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Consider a discrete time branching process $\left\{X_{n}\right\}$ with probability generating function $$ \varphi(s)=\frac{1-(b+c)}{1-c}+\frac{b s}{1-c s}, \quad 0<c<b+c<1 $$ where $(1-b-c) / c(1-c)>1$. Assume $X_{0}=1$. Determine the conditional limit distribution: $$ \lim \operatorname{Pr}\left\{X_{n}=k \mid X_{n}>0\right\} $$

Attempt. From the question, expected value of individuals in the $1$st generation $=\frac{b}{(1-c)^2}<1$ (obtained from the relation $(1-b-c)/c(1-c)>1$). Hence, probability of extinction i.e., $\lim_{n\rightarrow\infty}\text{Pr}(X_n=0)=1$. Now, for some $k\in\mathbb{N}$ $$\lim \operatorname{Pr}\left\{X_{n}=k \mid X_{n}>0\right\}=\lim\frac{\text{Pr}(X_n=k)}{\text{Pr}(X_n>0)}.$$ I'm can't figure how to proceed from here? Finding $\varphi_n(s)$ ie. $n$th convolution of $\varphi$ seems computationally huge with no pattern. If there were a pattern, finding the $k$th coefficient again, is computationally huge.

Also, I'm not sure if the following is useful at all: $$\varphi(s)=\frac{1-b-c}{1-c}+bs+bcs^2+bc^2s^2+bc^3s^3+...$$

zaira
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2 Answers2

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If $M$ is the matrix $[a,b;c,d]$ we write $h_M(s)=\frac{ax+b}{cx+d} $ and we have $$h_{MM'}(s)=h_M(h_{M'}(s)).$$Here your generating function is of that type but I do not like your parameterization by $b$ and $c$ and I will take $p=1-c$ and $r=b/(1-c)$. With these notations your $h_M$ is for

$$M=M(p,r)=\left[\begin{array}{cc}p+r-1&1-r\\p-1&1\end{array}\right]$$ the reason of this choice is that $p$ and $r$ are the eigenvalues of $M(p,r)$ which makes the power $M^n$ easy to compute. Actually if you denote

$$\lambda_n=\frac{p^n(1-r)-r^n(1-p)}{p-r},\ p_n=p^n/\lambda_n,\ r_n=r^n/\lambda_n$$ you can check by induction that $$h_{M^n(p,r)}(s)=h_{M(p_n,r_n)}(s)=E(s^{X_n})=\frac{(p_n+r_n-1)s+1-r_n}{1-(1-p_n)s}=1-r_n+\frac{p_nr_ns}{1-(1-p_n)s}.$$ As a consequence $$\Pr(X_n>0)=r_n,\ \ \Pr(X_n=k)=p_nr_n(1-p_n)^{k-1}.$$ The hypothesis $E(X_1)<1$ is equivalent to $p>r$ which implies $p_n\to \frac{p-r}{1-r}=p'$ This elementary method shows that the amswer to the question is $$\Pr(X_n=k|X_n>0)\to (1-p')^{k-1}p'$$ a geometric distribution with mean $1/p'$.

Gérard Letac
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It is known that in the subcritical case, the conditional limit distribution has p.g.f. $F(s)$ such that $F(\phi(s))=1-m(1-F(s))$. Then you can try to calculate the limiting probability. You can also look at the book of T. Harris.

kmitov
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