
        $A_4$& Total number of acts per time for a low-risk woman.   & 1/days&  $0.04$   \\
        $\bar p_{1}$ & Desired number of partners per time for a high-risk man.& people/days&  $0.14$ \\
         $\bar p_{2}$ & Desired number of partners per time for a low-risk man.& people/days&  $0.03$   \\
          $\bar p_{3}$ & Desired number of partners per time for a high-risk woman.& people/days&  $0.06$\\
           $\bar p_{4}$ & Desired number of partners per time for a low-risk woman. & people/days&  $0.02$ \\
            $\bar \rho_{13}$ & Desired fraction of high-risk partner for a high-risk man.& --&  $0.75$ \\
            $\bar \rho_{24}$ & Desired fraction of low-risk partner for a low-risk man.& --&  $0.80$ \\
            $\bar \rho_{31}$ & Desired fraction of high-risk partner for a high-risk woman.& --&  $0.75$ \\
            $\bar \rho_{42}$ & Desired fraction of low-risk partner for a low-risk woman.& --&  $0.80$ \\
            $\sigma_y^i~ (i=1,2,3,4)$ & Fraction of people in group $i$ randomly screened per year. & -- & $0.00$ \\
\bottomrule[1.5pt]
\end{tabular}}
\caption[\textbf{Variables and parameters of the  model}]{Table for the variables and parameters in the  model (\ref{multi_model}).
All the parameter values are based on our assumption and are not estimated or taken from literature.}
\label{tab:params_table}
\end{table}

\section{Basic Reproduction Number}

The basic  reproduction number, $\mathcal{R}_0$, is the number of new infections introduced if a newly infected person is introduced into a population at the ($S_k=N_k$, $I_k=0$).
\iffalse
\subsection{Branching process analysis}

The equation for the  infected population can be written as
\begin{equation}
\frac{d I_k}{d t}=\sum_j \alpha_{kj} I_j  -   \tau_k^{-1} I_k ~~.
\label{model2I}
\end{equation}
Here $\tau_k=1/(\gamma_k+\mu)$ is the average time that an infected person stays in the $k$-th infection compartment.
The force from infection, $\alpha_{kj}=\beta_k q_{kj} {S_k}$, is the rate (per day) that a typical infected person in group $j$ infects a susceptible one in group $k$.  Here the factor $q_{kj}$ is defined as $q_{kj}=p_{kj}/{N_j}$ and because we had $p_{kj}N_k=p_{jk}N_j$, therefore, $p_{kj}/{N_j}=p_{jk}/{N_k}$, and that means $q_{kj}=q_{jk}$  is
 the fraction of people in group $j$ someone in group $k$ has as a partner.

We define the group $j$-to-group $k$  reproduction number,  $\mathcal{R}_0^{j \rightarrow k}=\alpha_{jk} \tau_j$,  as the average number of  people that a new infected person in group $j$ will infect someone in group $k$ at  Disease Free Equilibrium, DFE.
Note that the hidden $\frac{S_j}{N_j}$ hidden in the definition of $\mathcal{R}_0^{j \rightarrow k}$ is equal to one and  if group $j$ and group $k$ are the same sex, then $\mathcal{R}_0^{j\rightarrow k}=0$.

Consider the situation where the first infection is in  group $j$, then the number of secondary infections over all possible groups is
$\mathcal{R}_0^{j \rightarrow *}= \sum_{k} \mathcal{R}_0^{j \rightarrow k}$.
Each of these secondary infections in group $k$ then infect  $\mathcal{R}_0^{k \rightarrow *}$ others.
That is, the total number of new infections is  a branching process and grows geometrically.

The reproduction numbers measure the average number of new infected cases over a \emph{single} infection cycle and
is defined as the square-root of the number of new infections over two cycles.  We can define the basic reproduction number for group $j$ as
\begin{equation*}
\mathcal{R}_0^j=
\sqrt{\sum_{k} \big( \mathcal{R}_0^{j \rightarrow k} \mathcal{R}_0^{k \rightarrow *} \big)} =
\sqrt{\sum_{k} \big( \mathcal{R}_0^{j \rightarrow k} \sum_{i}  \mathcal{R}_0^{k \rightarrow i}   \big)}
\end{equation*}
\fi

