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Consider a continuous time point process $\eta(t)$ representing the number of points in the interval $[0, t]$. Let $\eta(\infty)$ be distributed as the total number of children of a particle. Define $\mu(t)=\mathbb{E}[\eta(t)]$ and so $\mu(\infty)=\int_0^{\infty}\mu(dt)$.

The Malthusian parameter, if it exists, is defined as the solution $\alpha$ of

$$ 1=\int_0^{\infty} e^{-\alpha t}\mu(dt). $$

I want to show that if $\mu(\infty)\in(1,\infty)$ then the Malthusian parameter $\alpha$ satisfying the integral must exist but I'm completely unsure how to proceed.

It looks like it's presented as a part of Proposition $2.2$ on page $6$ of this paper but I don't see a proof included. Any hints, sources or other questions on this platform related to this are welcome.

elysian-peace
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1 Answers1

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That follows by continuity and intermediate value theorem. We have continuity since by monotonicity

$$|F(a)-F(a+\epsilon)|=\int_{0}^{\infty}e^{-at}(1-e^{-\epsilon t})\mu(dt)\leq \epsilon \int_{0}^{\infty}e^{-at}t\mu(dt)\to 0$$

and similarly for $|F(a)-F(a-\epsilon)|$. And since $F(0)=\mu(\infty)>1$ and $F(+\infty)=0$ we have some $a_{0}$ s.t. $F(a_{0})=1$.

Thomas Kojar
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    Small follow up question: shouldn't it read $\int_{0}^{\infty}e^{-at}(1-e^{\epsilon t})\mu(dt)$ after the first equality? – elysian-peace Feb 02 '24 at 14:24
  • Maybe we should consider $F(a+\varepsilon)-F(a)=\int_{0}^{\infty} e^{-at}(e^{-\varepsilon t}-1) \mu(dt)$. Then, for $t\geq 0$, $e^{-\varepsilon t}-1<x$, bc. $e^{-\varepsilon t}-1<0$, so we can write $F(a+\varepsilon)-F(a)\leq\varepsilon \int_{0}^{\infty}t e^{-at} \mu(dt) $. – toni_iva Feb 04 '24 at 14:55
  • @smoking_big_ole_doinks yes indeed. I updated. – Thomas Kojar Feb 04 '24 at 20:59