Determining number of randomly picked people

Question

Firstly I want to put big disclaimer here. This particular problem is a smaller part of my homework. Since even after discussion with my fellow classmates we are not sure how to handle it we decided to post a question here.

Basically we are supposed to simulate spreading of disease.

Every day every infected person will pick random number of people (possible infected candidates) with Poisson distribution where parameter is 5. Everyone of those selected people (=possible infected candidates) will be infected with probability of 1/2 and at the end of the day the infection takes effect. Next day this particular infected person will be also spreading infection and will again pick random number of people (another possible infected candidates) with Poisson distribution where parameter is again 5. Number of possibly infected candidates and event static whether person will be infected or not are both independent on each other.

Let's suppose on the first day there is only one infected person. We should simulate disease spreading for one week. And how many infected persons there will be at the end of 8th day.

Now finally to my question.

Confusing part for me is how are we suppose to determine the number of randomly picked people which represent possible infected candidates since we don't know how many people is there? I assume that with higher number of people the disease should spread faster? I also think that this number of randomly picked possible candidates should be somehow determined by given parameter and also by Poisson distribution itself but I am struggling how to make these things "work" together.

Answer 1

Consider the first day, in which there is one infected person to start. Let $C$ be the number of candidates selected for possible infection on the $1st$ day, and let $I$ be the number infected by "patient zero" on the $1st$ day. I assume that out of the $C$ candidates, each is infected independently with probability $0.5$. Then $C \sim Pois(5)$ and $I \mid C = c \sim Bin(c,0.5)$, $c = 0,1,2,\ldots$. You can check from this post that $I \sim Pois(2.5)$, where the $2.5$ is from $\lambda p = 5*0.5 = 2.5$. Thus the expected value of $I$ is 2.5, and adding 1 for patient zero we get that after one day, there are $3.5 = \frac{7}{2}$ patients expected to be infected.

How about day 2? For each infected patient at the end of day 1, we expect $\frac{7}{2}$ to be infected at the end of day 2. Thus the total expected number of infected after two days is $\left(\frac{7}{2}\right)^2$.

For eight days, we expect $$ \left(\frac{7}{2}\right)^8 $$ to be infected.

This was verified numerically with my MATLAB implementation:

% Start with 1 infected person.  At the start of each day, he picks
% a Pois(5) number of people as candidates to infect.  Each of these
% candidates is infected with probability 1/2, assuming each
% independently of the other.  Verifying my analytical solution to the
% expected number of people infected after 8 days.

%%% Simulation parameters %%%
runs = 10000;  % number of runs of simulation
lambda = 5;  % poisson parameter
p = 0.5;  % binomial success probability
N = 8;  % number of days to run each simulation


%%% Allocate arrys %%%
C = zeros(runs,N);  % number of candidates for infection each step
I = zeros(runs,N+1);  % number of infected after each step

%%% Init arrays %%%
I(:,1) = 1;  % one person infected to start each simulation

for r=1:runs
    for i = 1:N

        % draw number of candidates of infection for each infected
        for k = 1:I(r,i)
            C(r,i) =  C(r,i) + poissrnd(lambda);
        end

        % flip coin for each candidate.  <0.5 = infected, >0.5 = healthy
        x = rand(1,C(r,i));
        I(r,i+1) = sum(x<p) + I(r,i);  % count number infected
    end
end

numerical_avg = mean(I(:,N+1))

Answer 2

Let $x_i$ be the number of infected during day i, $S_i$ be the totle number of infected at the end of day i and $S_0$=0, $S_1=1$

By thinning property of Poisson process, $x_i$~Poisson($0.5\cdot 5E(S_i$)) and,

E($x_i$)=0.5$\cdot$5E($S_i$)=2.5E($S_i$)

Also we have a recursive relation that

E($S_i$)=E($S_{i-1}$)+E($x_{i-1}$)

=E($S_{i-1}$)+2.5E($S_{i-1}$)

=3.5E($S_{i-1}$)

Finally, E($S_i$) forms a geometric series

E($S_8$)=${1-3.5^8}\over{1-3.5}$

=9007.1016