We will derive the basic reproduction number $\mathcal{R}_0$  using the next generation approach  \cite{van2002reproduction, heffernan2005perspectives}
for situation  with two risk levels for men and women labeled: $1=$ high-risk men, $2=$ low-risk men, $3=$ high-risk women and $4=$ low-risk women.
We have  differential equations
\begin{equation}
\frac{d I_k}{d t}=\sum_j \alpha_{kj} I_j  -   \tau_k^{-1} I_k,
\label{model2I}
\end{equation}
where $\tau_k=1/(\gamma_k+\mu)$ is the average time that an infected person stays in the $k$-th infection compartment.
The force from infection, $\alpha_{kj}=\beta_k q_{kj} {S_k}$, is the rate (per day) that a typical infected person in group $j$ infects a susceptible one in group $k$.  Here the factor $q_{kj}$ is defined as $q_{kj}=p_{kj}/{N_j}$ and because we had $p_{kj}N_k=p_{jk}N_j$, therefore, $p_{kj}/{N_j}=p_{jk}/{N_k}$, and that means $q_{kj}=q_{jk}$  is
 the fraction of people in group $j$ someone in group $k$ has as a partner.
The  Equation (\ref{model2I}) for the infected populations,  $I=(I_1, I_2, I_3, I_4)^T$,  can
be written as a matrix equation for the rate of production of new infections, $F$,
minus the removal rate of individuals from that population class, $V$,
 \begin{equation}\label{nextgen}
\frac{dI}{dt} =  {\bf F} I-  {\bf V} I,
\end{equation}
where  the $kl$-th element of the matrix \textbf{F} is ${\bf F}_{kl}=\alpha_{kl}$, and $V$ is diagonal matrix  $V_{kk}=\tau_k^{-1}$, for $k,l=1,...,4$.
At DFE, $\alpha_{kj}=\beta_k q_{kj} {N_k} $ and the  Jacobian matrices, ${\bf J}_F $ and ${\bf J}_V^{-1}$, of $FI$ and $VI$ are
\begin{equation*}
J_F= \left( \begin{array}{cccc}
0 & 0 & \alpha_{13}& \alpha_{14}\\
0 & 0 & \alpha_{23} & \alpha_{23} \\
\alpha_{31} & \alpha_{32} & 0  & 0\\
\alpha_{41} & \alpha_{42} & 0 & 0
 \end{array} \right), ~~~
J_V^{-1}= \left( \begin{array}{cccc}
\tau_1 & 0 & 0 & 0\\
0 & \tau_2 & 0 & 0\\
0 & 0 & \tau_3 & 0 \\
0 & 0 &  0 &\tau_4
\end{array} \right).
\end{equation*}
We define $\mathcal{R}_0$ as spectral radius of $J_F J_V^{-1}$ or (equivalently) $J_V^{-1}J_F$,
         \begin{equation}
      J_F J_V^{-1}=\left(
         \begin{matrix}
           0 & 0 & \alpha_{13} \tau_3& \alpha_{14} \tau_4  \\
           0 & 0 & \alpha_{23} \tau_3& \alpha_{24}  \tau_4\\
           \alpha_{31} \tau_1 &   \alpha_{32}  \tau_2 & 0 & 0 \\
            \alpha_{41}  \tau_1 &  \alpha_{42} \tau_2 & 0& 0\\
         \end{matrix}
         \right),~~~
        J_V^{-1}J_F=\left(
         \begin{matrix}
           0 & 0 & \alpha_{13} \tau_1& \alpha_{14} \tau_1  \\
           0 & 0 & \alpha_{23} \tau_2& \alpha_{24} \tau_2\\
           \alpha_{31} \tau_3 & \alpha_{32} \tau_3 & 0 & 0 \\
           \alpha_{41} \tau_4 & \alpha_{42} \tau_4 & 0& 0\\
         \end{matrix}
         \right).
         \end{equation}
Note that $jk$-th element of $J_V^{-1}J_F $ is  $\mathcal{R}_0^{j \rightarrow k}=\alpha_{jk} \tau_j$.
Based on a result of  Sylvester's inertia theorem \cite{sylvester1852xix}\footnote{Let $K$ be a hermittian matrix. We define $e^+(K)$ as the number of positive eigenvalues, $e^-(K)$ as the number of negative eigenvalues, and $e^0(K)$ as the number of zero eigenvalues. Inertia of $K$ is a tuple $(e^+(K), e^-(K)), e^0(K)$. If $A$ is an invertible matrix then Sylvester inertia theorem states:
	$inertia(K)=inertia(A^{-1}KA).$ },  if a matrix $K$ can be factored into the product of a diagonal  positive definite matrix $A$ and a symmetric matrix $B$, then eigenvalues of  $K$ are the same as eigenvalues of  $A^{\frac{1}{2}}  B  A^{\frac{1}{2}}$.
To apply this result, we  rewrite $J_V^{-1}J_F=AB$ where $A$ is a diagonal  positive definite matrix and $B$ is symmetric,
        \begin{equation}
        A=\left(
                \begin{matrix}
                  \beta_1N_1 \tau_1  & 0 & 0& 0  \\
                   0 & \beta_{2} N_2 \tau_2  & 0& 0 \\
                 0 & 0 & \beta_{3} N_3 \tau_3 & 0 \\
                  0 & 0 & 0& \beta_{4}  N_4 \tau_4\\
                \end{matrix}
                \right),~~~
                B= \left(
                         \begin{matrix}
                           0 & 0 &q_{13}&q_{14}  \\
                           0 & 0 & q_{23}& q_{24} \\
                           q_{31} & q_{32} & 0 & 0 \\
                           q_{41} & q_{42} & 0& 0\\
                         \end{matrix}
                         \right).
                            \end{equation}
   Therefore, eigenvalues of  $J_V^{-1}J_F$ are the same as eigenvalues of symmetric  block anti-diagonal generation matrix
  \small

