TBan-IV

Bienaymé Galton Watson process

A time $t=0$ appears as infected individual which dies with a rate $a$ and branches with a rate $b$. On average, each infection generates in average $R_0=b/a$ new ones. Real epidemics corresponds to $R_0>1$.

At time $t$, the infected population is $n(t)$, while the total infected population is

$N(t)=\underset{0}{\overset{t}{\int }}n(t^{\prime })d{t}^{\prime }$

Our goal is to compute $P(N(t))$ and we introduce its Laplace Transform:

$Q_s(t)=\underset{0}{\overset{\mathrm{\infty }}{\int }}P(N)e^{-sN}dN=⟨e^{-s\underset{0}{\overset{t}{\int }}n(t^{\prime })d{t}^{\prime }}⟩$

. Note that the normalization imposes

$Q_0(t)=1$

.

• Evolution equation: Consider the evolution up to the time $t+dt$ as a first evolution from $0$ to $dt$ and a following evolution from $dt$ to $t+dt$. Derive the following equation for $Q_s(t)$

$Q_s(t+dt)=(1-(a+b)dt)e^{-sdt}{Q}_s(t)+adt+bdt{Q}_s^2(t)+O(d{t}^2)$

which gives

$\frac{d{Q}_s(t)}{dt}=-(a+b+s)Q_s(t)+a+b{Q}_s^2(t)$

• Critical case: the stationary solution: Let's set $b=a$ and $a=1$ to recover the results of the mean field cellular automata. In the limit $t\to \mathrm{\infty }$ we are interested to the total population size $S=N(t\to \mathrm{\infty })$. The Laplace transform of $P(S)$ is

$0=-(2+s)Q_s^{\text{stat}}+1+(Q_s^{\text{stat}}{)}^2$

which gives

$Q_s^{\text{stat}}=\frac{(2+s)-\sqrt{s^2+4s}}{2}\sim 1-\sqrt{s}+O(s)$

with

$\underset{0}{\overset{\mathrm{\infty }}{\int }}dSP(S)e^{-sS}=Q_s^{\text{stat}}$

• Critical case: Asymptotics: We want to predict the power law tail of the distribution $P(S)\sim A\cdot S^{-\tau }$. Taking the derivative with respect to $s$ we have

$A\underset{0}{\overset{\mathrm{\infty }}{\int }}dS{S}^{1-\tau }{e}^{-sS}=\frac{1}{2\sqrt{s}}$

and conclude that $\tau =3/2$ and

$A=\frac{1}{2\underset{0}{\overset{\mathrm{\infty }}{\int }}dz{e}^{-z}/\sqrt{z}}=\frac{1}{2\sqrt{\pi }}$

Hence we find back our previous result

$P(S)\sim \frac{1}{2\sqrt{\pi }}\frac{1}{S^{3/2}}$