I am reading my textbook and I don't seem to understand some stuff. Here is what is written in the textbook:
Consider a population of $N$ people. There are $3$ different classifications of each person:
$1)$ Susceptibles. $S_n$ denotes the number of susceptibles in the population at time $n$
$2)$ Infected. $I_n$ denotes number of infected in population at time $n$
$3)$ Recovered/Removed. $B_n$ denotes number of people recovered/removed at time $n$
Let $X_n=(S_n,I_n)$.
We will only consider a population that is closed, i.e no migration.
Assumptions:
$N$ is fixed
In between time steps $n$ and $n+1$, the probability the $i$-th susceptible avoids contact with any given infective is $P$ (independent of all others)
Upon contact we assume that a susceptible becomes infected
$\Bbb P($ $i$ th susceptible avoids the $I_n$ infectives at time $n$ $)= P^{I_n}$
The infection period is distributed accordingly to some RV $T_I$
$T_I \sim $Geom $(\lambda)$ $\Rightarrow \Bbb P(T_I =1)=\lambda$ .($(T_I=1)$ Basically denotes the time to recover.)
We have 2 basic models: SIS and SIR model
SIS MODEL
In this model, the individual is either infective or susceptible.
$S_{n+1}$ is the number of susceptibles at time $n+1$ in population
$S_{n+1}=$ Bin$(S_n,P^{I_n})$ $+$ Bin$(I_n, \lambda)$
Where:
- $A_{n+1}=$ Bin$(S_n,P^{I_n})$ is the number of susceptibles at time $n$ who avoid infection in the next time step
- $R_{n+1}=$Bin$(I_n, \lambda)$ is the number of infectives from time $n$ who recover over the next time step
Now, since we're in a closed population and no removed category ($B_n$) $\Rightarrow I_{n+1}=N- S_{n+1}$
The transition probabilities which define the $P$-matrix are :
$$\Bbb P(S_{n+1} = v | S_n=w)= \Bbb(A_{n+1} + R_{n+1}=v | S_n = w)=$$ $$ \sum^w_{k=0} \Bbb P(A_{n+1} =k | S_n =w) \Bbb P(R_{n=1} =v-k | S_n = w)= $$ $$\sum^w_{k=1} {w \choose k} \Bigl(1-P^{N-w}\Bigr)^{w-k}\Bigl(P^{N-w}\Bigr)^k {n-w \choose v-w} \lambda^{v-k}(1- \lambda)^{N-w+k} \tag{1}$$
Could somebody please explain me how do they get $(1)$. I am completely lost. If every term could be explained, that would help a lot.
Now we continue to the second model
SIR MODEL
Now we have 3 states for an individual: Susceptible, Infected, removed/recovered
$X_n=(S_n, I_n)$ . Note : $B_n= N-S_n-I_n$
$S_n=$ Bin$(S_n, P^{I_n})$
$I_{n+1}=$ Bin$(I_n, 1- \lambda)+(S_n-S_{n-1})$
Where:
Bin$(I_n, 1- \lambda)$ is the number of infectives still infected from time $n$
$(S_n-S_{n-1})$ is the number of newly infected individuals from time $n$
The transition probabilities which define the $P$-matrix are :
$$\Bbb P(X_{n=1}= (v,x) | X_n = (w,y))=$$
$$\Bbb P(S_{n+1}= v , I_{n+1} =x | S_n = w , I_n=y)=$$
$$\Bbb P(S_{n=1} =v | X_n =(w,y)) \Bbb P( I_{n=1} =x | X_n=(w,y))=$$
$$ {w \choose v} \Bigl(P^y\Bigr)^v \Bigl(1-p^y\Bigr)^{w-v} x {y \choose x-(w-v)} \Bigl(1- \lambda \Bigr)^{x-(w-v)}\lambda^{y-x+(w-v)} \mathbb 1_{\{w \geq v \}} \Bbb 1_{\{y \geq x-(w-v)\}} \tag{2}$$
Could somebody please explain me how do they get $(2)$. I am completely lost. If every term could be explained, that would help a lot.
Now we look at a special case of the SIR model:
(SIR) REED FROST MODEL
Here, $\lambda=1$ . i.e recovery happens deterministically in $1$ timestep.
So, $$\Bbb P(I_{n+1}=x | X_n = (v,x)= {v \choose x} \Bigl(P^y\Bigr)^{v-x}\Bigl(1-P^y\Bigr)^x$$
Ok, I think I understand this:
- {v \choose x} means how many ways we can choose $x$ amount of susceptibles, out of $v$, to be infected.
$\Bigl(P^y\Bigr)^{v-x}$ means until the $n$-th step, we got the individuals to survive $y$ infectives around them (avoiding $y$ infected). So that's $P^y$. We put $P^y$ to the power of $v-x$ because from the $n$ to $n+1$-th step, $v-x$ susceptibles avoid being infected. Hence, $\Bigl(P^y\Bigr)^{v-x}$
And then $\Bigl(1-P^y\Bigr)^x$ is because the people who survived until the $n$-th step - $P^y$ are infected $x$ amount of times.
I know this is a binomial distribution. If there are any errors in my explanation please correct me
So we give 2 deifinitions for the Reed-Frost model:
$1)$ The basic reproduction number, "$R_0$", is the average number of new infectives caused by a typical infective during the early stage of the epidemic.
$2)$ The Final Size of the epidemic is defined as the number of initially susceptible individuals who become infected during the epidemic- Denoted $Z$
So,
$$\Bbb P(Z=z | S_0 =n , I_0=m) = \sum_{\underline{y} : |y| =z} \Bbb P\bigcap^m_{j=1}(\{I_j=y_i\} | S_0=n, I_0=m)$$
So here, I'm not exactly sure why exactly $\underline{y} : |y| =z$. So this means $I_1=z, I_2=z,...,I_m=z$ ? I can't really make sense of this. How I interpret this is: the intersection of all events where the number of infected is equal to the $z$ as defined above. Not exactly sure how this makes sense..
So here's an example the textbook gives:
Suppose $S_0=2$ and $I_0=1$
Then
$\Bbb P(Z=0 | X_0=(2,1))= P^2$ - This I understand. The final size of infected orgininally-susceptibles is $0$, and this means that the $2$ susceptibles avoided the infection- hence $P^2$
$\Bbb P(Z=1 | X_0 = (2,1))= \Biggl({2 \choose 1}P(1-P)\Biggr)P$ - Now this i'm not exactly sure how to interpret. Out of 2 originally-susceptibles, 1 out of the 2 don't avoid the infection, hence the $(1-P)$. And $1$ avoids it, hence the $P$. But i'm not sure where does the other $P$ come from?
$\Bbb P(Z=2 | X_0 =(2,1)) =\Bbb P(I_1=2 , I_2=0 | X_0=(2,1) +\Bbb P(I_1=1, I_2=2 | X_0=(2,1)) = (1-P)^2+ \Biggl({2 \choose 1}(1-P)P \Biggr)(1-P)$ . Here I understand why they partition this into two cases where $I_1=2 , I_2=0 $ and $I_1=1, I_2=2$ - since we can have the first case with prob $(1-P)^2$ -which i understand how is derived, but the second case I have the same confusion as the above.
Any help is appreciated!