   \begin{align}
   & A^{\frac{1}{2}}  B  A^{\frac{1}{2}} = \left( \begin{matrix}
                                  0 & 0 & \sqrt{\alpha_{13} \tau_1  \alpha_{31}  \tau_3 }&
                                  \sqrt{\alpha_{14} \tau_1 \alpha_{41} \tau_4} \\
                                         0 & 0 & \sqrt{\alpha_{23} \tau_2 \alpha_{32} \tau_3 }&
                                         \sqrt{\alpha_{24} \tau_2 \alpha_{42} \tau_4} \\
                                  \sqrt{\alpha_{31} \tau_3 \alpha_{13} \tau_1} &
                                  \sqrt{\alpha_{32} \tau_3 \alpha_{23} \tau_2 } & 0 & 0 \\
                                  \sqrt{\alpha_{41} \tau_4 \alpha_{14} \tau_1 } &
                                  \ \sqrt{\alpha_{42} \tau_4 \alpha_{42} \tau_2} & 0& 0\\
                                \end{matrix} \right)
                                = \left(
  	\begin{matrix}
  	0_{4\times 4} & M   \\
  	M^T & 0_{4\times 4}
  	\end{matrix}
  	\right),
  \label{matrix1}\end{align}
  where $M^T$ is transpose of $M$ and
	\begin{equation}
              M =
               \left(
                 \begin{matrix}
                                   \sqrt{\alpha_{13} \tau_1  \alpha_{31}  \tau_3 }&
                                  \sqrt{\alpha_{14} \tau_1 \alpha_{41} \tau_4} \\
                                  \sqrt{\alpha_{23} \tau_2 \alpha_{32} \tau_3 }&
                                   \sqrt{\alpha_{24} \tau_2 \alpha_{42} \tau_4} \\
                 \end{matrix}
                 \right)~~~ =~~~\left(
                 \begin{matrix}
                                  r_{13}&
                                   r_{14} \\
                                  r_{23}&
                                    r_{24} \\
                 \end{matrix}
                 \right),
              \end{equation}
where  $r_{jk} =  \sqrt{\mathbb{R}_0^{j \rightarrow {k}}\mathbb{R}_0^{{k}\rightarrow j}}$ is the geometric average of group j-to-group $k$ and group $k$-to-group j reproduction numbers.
The basic reproduction number  is spectral radius of $J_V^{-1}J_F$, therefore,
$\mathcal{R}_0=\rho({A^{\frac{1}{2}}  B  A^{\frac{1}{2}}})=\sqrt{\rho(M^TM)}$ where
 $$
 M^TM=
 \left( \begin{matrix}
r_{13}^2 + r_{23}^2& r_{13}r_{14} + r_{23}r_{24}\\
 r_{13}r_{14} + r_{23}r_{24}&     r_{14}^2 + r_{24}^2
\end{matrix}   \right),
 $$
 and
   \begin{equation} \label{R0}
   \mathcal{R}_0=\frac{1}{2}((r_{13}^2+r_{23}^2+r_{14}^2+r_{24}^2)
   +\sqrt{(r_{13}^2+r_{23}^2-r_{14}^2-r_{24}^2)^2+4(r_{13}r_{14}+r_{23}r_{24})^2 }).
   \end{equation}

\section{Sensitivity Analysis}
We  use sensitivity analysis  to  quantify the change in model output quantities of interest (QOI), such as the basic reproduction number $\mathcal{R}_0$ and endemic equilibrium point,  due to variations in  the model input  parameters of interest (POI), such as the average time to recovery after infection \cite{arriola2007being, arriola2009sensitivity, manore2014comparing}.

Consider the situation where the baseline value of the input POI is $p_{b}$ and generates the baseline output QOI $q_b=q(p_b)$.
Sensitivity analysis is used to address what happens if $p_b$ is changed by the fraction $\theta_p$, $p_{new}=p_b(1+\theta_p)$. Our goal is to find resulting fractional change in the output variable $q_{new}=q_b(1+\theta^q_p)$.
That is, the normalized  sensitivity index measures the relative change in the input variable  $p$, with respect to the output variable $q$
and  can be estimated by the Taylor series
\begin{equation}
q_{new}=q(p_b+\theta_p p_b) \approx q_b+\theta_p p_b \frac{\partial q}{\partial p}\biggr\vert_{p=p_b}
=q_b(1+\theta^q_p)~.
\end{equation}
We define the normalized  sensitivity index as
\begin{equation}\label{E:defsenind}
\mathbb{S}_p^q := \frac{p_b}{q_b} \frac{\partial q}{\partial p}\biggr\vert_{p=p_b} = \frac{\theta^q_p}{\theta_p}.
\end{equation}
That is, if the input $p$ is changed by $\theta_p$ percent, then the output $q$ will change by $\theta^q_p=\mathbb{S}_p^q \theta_p$ percent.
The sign of $\mathbb{S}_p^q$  determines the direction of changes, increasing (for positive $\mathbb{S}_p^q$) and decreasing (for negative $\mathbb{S}_p^q$). Note that this local  sensitivity index is valid only in a small neighborhood of the baseline values.

\subsection{Sensitivity indices of $\mathcal{R}_0$}

The ability of Ct to become established in a population and its early growth rate is characterized by $\mathcal{R}_0$, Equation (\ref{R0}).
Sensitivity analysis of $\mathcal{R}_0$ can quantify the relative importance of the  different social
and epidemiological parameters in reducing the ability of the STI to become established in a new population.

Table (\ref{table:sens})  of the sensitivity indices of $\mathcal{R}_0$ shows that it is most sensitive to the
probability of transmission per act $\beta$ with $\mathbb{S}^{\mathcal{R}_0}_{\beta} = 1.95.$
That is, if the probability of infection per act decreases- say by increasing the condom-use - by $15\%$
then $\theta_{ \beta} =-0.15$, then $\mathcal{R}_0$ will decrease by $30\%$ from $4.01$ to $2.84$:
$$
{\mathcal{R}_0}_{new}= \mathcal{R}_0 ( 1 + \theta_{\beta} \mathbb{S}^{\mathcal{R}_0}_{{\beta}} ) =4.01( 1 - 0.15 \times 1.95) = 2.84.
$$
That is, sensitivity analysis can quantify the amount of behavior change that would be needed to keep an epidemic from becoming established in a new population.

A negative sensitivity index indicates that $\mathcal{R}_0$ is a decreasing function of correspondent parameter, while the positive ones show $\mathcal{R}_0$ increases when the parameter increases.
The second most important model parameters for the early growth rate are the recovery rates of the high-risk men and women $\gamma_1$ and $\gamma_3$.
Since, $\mathbb{S}^{\mathcal{R}_0}_{\gamma_1}= \mathbb{S}^{\mathcal{R}_0}_{\gamma_3}=-0.95$, a $10\%$ increase in the screening rate
would result in a $9.5\%$ decrease in $\mathcal{R}_0$, which this supports the need to actively screen both men and women for Ct infection.

