Population growth: Galton-Watson process. Population growth: Galton-Watson process. A simple branching process. A simple branching process

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1 Population growth: Galton-Watson process Population growth: Galton-Watson process Asimplebranchingprocess September 27, 2013 Probability law on the n-th generation 1/26 2/26 A simple branching process A simple branching process Let us consider a simple stochastic model for the evolution of the size of a population. (a) The population evolves in generations. Let Z k be the number of members of the k-th generation. By assumption, Z 0 =1. (b) Each member of the k-th generation gives birth to a family, possibly empty, of members of the (k + 1)-th generation. I I The family sizes form a collection of independent and identically distributed random variables. The size of each of these families (i.e., the number of descendants of an individual) has probability function: Such a branching process {Z 0, Z 1,...,Z n,...} is called a Galton-Watson process. P( = i) =p i, i 0 3/26 4/26

2 Example Example Suppose, for instance, that Z 1 = 2. Then For instance, suppose that Then p 0 = 1 2, p 1 = 1 4, p 2 = 1 4 P(Z 2 =0 Z 1 = 2) = P( = 0) = p 2 0 = 1 4 P(Z 2 =1 Z 1 = 2) = P( = 1) = p 0 p 1 + p 1 p 0 = 1 4 P(Z 2 =2 Z 1 = 2) = P( = 2) = p 0 p 2 + p 1 p 1 + p 2 p 0 = 5 16 P(Z 2 =3 Z 1 = 2) = P( = 3) = p 1 p 2 + p 2 p 1 = 1 8 P(Z 1 = 0) = 1 2, P(Z 1 = 1) = 1 4, P(Z 1 = 2) = 1 4 P(Z 2 =4 Z 1 = 2) = P( = 4) = p 2 2 = 1 16 Of course, we have P(Z 2 = n Z 1 = 2) = 1 n 0 5/26 6/26 Example Probability distribution of Z k+1 Performing similar calculations and taking into account that P(Z 2 = n) = 2 P(Z 2 = n Z 1 = r)p(z 1 = r) r=0 we obtain the probability function of Z 2 : P(Z 2 = 0) = 11 16, P(Z 2 = 1) = 2 16, P(Z 2 = 2) = 9 64, P(Z 2 = 3) = 1 32, P(Z 2 = 4) = 1 64 In general, P(Z k+1 = n Z k = r) =P( r = n), where the r.v. i are independent and identically distributed as. Of course, P(Z k+1 =0 Z k = 0) = 1 By the TPT, P(Z k+1 = n) = P(Z k+1 = n Z k = r)p(z k = r) r 0 7/26 8/26

3 Extinction probabilities Extinction probabilities Let us calculate the probability that the population disappears in a number of generations apple m, d m P(Z m = 0) Let D m = {Z m =0}. Then d m = P(D m ) = P(D m Z 1 = k)p(z 1 = k) = m 1 ) k 0 k 0(d k P( = k) In the previous example, d 0 =0 d 1 = 1 2 d 2 = Hence, if G (s) = s k P( = k) k 0 is the probability generating function of,then d m = G (d m 1 ) 9/26 10 / 26 Extinction probabilities In our example, Hence d 0 =0 G (s) = s s2 d 1 = G (d 0 )=G (0) = d 2 = G (d 1 )=G = = d 3 = G (d 3 )=G = = Notice that d 0 apple d 1 apple d 2 apple apple d m apple apple 1 Hence the following limit exists: What is its meaning? d lim m!1 d m 11 / / 26

4 We have D 0 D 1 D m D m+1 Therefore we can consider the limit event D = lim m!1 D m = 1[ D k, k=0 which corresponds to the ultimate extinction of the population. Hence d = lim d m = lim P(D m)=p lim D m m!1 m!1 m!1 = P(D) From d m = G (d m 1 ) we get that (m!1): d = G (d) I So, the extinction probability d is a solution of the equation s = G (s) I Notice that s =1is always a root. is the probability of that ultimate extinction. 13 / / 26 Returning to the example, s = G (s) () s = p 0 + p 1 s + p 2 s 2 (1) Case m < 1 () p 0 > p 2 () s 2 > Hence The two roots are p 2 s 2 (p 0 + p 2 ) s + p 0 =0 s 1 =1, s 2 = p 0 p Let m be the expected number of descendants per individual: m =E( )=p 1 +2p 2 =1 p 0 + p 2 Figure: p 0 =1/2, p 1 =1/4, p 2 =1/4 The iterations d m = G (d m 1 ) converge to d = / / 26

5 (2) Case m =1() p 0 = p 2 () s 1 = s 2 =1 (3) Case m > 1 () p 0 < p 2 () s 2 < Figure: p 0 =1/4, p 1 =1/2, p 2 =1/4 Figure: p 0 =1/4, p 1 =1/4, p 2 =1/2 The iterations d m = G (d m 1 ) also converge to d = 1. The iterations d m = G (d m 1 ) converge to d = s 2 < / / 26 General case General case The same analysis can be done in a more general situation. Notice that if s 0, then G (s) = p 0 + p 1 s + p 2 s 2 + p 3 s G 0 (s) = p 1 +2p 2 s +3p 3 s G 00 (s) =2p 2 +6p 3 s + 0 So, the plot of G (s) is as in the case of a polynomial of degree 2. I In general, the plot of G (s) will intersect in two points the line (with equation) s. So, the equation s = G (s) willhave two solutions. I The iterations d m = G (d m 1 ) will converge to the smallest of the two solutions. I Moreover, m =E( )=G 0 (1) is the slope of the tangent line of G (s) ats = 1. I Hence we have again the three cases considered above for G a polynomial of degree / / 26

6 Probability generating function of Z n Theorem Let d be the smallest root of the equation s = G (s) and let m =E( ). Then d is the probability of ultimate extinction of the population. Moreover, I m < 1 ) d =1. The extinction is sure. I m =1) The line s is tangent at s =1to G (s) (double root d =1). The extinction is sure. I m > 1 ) d < 1. In this case, there is a positive probability 1 d > 0 of non-extinction. Let G n (s) the probability generating function of Z n. We have G n+1 (s) =E Moreover, E s Z n+1 = E s Z n+1 Z n = k P(Z n = k) k 0 s Z n+1 Z n = k =E s k Z n = k =E s k = E s k =(G (s)) k 21 / / 26 Probability generating function of Z n Probability generating function of Z n Therefore Recall that Then G n+1 (s) = k 0 (G (s)) k P(Z n = k) G n (z) = k 0 z k P(Z n = k) G n+1 (s) =G n (G (s)) So, we have G 1 (s) =G (s) G 2 (s) =G 1 (G (s)) = G (G (s)) = G (2) (s) G 3 (s) =G 2 (G (s)) = G (G (G (s))) = G (3) (s) G n (s) =G n 1 (G (s)) = G (G ( G (s) )) = G (n) (s) 23 / / 26

7 Probability generating function of Z n Expected number of individuals In the example, G (s) = s s2 Let m n =E(Z n ). The derivative of G n+1 (s) =G n (G (s)) is Then G 2 (s) =G (G (s)) = G (s)+ 1 4 (G (s)) 2 = s s s s2 = s s s s4 Gn+1(s) 0 =Gn 0 (G (s)) G 0 (s) Taking s =1wehave Gn+1(1) 0 = Gn 0 (G (1)) G 0 (1) = G n 0 (1) G 0 (1) Hence m n+1 = m n m.therefore m n = m n 25 / / 26