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I need to prove that for any Branching Process with $\mu > 1$ and extinction probability $a < 1$ with generating function $\phi$ , that $$\phi '(a) < 1 $$

With the assumption that $X_0 = 1$

I am having trouble with this problem as my knowledge of differential equations, which to my understanding is necessary to arrive at the solution, is somewhat limited. After fumbling around with this I have made the observation that $\phi '(a) = \sum_{i=1}^n ip_ia^{i-1}$ which is $\mu$ if $a = 1$.

Nils F
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  • It's not clear to me what your question is, or whether you think you have proved the desired result. And, most of all, how much you know about branching processes. Do you know that the extinction probability $a$ satisfies the equation $a=\phi(a)$? Do you know that the generating function $\phi$ is convex on $[0,1]$? – kimchi lover Dec 04 '17 at 02:50
  • I know that $a$ is the smallest positive solution to $\phi (a) = a$, this fact is about the extent of my knowledge on the subject, though I didn't know that $\phi$ is convex on $[0,1]$. I have not proved the desired result and have made very little progress in doing so, I'm just not sure how to approach this. – Nils F Dec 04 '17 at 02:57

1 Answers1

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There is a derivative in this problem, but differential equations aren't relevant here. But general analytical facts are relevant. Here is one way to approach your problem.

Assume $p_i>0$ for some $i>1$, so the function $\phi$ is strictly convex on $[0,1]$. (Because it is a convex combination of monomials $x^i$, all of which are strictly convex, save $x^0$ and $x^1.$) Then the graph of $\phi$ lies above each tangent line of $\phi$, in particular the one passing through $(a,\phi(a))=(a,a)$. That is to say, for all $x\in[0,1]$, $$ \phi(x)\ge \phi(a) + \phi'(a)(x-a),$$ with, because of the strict convexity, equality only if $x=a$. We are told $a<1$, so in the case $x=1$, we have $$ 1 = \phi(1)>\phi(a)+\phi'(a)(1-a) = a +\phi'(a)(1-a)$$ and hence $1>a+\phi'(a)(1-a)$ so of course $\phi'(a)<1$.

kimchi lover
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