The number of acts for high-risk men, $A_1$, is also an important parameter for controlling the early growth of the Ct. Because local-sensitivity analysis is valid in a small neighborhood of the baseline case, sometimes it is useful to plot the change in the QOI over a wide range of possible values.  The sensitivity index is then the slope of the response curve at the baseline  values.
The Figure  (\ref{fig:sensitive}) shows how $\mathcal{R}_0$ changes as  parameters $\beta$ and $A_1$ are varied over a broad range.

\begin{table}[H]
  \centering
\begin{tabular}{@{}c d{4} d{4} d{4} d{4} d{4} d{4} @{}}
    \toprule

    & \multicolumn{6}{c}{\textbf{Sensitivity index of $\mathcal{R}_0$ for all parameters in the model}}
    \\
    \cmidrule(rl){2-5} \cmidrule(rl){5-7}

    & \mlc{Parameter p} & \mlc{Baseline } & \mlc{$\mathbb{S}^{\mathcal{R}_0}_p$}
    & \mlc{Parameter p} & \mlc{Baseline } & \mlc{$\mathbb{S}^{\mathcal{R}_0}_p$}\\
    \midrule
     & {\beta}  & 0.11   & 1.95   & \bar{p}_1   & 0.14    & 0.05 \\
     & A_1  & 0.14  & 0.76   &  \bar{p}_2 & 0.03  & -0.06\\
     & A_2 & 0.07 & 0.15 &  \bar{p}_3 & 0.06& 0.24\\
     & A_3   & 0.07   & 0.71   & \bar{p}_4  &  0.02  & -0.07\\
     & A_4 & 0.04& 0.22   & \gamma_1& 0.003 & -0.95\\
     & \bar{\rho}_{13} & 0.75 & 0.35 &\gamma_2& 0.003  & -0.48\\
      & \bar{\rho}_{31} & 0.75 & 0.58 & \gamma_3 & 0.003  & -0.95\\
       & \bar{\rho}_{24}& 0.8 & 0.04 & \gamma_4 & 0.003  & -0.48\\
        & \bar{\rho}_{42} & 0.8 & 0.12 & \mu & 0.00  & -0.09\\
   \bottomrule

  \end{tabular}
  \caption[\textbf{Sensitivity index of} $\pmb{\mathcal{R}_0}$ \textbf{with respect to parameters of the model}]{The sensitivity index of $\mathcal{R}_0$ with respect to parameters of the model at the baseline parameter values where $\mathcal{R}_0=4.01$. The most sensitive parameter is the  probability of transmission per act, $\beta$, followed by the recovery (screening) rates of the high-risk men and women $\gamma_1$ and $\gamma_3$.
}
\label{table:sens}
\end{table}

\begin{figure}[htp]
\begin{center}
  \includegraphics[width=90mm]{Figures/sens.pdf}\\
 \caption[\textbf{Sensitivity of basic reproduction number}]{ The sensitivity of $\mathcal{R}_0$ with respect to $\beta$ (solid line), and $A_1$ (dashed line).
The sensitivity index is then the slope of the response curve at the baseline  values, indicated by $*$.
The response is approximately linear near the baseline case and, therefore, the local sensitivity analysis is actually valid over a  broad range of parameters.}

  \label{fig:sensitive}
  \end{center}

\end{figure}

\subsection{Sensitivity indices of  endemic equilibriums}

The current Ct epidemic is  established in many cities, therefore,  to evaluate the relative impact of the model
parameters in bringing it under control requires that the sensitivity analysis be preformed
about the current state of the system, the steady-state endemic equilibrium.
We will investigate the impact the mitigation efforts on the relative change in the number of infected people
as a function of the relative change in the model parameters.  This is best done in terms of
the nondimensional variables defined by dividing each variable by the
steady-state zero-infection equilibrium total population for that sex.
That is, $i_m=I_m/N^o_m$, $i_w=I_w/N^o_w$,
$n_1=N_1/N^o_m$,  $n_2=N_2/N^o_m$,  $n_3=N_3/N^o_w$,  and $n_4=N_4/N^o_w$,
where,  $N^o_m=N^o_1+N^o_2$, $N^o_w=N^o_3+N^o_4.$ The Table (\ref{table:sie}) shows that the  sensitivity indices for endemic (steady-state) equilibrium infected populations, $i_j$,
 as a function of the model parameters.  Note that the magnitudes (relative importance) of sensitivity indices have
the same order as they did for $\mathcal{R}_0$, although the magnitudes are different.
\begin{table}[H]
  \centering
\begin{tabular}{@{}c d{4} d{4} d{4} d{4} d{4} d{4} @{}}
    \toprule

    & \multicolumn{6}{c}{\textbf{Sensitivity of equilibriums for all parameters in the model}}
    \\
    \cmidrule(rl){2-5} \cmidrule(rl){5-7}

