# Mathematics of suffering:Spread of Aids in a hypothetical community

For the last few days I have been thinking about this problem which I posed to myself but I have no idea how to make a start.Suppose in a Closed community consisting of N men and M women ,n and m members respectively are intitially infected with Aids.Each day any randomly chosen couple from MN couples(consisting of a man and a woman) has a probabily p of unprotected sexual intercourse.How can we find the probabilty distribution of the infected persons on the ith day and the expected number of infected persons .Lot of Thanks for any help in advance.

• Can multiple instances of intercourse occur on the same day? If so, how can we distinguish the order of the intercourse on that day? Can we model this problem as a series of intercourse events spaced out by the exponential distribution? – Vortico Nov 28 '14 at 18:47
• yes multiple intercourses can happen.As stated in the problem every possible heterogeneous couple has an independent probability p of intercourse but any particular couple can have only one intercourse on a single day – sajjad veeri Nov 28 '14 at 18:54

Introduce some notation:

\begin{eqnarray*} m_i &=& \text{the total number of women infected after day $i$} \\ n_i &=& \text{the total number of men infected after day $i$} \\ && \\ q_i &=& \text{probability of a given woman becoming infected on day $i$} \\ r_i &=& \text{probability of a given man becoming infected on day $i$} \\ && \\ q_i,j &=& \text{probability of $j$ women becoming infected on day $i$} \\ r_i,j &=& \text{probability of $j$ men becoming infected on day $i$}. \end{eqnarray*}

Since the probability of a given woman not having sex will all $n_i$ infected men is $(1-p)^{n_i}$, we have,

\begin{eqnarray*} q_{i+1} &=& 1-(1-p)^{n_i} \\ r_{i+1} &=& 1-(1-p)^{m_i}. \end{eqnarray*}

Using the binomial distribution,

\begin{eqnarray*} q_{i+1,j} &=& \binom{M-m_i}{j} q_{i+1}^j(1-q_{i+1})^{M-m_i-j} \\ r_{i+1,j} &=& \binom{N-n_i}{j} r_{i+1}^j(1-r_{i+1})^{N-n_i-j}. \end{eqnarray*}

This is a bivariate Markov chain (a Markov chain with two random variables, $m_i, n_i$ instead of the one in a standard (univariate) Markov chain): the probability of moving to "state" $(m_{i+1},n_{i+1})$ in one step (i.e. one day, here) is dependent only on the previous state $(m_i,n_i)$.