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Consider a pure-birth process $X(t)$ with rates $\lambda_i$ that satisfies $$\sum_{i=0}^\infty \frac{1}{\lambda_i} = \infty.$$ By Reuter's criterion this is sufficient for $X(t)$ to be regular, ie $X(t) < \infty$ for all $t \ge 0$ holds a.s.

For $\lambda > 0$ let $$\hat{X}(\lambda) := \int_0^\infty \lambda e^{-\lambda t} X(t) dt$$ be the formal Laplace-Transform of $X(t)$.

Suppose there is a $\lambda^* > 0$ so that \begin{align} E\hat{X}(\lambda^*) &= 1 \\ E\int_0^\infty t e^{-\lambda^* t} X(t) dt & < \infty \end{align}

holds. Is the expected number of jumps $EX(a) \le 1$ for some $a$?

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If $\mathrm E(X(t))\gt c$ for every $t\gt0$, then for every $\lambda\gt0$, $$ \mathrm E(\hat X(\lambda))\gt\int_0^{+\infty}\lambda\,\mathrm e^{-\lambda t}\,c\,\mathrm dt=c. $$ Apply this to $c=1$.

(Hence, the second integrability hypothesis is not needed.)

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Oh yes of course, that was painfully obvious, sorry. Can you actually show that the second integrability hypothesis follows from $E\hat{X}(\lambda^*)=1$? Can you assume that $\mu(t) := EX(t)$ has a density $f$? Then $E\hat{X}(\lambda^*) = \int_0^\infty e^{-\lambda^* t} f(t) dt$ is a Laplace transform and the hypothesis would then follow from derivative of $E\hat{X}(\lambda^*)$. –  Haderlump Aug 3 '12 at 15:07
    
Note that I did not write that the second integrability condition follows from the condition $E(\hat X(\lambda^*))=1$ (it does not). But, if $E(\hat X(\lambda^*))$ is finite for some $\lambda^*\gt0$, then the second integrability condition holds for every $\lambda\gt\lambda^*$ (and $E(\hat X(\lambda))$ is finite for every $\lambda\gt\lambda^*$). –  Did Aug 3 '12 at 16:42
    
Oh, I forgot: sorry but your question about a density of $E(X(t))$ is opaque to me. –  Did Aug 3 '12 at 16:43
    
I meant "Radon-Nikodym derivative" $d\mu/dL$ where $L$ is the Lebesgue measure, but never mind. The scond integrability condition for $\lambda > \lambda^*$ is all i need. Could you elaborate on the proof a little? I don't see it. –  Haderlump Aug 3 '12 at 17:04
    
Hint: if $\lambda\gt\lambda^*$, there exists some finite $C$ such that $t\mathrm e^{-\lambda t}\leqslant C\mathrm e^{-\lambda^* t}$ for every nonnegative $t$. –  Did Aug 3 '12 at 17:46

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