    & \mlc{Parameter p} & \mlc{Baseline } &i_1& i_2 &i_3 & i_4\\
    \midrule
     & \beta  & 0.11   & 2.61  &4.05&2.7& 4.02\\
     & A_1  & 0.14  & 1.7   &0.58&0.71&0.66\\
     & A_2 & 0.07 & 0.31 & 1.5 &0.33&1\\
     & A_3   & 0.07   & 0.65   & 0.56&1.13&0.53\\
     & A_4 & 0.04 & 0.42   &1.18&0.37&1.61\\
     & \bar{\rho}_{13} & 0.75&0.36 & -0.21&  0.24  & -0.36 \\
      & \bar{\rho}_{31} & 0.75 &0.41 & -0.15 & 0.56 & 0.03 \\
       & \bar{\rho}_{24}& 0.8 & 0.02& -0.13 & 0.08 & -0.10\\
        & \bar{\rho}_{42} & 0.8 &  0.15&-0.21 & 0.08 & -0.33\\
        &  \bar{p}_1   & 0.14    &0.04 & 0.17 &  0.08 &  0.23 \\
     &  \bar{p}_2 & 0.03  & 0.27 & 0.31 & 0.21 & 0.26 \\
     &  \bar{p}_3 & 0.06&  0.27 & 0.31 & 0.21 & 0.26 \\
     &  \bar{p}_4  &  0.02  &  -0.05 & 0.03 & -0.03 & 0.04 \\
     &  \gamma_1& 0.003  & -1.10&    -0.60 &  -0.73&  -0.68\\
     & \gamma_2& 0.003  & -0.35 & -1.66&  -0.37&  -1.11\\
      &  \gamma_3 & 0.003  &-0.66 &  -0.56 &  -1.13& -0.53 \\
       & \gamma_4 & 0.003  &-0.47 & -1.31 &  -0.41 &  -1.78\\
        & \mu & 0.00  &  -0.13& -0.20 & -0.13 &-0.20\\
   \bottomrule

  \end{tabular}
  \caption[\textbf{Sensitivity index of endemic point with respect to parameters of the model}]{Local sensitivity indices of the endemic equilibrium points. At this baseline $\mathcal{R}_0\ge 1$, so this endemic point is a solution of model at steady state. }
\label{table:sie}
\end{table}
 The prevalence of infection, $i_j$,  is most sensitive to probability of transmission per act $\beta$, i.e increasing $\beta$ increases $i_j$s more than other parameters. Then $A_j$s and $\gamma_j$s  have the second most effect on $i_j$s in positive and negative direction, correspondingly.

 Prevalence in high-risk men, $i_1$, is sensitive to the total number of acts for the high-risk men $A_1$ and $\gamma_1$ more than the other $A_j$s and $\gamma_j$s for $j\neq 1$. Prevalence in high-risk women, $i_3$, is also sensitive to $A_3$ and $\gamma_3$ more than the other $A_j$s and $\gamma_j$s for $j\neq 3$. It means when high-risk people increase their number of acts, regardless of  what others do, the fraction of infected people between high-risk people increases, because they have many partners.  On the other hand, when infection period for high-risk people increases, the prevalence in high-risk population increases.

For low-risk men, the prevalence, $i_2$,  has the same sensitivity to $A_2$ and $A_4$. It means when  low-risk people increase their act, the prevalence in low-risk men increases, and we have the same story for low-risk women. It  is reasonable, because  low-risk people do not have  many partner, therefore, more acts for them and their partners plays an important role.
Prevalence in low-risk men, $i_2$,  has also the same sensitivity to $\gamma_2$ and $\gamma_4$. It means when we decrease $\gamma_2$ and $\gamma_4$-infected  people in low-risk men stay in infection category for a longer time and also  infected women in low-risk group stay in infection category for a longer time- we see  increment in the value of  $i_2$ more than the other parameters. There is a similar  analysis for low-risk group $i_4$: low-risk group $i_4$ is sensitive to $\gamma_2$ and $\gamma_4$ with the same magnitude and more than  the other $\gamma_j$s.

Another interesting result is that the endemic equilibrium points are more sensitive, than $\mathcal{R}_0$, to most of the parameters. This result says,  controlling parameters to have a low fraction of  infected population is easier than adjusting the parameters to have smaller  $\mathcal{R}_0$.

 \section{Screening Scenarios}
 The goal of this Section is to study the impact of  different screening strategies  on prevalence of Ct among different groups. In  all the simulations, the parameters are fixed with the baseline values given in Table (\ref{tab:params_table}), unless specifically defined otherwise.
In the first simulation, we assume that fraction of people who can be screened each year, $\sigma_y$, is limited by a budget, or other factors.
We also assume that if an infected person is screened for Ct, then there is a $100\%$  probability that infection  will be detected.

We will compare the fraction of the population that is infected  as a function of the screening rate  $\sigma_y^k$ people from different subgroups $k$.
We will also optimize the $\sigma_y$, for a fixed budget, that will minimize the fraction of infected people at steady state.
That is, if $(i^*_1,i^*_2,i^*_3,i^*_4)$ are the fraction  of infected people at steady state, we find the optimal screening rates that  solve the optimization problem:
  \begin{equation*}
\begin{aligned}
& \underset{\sigma_y^k}{\text{minimize}}
& & \sum_{j=1}^4N_ji^*_j(\sigma_y^1,\sigma_y^2,\sigma_y^3,\sigma_y^4),\\\
& \text{subject to}
& & \sum_{j=1}^{4} N_j\sigma_y^j=N\sigma_y=0.2N=400,
\end{aligned}
\end{equation*}
 where $N_ji^*_j$ is the number of infected people in group j and in steady state. The Figure. (\ref{fig:senario}) shows  the result for six different scenarios defined in Table (\ref{table:R0}).
\begin{table}[htp]
\centering
\begin{tabular}{llll}
\hline
\cline{1-4}
\textbf{Scenario} & $\mathcal{R}_0$&\textbf{Scenario} & $\mathcal{R}_0$\\
\hline
 (1) No screening & $4.01$& (2) Screen high-risk men& $0.86$\\
 \hline
(3) Screen high-risk women&  $0.91$& (4) Screen low-risk men& $1.26$   \\
 \hline
(5) Screen low-risk women&  $1.22$ &(6) Optimized Screening& $0.12$\\
\hline
\cline{1-4}
\end{tabular}
\caption[\textbf{ $\pmb{\mathcal{R}_0}$ for different scenarios}]{Basic reproduction number, $\mathcal{R}_0$, for different scenarios with respect to parameters of the model at the baseline parameter values. Implementing optimized screening decreases $\mathcal{R}_0$ to the order of $-1$.}
\label{table:R0}
\end{table}

