TBan-IV - Revision history

Rosso: /* Bienaymé Galton Watson process */

2025-09-16T13:37:38Z

Bienaymé Galton Watson process

← Older revision		Revision as of 15:37, 16 September 2025
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	and conclude that <math> \tau=3/2 </math> and		and conclude that <math> \tau=3/2 </math> and
	<center> <math> A =\frac{1}{2 \int_0^\infty d z e^{-z}/\sqrt{z}}= \frac{1}{2 \sqrt{\pi}} </math></center>		<center> <math> A =\frac{1}{2 \int_0^\infty d z e^{-z}/\sqrt{z}}= \frac{1}{2 \sqrt{\pi}} </math></center>
	~~Hence we~~ find ~~back our previous~~ result		We find the result
	<center><math> P(S) \sim \frac{1}{2 \sqrt{\pi}}\frac{1}{S^{3/2}} </math> </center>		<center><math> P(S) \sim \frac{1}{2 \sqrt{\pi}}\frac{1}{S^{3/2}} </math> </center>

Rosso: /* Bienaymé Galton Watson process */

2025-09-16T13:36:54Z

Bienaymé Galton Watson process

← Older revision		Revision as of 15:36, 16 September 2025
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	<center> <math> \frac{d Q_s(t)}{d t}= -(a+b+s) Q_s(t)+a+ b Q_s^2(t) </math></center>		<center> <math> \frac{d Q_s(t)}{d t}= -(a+b+s) Q_s(t)+a+ b Q_s^2(t) </math></center>

	* <Strong> Critical case: the stationary solution</Strong>: Let's set <math> b=a</math> and <math> a=1</math> to recover the results of the mean field cellular automata. In the limit <math> t \to \infty</math> the total population ~~coincides with the avalanche~~ size, <math> N(t\to \infty) =S</math>. The Laplace transform of <math> P(S)</math> is		* <Strong> Critical case: the stationary solution</Strong>: Let's set <math> b=a</math> and <math> a=1</math> to recover the results of the mean field cellular automata. In the limit <math> t \to \infty</math> we are interested to the total population size <math> S= N(t\to \infty) </math>. The Laplace transform of <math> P(S)</math> is
	<center> <math> 0= -(2+s) Q_s^{\text{stat}}+1+ (Q_s^{\text{stat}})^2 </math></center>		<center> <math> 0= -(2+s) Q_s^{\text{stat}}+1+ (Q_s^{\text{stat}})^2 </math></center>
	which gives		which gives
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	<center> <math>\int_0^\infty d S P(S) e^{-sS}= Q_s^{\text{stat}} </math></center>		<center> <math>\int_0^\infty d S P(S) e^{-sS}= Q_s^{\text{stat}} </math></center>

	* <Strong> Critical case: Asymptotics</Strong>: We want to predict the power law tail of the ~~avalanche~~ distribution <math> P(S) \sim A \cdot S^{-\tau} </math>. Taking the derivative with respect to <math> s </math> we have		* <Strong> Critical case: Asymptotics</Strong>: We want to predict the power law tail of the distribution <math> P(S) \sim A \cdot S^{-\tau} </math>. Taking the derivative with respect to <math> s </math> we have
	<center> <math> A \int_0^\infty d S S^{1-\tau} e^{-sS}=\frac{1}{2 \sqrt{s}} </math></center>		<center> <math> A \int_0^\infty d S S^{1-\tau} e^{-sS}=\frac{1}{2 \sqrt{s}} </math></center>

Rosso: Created page with "== 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 $N(t) = \int_0^t n(t') d t'..."$

2025-09-09T07:11:50Z

Created page with "== Bienaymé Galton Watson process== A time <math> t=0 </math> appears as infected individual which dies with a rate <math> a </math> and branches with a rate <math> b </math>. On average, each infection generates in average <math> R_0 = b/a </math> new ones. Real epidemics corresponds to <math> R_0>1 </math>. At time <math> t </math>, the infected population is <math> n(t) </math>, while the total infected population is <center> <math> N(t) = \int_0^t n(t') d t'..."

New page

== Bienaymé Galton Watson process==

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

At time <math> t </math>, the infected population is <math> n(t) </math>, while the total infected population is
<center> <math> N(t) = \int_0^t n(t') d t' </math> </center>
Our goal is to compute <math> P(N(t)) </math> and we introduce its Laplace Transform:
<center> <math> Q_s(t)=\int_0^\infty P(N) e^{-s N} dN=\left\langle e^{-s\int_0^t n(t') dt'}\right\rangle
</math></center>. Note that the normalization imposes <math> Q_0(t)=1 </math>.

* Evolution equation: Consider the evolution up to the time <math> t+dt</math> as a first evolution from <math> 0</math> to <math> dt</math> and a following evolution from <math> dt</math> to <math> t+ dt</math>. Derive the following equation for <math> Q_s(t)</math>
<center> <math> Q_s(t+dt) = (1-(a+b) d t) e^{-s dt} Q_s(t) +a dt + b dt Q_s^2(t) +O(dt^2) </math></center>
which gives
<center> <math> \frac{d Q_s(t)}{d t}= -(a+b+s) Q_s(t)+a+ b Q_s^2(t) </math></center>

* Critical case: the stationary solution: Let's set <math> b=a</math> and <math> a=1</math> to recover the results of the mean field cellular automata. In the limit <math> t \to \infty</math> the total population coincides with the avalanche size, <math> N(t\to \infty) =S</math>. The Laplace transform of <math> P(S)</math> is
<center> <math> 0= -(2+s) Q_s^{\text{stat}}+1+ (Q_s^{\text{stat}})^2 </math></center>
which gives
<center> <math>Q_s^{\text{stat}}= \frac{(2+s) -\sqrt{s^2 +4 s}}{2} \sim 1 - \sqrt{s} +O(s) </math></center>
with
<center> <math>\int_0^\infty d S P(S) e^{-sS}= Q_s^{\text{stat}} </math></center>

* Critical case: Asymptotics: We want to predict the power law tail of the avalanche distribution <math> P(S) \sim A \cdot S^{-\tau} </math>. Taking the derivative with respect to <math> s </math> we have
<center> <math> A \int_0^\infty d S S^{1-\tau} e^{-sS}=\frac{1}{2 \sqrt{s}} </math></center>

and conclude that <math> \tau=3/2 </math> and
<center> <math> A =\frac{1}{2 \int_0^\infty d z e^{-z}/\sqrt{z}}= \frac{1}{2 \sqrt{\pi}} </math></center>
Hence we find back our previous result
<center><math> P(S) \sim \frac{1}{2 \sqrt{\pi}}\frac{1}{S^{3/2}} </math> </center>