Unknown to public health officials, a person with a highly contagious disease enters the population. During each period he either infects a new person which occurs with probability $p$, or his symptoms appear and he is discovered by public health officials, which occurs with probability $1 - p$. Compute the probability distribution of the number of infected but undiscovered people in the population at the time of first discovery of a carrier. Assume each infective behaves like the first.

Here is my attempt to solve this:

Let $t$ denote the time of discovery of an infected person, and let $X$ denote the number of infected by undiscovered persons at time $t$

At $t = 1$:

$P(X = 0) = 1 - p$

At $t = 2$:

$P(X = 1) =2p(1 - p)$

At $t = 3$:

$P(X = 3) = 4p^{3}(1 - p)$

Then for $t = n$

$P(X = 2^{n -1} - 1) = 2^{n-1}p^{2^{n-1} - 1}(1 - p)$

What do you guys think?

Edit: Note that at time $t$, only 2 things can happen for each person: either he infects another person, or he is discovered.

Edit: Edited to take into account the probability of infecting another person, for each undetected person.

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    $\begingroup$ When he is discovered by the public health officials, is he removed from the population so he cannot infect more people? That would change things somewhat, I think. $\endgroup$ – Ian Coley Mar 18 '13 at 19:29
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    $\begingroup$ @FrankMcGovern Is that relevant?, from the text of the problem it seems that we can assume that once an infected person is detected the process stops, or at least that we can ignore what happens after. $\endgroup$ – Mario Mar 18 '13 at 20:03

Denote the $i$th affected person as $c_i$. Let's make two assumptions:

(1) The number of affected people is counted only at the end of each period;

(2) A newly affected person cannot do anything during the period in which he is affected, i.e., cannot affect another person or be discovered. Without this assumption, explosive effect can happen in just one period. $c_1$ affects $c_2$, $c_2$ immediately affects $c_3$, $c_3$ immediately affects $c_4$, ... Both affecting another person and being discovered take a whole period. A newly affected person has time less than one period, so he can do nothing.

Denote the number of affected but undiscovered people at the end of period $n$ as $X_n$.

End of period 1:

If $c_1$ is discovered during period 1, $P(X_1=0)=1-p$.

If $c_1$ affects $c_2$ during period 1, $P(X_1=2)=p$.

There is no way of $X_1=1$ here. Note that $c_2$ can do nothing during his newly affected period by assumption 2.

End of period 2:

Then $c_1$ must affect $c_2$ during period 1. Otherwise there is no period 2.

If both $c_1$ and $c_2$ are discovered during period 2, $P(X_2=0)=(1-p)^2$.

If both $c_1$ and $c_2$ affect other people, producing $c_3$ and $c_4$ during period 2, $P(X_2=4)=p^2$.

If one disease carrier is discovered and the other affects another person, producing $c_3$, $P(X_2=2)=2p(1-p)$. The two undiscovered carriers are either ($c_1$, $c_3$) or ($c_2$, $c_3$).

There is no way of $X_2=1$ or $X_2=3$ here. Note that $c_3$ and $c_4$ can do nothing during their newly affected period by assumption 2.

End of period $n+1$:

Then no affected carrier were discovered during any period before $n+1$. Otherwise there is no period $n+1$. $X_{n+1}$ only depends on $X_n$. This is a Markov chain. Suppose $X_n=j$.

If none of the $j$ carriers are discovered, then each one of them produces a new carrier, $P(X_{n+1}=2j)=p^j$.

If all of the $j$ carriers are discovered, $P(X_{n+1}=0)=(1-p)^j$.

If $m$ of the $j$ carriers produce a new carrier and $j-m$ of them are discovered, $P(X_{n+1}=2m)=\binom{j}{m}p^m(1-p)^{j-m}$.

We supposed $X_n=j$. Now we know $X_n=2X_{n-1}=2^2X_{n-2}=...2^{n-1}X_1=2^n$ since there is no affected discovered to arrive perior $n+1$ and it must be a maximum number of affected.

Thus $P(X_{n+1}=2m)=\binom{2^n}{m}p^m(1-p)^{2^n-m},m=0,1,2,..,2^n$.

Now we can calculate $P(X=2m)=\sum P(X_{n+1}=2m)$. The summation should up to infinity. Where should it start? Given $m$, not all $n$ has $X_n=2m$. For example, $n=1$ does not have $X_n=2\times 2$. What is the smallest $n$ such that $X_n=2m$ is possible? The largest $m$ that $X_{n+1}$ can take is $2^n$. Thus $m<2^n$, i.e., $n\geq\lfloor1+\text{log}_2m\rfloor$. Note that $m$ cannot be $2^n$ since no affected is discovered and the process is not ended.

Thus we should calculate $P(X=2m)=\sum^\infty_{n=\lfloor1+\text{log}_2m\rfloor} P(X_{n+1}=2m)=\sum^\infty_{\lfloor1+\text{log}_2m\rfloor}\binom{2^n}{m}p^m(1-p)^{2^n-m}$.

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