\begin{figure}[htp]
\centering
\subfloat[\textbf{Different scenarios}][]{\label{fig:scenario_a}\includegraphics[width=0.45\textwidth]{Figures/policies.pdf}}
\subfloat[\textbf{Optimized scenario}][]{\label{fig:scenario_b}\includegraphics[width=0.45\textwidth]{Figures/zoom.pdf}}
 \caption[\textbf{Prevalence of Ct for different screening scenarios}]{\ref{fig:scenario_a}: the fraction of infected people after implementing different scenarios: no screening (solid line), screen $\sigma_yN$ people per year for: high-risk men(dash line), high-risk women(dash-dot line), low-risk men (dash-star), low-risk women (dash-circle), and optimized screening(dotted line). \ref{fig:scenario_b}: zooms on the optimized screening, by optimized screening the infection dies out very fast.}
            \label{fig:senario}
\end{figure}

We observe that in case of  no screening the epidemic goes up to its original endemic equilibrium point.
The effectiveness of screening is seen by the dramatic reduction in the fraction of infected people.  However, between all scenarios screening high-risk people and optimized screening cause that epidemic dies out and for  optimal choice it dies out much faster than the other two  cases.

We also list the value of $\mathcal{R}_0$ for different scenarios in Table (\ref{table:R0}).  The Figure (\ref{fig:senario}) and  Table (\ref{table:R0}) show that  $\mathcal{R}_0>1$ implies a persistent infection, though not a macroscopic outbreak in screening cases,  and when $\mathcal{R}_0 <1$ epidemic goes to DFE. Also, for the optimized scenario, which  its $\mathcal{R}_0$ is the lowest one, the epidemic dies out faster than the other scenarios.
Therefore, optimized screening was the most effective scenario among all six scenarios.

To push our understanding of  the effects of optimized screening further,  we  do sensitivity analysis of equilibrium points with respect to screening rates at their optimized values.
In this case sensitivity index become a matrix like:
\[ \mathbb{S}= \left( \begin{array}{cccc}
i^*_1& 0 & 0& 0\\
0 & i^*_2& 0& 0\\
0 & 0& i^*_3& 0\\
0 & 0& 0& i^*_4
\end{array} \right)^{-1}

\times J_{i^*}(\sigma_y)\times
\left( \begin{array}{ccc}
\sigma_y^1& 0 &0\\
0 & \sigma_y^2&0\\
0 & 0&\sigma_y^3
\end{array} \right),
\]
where $J_{i^*}(\rho_y)$ is jacobian matrix. Each column $k$ of $\mathbb{S}$ represents sensitivity index of equilibrium points with respect to  screening rate $\sigma_y^k$. Therefore, $(k,j)th$ element of $\mathbb{S}$ is sensitivity index of $i^*_k$ with respect to $\sigma_y^j$. Table (\ref{table:sa_mat})  lists the elements of this matrix: all the values in table are negative, it means there is a inverse pattern between equilibrium points and screening rate: when we increase screening rates the fraction -therefore, the number- of infected people at steady state will decrease.
Among all, $i^*_1$ is the most sensitive one, it means changing screening rates affects the  high-risk men more than the others.
\begin{table}[htp]
\centering
\begin{tabular}{lllll}
\hline
\cline{1-5}
 & $i^*_1=0.00$ & $i^*_2=0.00$ & $i^*_3=0.00$ & $i^*_4=0.00$  \\
\hline
$\sigma_y^1=0.06$& $ -0.13$ & $-0.00 $& $-0.01$ & $-0.13$ \\
 \hline
$\sigma_y^2=0.20$& $-0.56$ & $-0.37$ & $-0.27$ & $-0.41$  \\
\hline
 $\sigma_y^3=0.05$& $-0.08$ & $-0.09$ & $-0.09$ & $-0.09$  \\
\hline
\cline{1-5}
\end{tabular}
\caption[\textbf{Sensitivity indices of equilibrium points with respect to screening rates }]{Sensitivity indices of equilibrium points with respect to screening rates at optimized baseline values. The most sensitive output parameter is fraction of infected high-risk men.}
\label{table:sa_mat}
\end{table}

\iffalse
 \begin{table}[H]
\caption{Sensitivity indices of equilibrium points w.r.t screening rates at optimized baseline values. The most sensitive output parameter is number of infected high-risk men.}

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
\backslashbox{Screening rates}{Equilibrium points}& $I^*_1=0.000$ & $I^*_2=0.000$ & $I^*_3=0.000$ & $I^*_4=0.000$   \\
\hline
$\rho^s_1=0.0600$& $ -0.1294$ & $-0.0015 $& $-0.0065$ & $-0.1319$  \\
\hline
$\rho^s_2=0.2033$& $-0.5576$ & $-0.3706$ & $-0.2728$ & $-0.4056$  \\
\hline
 $\rho^s_3=0.0475$& $-0.0805$ & $-0.0918$ & $-0.0946$ & $-0.0941$  \\
 \hline\end{tabular}
\end{center}
\label{sa_mat}
\end{table}
\fi

\section{Discussion and Conclusion}
In this Chapter we created a multi-risk  heterosexual SIS transmission model for the spread of Ct with biased mixing partnership
selection to investigate the impact that screening for the disease can have in controlling its spread.
We derived the threshold conditions for the early spread of the disease and defined the basic reproductive number,
$\mathcal{R}_0$, using the next generation matrix approach.  The analysis of $\mathcal{R}_0$ identified a new
approach to reduce the size of the next generation matrix for a heterosexual Ct model with $n$ risk groups from
an $2n \times 2n$ nonsymmetric sparse matrix to an $n \times n$ symmetric full matrix. This approach can be used
in similar heterosexual STI models to greatly simplify the threshold analysis.

