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

$N(t)=\int _{0}^{t}n(t')dt'$

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

$Q_{s}(t)=\int _{0}^{\infty }P(N)e^{-sN}dN=\left\langle e^{-s\int _{0}^{t}n(t')dt'}\right\rangle$

. 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 $Q_{s}(t)$

$Q_{s}(t+dt)=(1-(a+b)dt)e^{-sdt}Q_{s}(t)+adt+bdtQ_{s}^{2}(t)+O(dt^{2})$

which gives

${\frac {dQ_{s}(t)}{dt}}=-(a+b+s)Q_{s}(t)+a+bQ_{s}^{2}(t)$

• 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 $t\to \infty$ the total population coincides with the avalanche size, $N(t\to \infty )=S$. 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

$\int _{0}^{\infty }dSP(S)e^{-sS}=Q_{s}^{\text{stat}}$

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

$A\int _{0}^{\infty }dSS^{1-\tau }e^{-sS}={\frac {1}{2{\sqrt {s}}}}$

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

$A={\frac {1}{2\int _{0}^{\infty }dze^{-z}/{\sqrt {z}}}}={\frac {1}{2{\sqrt {\pi }}}}$

Hence we find back our previous result

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