We used the  sensitivity analysis of $\mathcal{R}_0$ and endemic
equilibrium steady-state solutions to quantify the relative effectiveness of different intervention strategies in mitigating
the disease.  The analysis identified the probability of transmission per act (related to condom-use) is
the most sensitive parameter in controlling the epidemic.  The second most effective control mechanism
was the screening, and treating infections, of both  high-risk men and women.  Currently, most mitigation programs
only target screening high-risk women. The model indicates that it is equally important to identify infections and treat
high-risk men.  We confirmed that in the model the higher-risk groups are driving the epidemic and that
$\mathcal{R}_0$ is most sensitive to the behavior of these higher-risk people.

We implemented different screening scenarios consist of screening only high-risk men, only high-risk women,
only low-risk men, only low-risk women, and optimized screening.
We then solved  for an optimal screening strategy for infection mitigation when there are limited resources
and then determined the best screening approach to minimize the endemic steady state infection prevalence, optimized screening.
Not surprisingly, we found that this same strategy also  minimizes $\mathcal{R}_0$.
In the next Chapter, we generalize our multi-risk model to continuous-risk, when individuals can take as many number of partners as they want and we focus on impact of condom-use to control the epidemic of Ct.

config: arxiv


TITLE: Conversion of Sizes In Photos
QUESTION [0 upvotes]: Let us suppose that I've taken a photo of a person, the size of this person in the photo is two centimeters
The question is how do I convert his size in the photo into his real size in the real life
Until now I guess that I must subtract the distance between him and the camera and multiply his size by 100, is that right?!

REPLY [0 votes]: There is no way to determine the height of the person without comparing to the height of some reference object also in the same photo.
The same photo can be enlarged, to make the person appear as large or small as desired.    Measuring distances on the printed photograph gives no useful information.

If you know the focal length of your camera lens, and you know the distance of the subject from the camera, and you know the fraction of the viewing angle occupied by the subject, you can estimate the size of the subject, but this will not be accurate enough to distinguish a 5'9" person from a 5'10" person (for instance).


TITLE: Portfolio VaR with Copula?
QUESTION [4 upvotes]: Let the portfolio be given by:
$$X=X_1+X_2$$
$(X_1,X_2)$ are dependent through a Copula function $C(u_1,u_2)$, such that the joint distribution is given by:
$$F(x_1,x_2)=C(F(x_1),F(x_2))$$
What is the VaR of this portfolio?
Usually VaR is the inverse quantile: $VaR_\alpha=F^{-1}(\alpha)$.
I am not sure how to determine it in this multivariate case?

REPLY [3 votes]: It may not be possible to determine VaR analytically and you will need to run a Monte Carlo simulation.
As you indicate, for a univariate distribution $F$, VaR is specified as the $\alpha$-percentile:
$$P(X \leq \text{VaR}) = \alpha \implies  \text{VaR}=F^{-1}(\alpha).$$
In the multivariate case, with $X = X_1 + X_2$, you need the distribution $F_{X_1+X_2}(x).$  This may not be possible in closed form for a general copula. However, in the case of a Gaussian copula with Gaussian marginals, you can calculate VaR analytically because the sum of normally distributed random variables is also normally distributed.
As an example, consider the determination of VaR using a Student's t copula and a correlation matrix $\Sigma$. You can simulate and sample from a joint distribution, specified by the copula, as follows:
(1) Using a standard random number procedure, generate a vector with a correlated multivariate normal distribution: $\mathbf{Z} \sim N(0,\Sigma)$, and an independent random variable with a chi-squared  distribution with $\nu$ degrees of freedom $\chi^2(\nu).$
(2) Let $S = \sqrt{\chi^2(\nu)/\nu}$. Then $\mathbf{X} = S^{-1}\mathbf{Z}$ will have a multivariate Student's t distribution.
(3) For each component $x_i$ of $\mathbf{X}$, calculate $y_i = N^{-1}[F_{t_\nu}(x_i)]$ where $N^{-1}$ is the inverse standard normal distribution function and $F_{t_\nu}$ is the distribution function for a $\chi$-squared distribution with $\nu$ degrees of freedom.
At this point, the vector $\mathbf{Y}$ is a sample from  a joint distribution specified by a Student's t copula. The marginal distributions for the components are standard normal. The degree of dependence and kurtosis is controlled through the parameters $\Sigma$ and $\nu$.
To estimate VaR for a portfolio $P = y_1 + \ldots + y_n$, generate many sample vectors  $\mathbf{Y}^{(1)}, \mathbf{Y}^{(2)},...$ and find the corresponding portfolio values $P^{(1)}, P^{(2)}, ...$  Sort the values and find the desired quantile estimate.
The actual copula is
$$C_{\nu,\Sigma} (x_1,\ldots,x_n)  = F_{\nu,\Sigma}[F_{t_\nu}^{-1}(x_1),\ldots,F_{t_\nu}^{-1}(x_n)],$$
where $F_{\nu, \Sigma}$ is the multivariate Student's t distribution function.
You can use the copula to impose other marginal distributions with the same dependence structure by altering step (3):
$$y_i = \hat{F}^{-1}[F_{t_\nu}(x_i)], $$
where $\hat{F}$ is the desired marginal distribution function.


text:
(*  Title:      CoW_Lyndon.Lyndon_Addition
    Author:     Štěpán Holub, Charles University
    Author:     Štěpán Starosta, CTU in Prague
*)

theory Lyndon_Addition
  imports Lyndon Szpilrajn.Szpilrajn

begin

subsection "The minimal relation"

text \<open>We define the minimal relation which guarantees the lexicographic minimality of w compared to its
nontrivial conjugates.\<close>

inductive_set rotate_rel :: "'a list \<Rightarrow> 'a rel" for w
  where  "0 < n \<Longrightarrow> n < \<^bold>|w\<^bold>| \<Longrightarrow> (mismatch_pair w (rotate n w)) \<in> rotate_rel w"

text\<open>A word is Lyndon iff the corresponding order of letters is compatible with  @{term "rotate_rel w"}.\<close>

lemma (in linorder) rotate_rel_iff: assumes "w \<noteq> \<epsilon>"
  shows  "Lyndon w \<longleftrightarrow> rotate_rel w \<subseteq> {(x,y). x < y}" (is "?L \<longleftrightarrow> ?R")
proof
  assume "Lyndon w" show "rotate_rel w \<subseteq> {(x,y). x < y}"
  proof
    fix  x assume "x \<in> rotate_rel w"
    then obtain n where "x = mismatch_pair w (rotate n w)" and "0 < n" and "n < \<^bold>|w\<^bold>|"
      using rotate_rel.cases by blast
    have "w <lex rotate n w"
      using LyndonD[OF \<open>Lyndon w\<close> \<open>0 < n\<close> \<open>n < \<^bold>|w\<^bold>|\<close>].
    from this[unfolded lexordp_conv_lexord]
      prim_no_rotate[OF Lyndon_prim[OF \<open>Lyndon w\<close>] \<open>0 < n\<close>  \<open>n < \<^bold>|w\<^bold>|\<close>]
    show "x \<in> {(a, b). a < b}"
      using lexord_mismatch[of w "rotate n w" "{(a,b). a < b}", folded \<open>x = mismatch_pair w (rotate n w)\<close>]
        \<open>rotate n w \<noteq> w\<close> rotate_comp_eq[of w n]
      unfolding irrefl_def by blast
  qed
next
  assume "?R"
  show "?L"
    unfolding Lyndon.simps
  proof(simp add: assms)
    have "w <lex rotate n w" if "0 < n"  "n < \<^bold>|w\<^bold>|" for n
    proof-
      have "\<not> w \<bowtie> rotate n w"
        using rotate_comp_eq[of w n] subsetD[OF \<open>?R\<close>, OF rotate_rel.intros[OF \<open>0 < n\<close> \<open>n < \<^bold>|w\<^bold>|\<close>]]
          mismatch_pair_lcp[of w "rotate n w"] by fastforce
      from mismatch_lexord_linorder[OF this subsetD[OF \<open>?R\<close>, OF rotate_rel.intros[OF \<open>0 < n\<close> \<open>n < \<^bold>|w\<^bold>|\<close>]]]
      show "w <lex rotate n w".
    qed
    thus "\<forall>n. 0 < n \<and> n < \<^bold>|w\<^bold>| \<longrightarrow> w <lex rotate n w"  by blast
  qed
qed

text\<open>It is well known that an acyclic order can be extended to a total strict linear order. This means that
a word is Lyndon with respect to some order iff its @{term "rotate_rel w"} is acyclic.
\<close>
lemma Lyndon_rotate_rel_iff:
  "acyclic (rotate_rel w) \<longleftrightarrow> (\<exists> r. strict_linear_order r \<and> rotate_rel w \<subseteq> r)" (is "?L \<longleftrightarrow> ?R")
proof
  assume "?R" thus "?L"
    unfolding strict_linear_order_on_def acyclic_def irrefl_def
    using trancl_id trancl_mono by metis
next
  assume "?L" thus "?R"
    using acyclic_order_extension by auto
qed

lemma slo_linorder: "strict_linear_order r \<Longrightarrow> class.linorder (\<lambda> a b. (a,b) \<in> r\<^sup>=) (\<lambda> a b. (a,b) \<in> r)"
    unfolding strict_linear_order_on_def strict_partial_order_def irrefl_def trans_def total_on_def
    by unfold_locales blast+

text\<open>Application examples\<close>

lemma assumes "w \<noteq> \<epsilon>" and "acyclic (rotate_rel w)" shows "primitive w"
proof-
  obtain r where "strict_linear_order r" and "rotate_rel w \<subseteq> r"
    using Lyndon_rotate_rel_iff assms by auto

  interpret r: linorder "\<lambda> a b. (a,b) \<in> r\<^sup>=" "\<lambda> a b. (a,b) \<in> r"
    using slo_linorder[OF \<open>strict_linear_order r\<close>].

  have "r.Lyndon w"
    using r.rotate_rel_iff[OF \<open>w \<noteq> \<epsilon>\<close>] \<open>rotate_rel w \<subseteq> r\<close> by blast

  from r.Lyndon_prim[OF this]
  show "primitive w".

qed

lemma assumes "w \<noteq> \<epsilon>" and "acyclic (rotate_rel w)" shows "\<not> bordered w"
proof-
  obtain r where "strict_linear_order r" and "rotate_rel w \<subseteq> r"
    using Lyndon_rotate_rel_iff assms by auto

  interpret r: linorder "\<lambda> a b. (a,b) \<in> r\<^sup>=" "\<lambda> a b. (a,b) \<in> r"
    using slo_linorder[OF \<open>strict_linear_order r\<close>].

  have "r.Lyndon w"
    using r.rotate_rel_iff[OF \<open>w \<noteq> \<epsilon>\<close>] \<open>rotate_rel w \<subseteq> r\<close> by blast

  from r.Lyndon_unbordered[OF this]
  show "\<not> bordered w".
qed

end

subset_name: curated