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I'm reading the proof of Corollary 3.13 of this paper. Let $(X(t))_{t\ge0}$ be a Markov branching process associated with a replacement matrix $R$, with eigenvalues $\lambda_i$ associated with left eigenvectors $v_i$ and right eigenvectors $u_i$. We order the eigenvalues s.t. $\lambda_1$ is simple and positive, and $\lambda_1>\Re(\lambda_2)\ge\Re(\lambda_i)$, for any other $i$. Define $P_i(X(t))=v_iu_i^TX(t)$ and assume that $\lambda_2$ is also real. I would like to know why, if $\Re(\lambda_i)\not=\lambda_2$, then, for any $i=3,\dots,n$, $$z^{-\frac{\lambda_2}{\lambda_1}}P_i(X(\tau_b(z)))\xrightarrow{p}0,$$ (convergence in probability) where $\tau_b(z)=\min\{t\ge0:b\cdot X(t)\ge z\}$.

Actually, this should follow from the following theorem:

  • If $\Re(\lambda_i)<\frac{\lambda_1}2$, $$z^{-1/2}\sum_{i=2}^n P_i(X(\tau_b(z)))\xrightarrow{d} c^{-1/2} V_1.$$
  • If $\Re(\lambda_i)=\frac{\lambda_1}2$, $$(z \ln(z))^{-1/2} P_i(X(\tau_b(z)))\xrightarrow{d} c^{-1/2} V_2$$
  • If $\Re(\lambda_i)>\frac{\lambda_1}2$, $$z^{-\lambda_i/\lambda_1} P_i(X(\tau_b(z)))\xrightarrow{a.s.} c^{-\lambda_i/\lambda_1} W,$$ where $c$ is a constant, $V_1,V_2$ are vector-valued random variables jointly Gaussian, $W$ is a vector-valued random variable.

Intuitively, I would understand that, in $z^{-\frac{\lambda_2}{\lambda_1}}P_i$, $P_i$ grows slower for $i\ge3$, but why is this enough to assert convergence in probability?

Kolakoski54
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Dada
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1 Answers1

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The intuition for your question is that (omitting $\text{log}$ factors which are a technical complication but no essential difficulty, and also omitting taking real parts of the eigenvalues):

$X(t)$ is of order $e^{\lambda_1 t}$, $b \cdot X(t)$ is also of order $e^{\lambda_1 t}$, but $P_i(X(t))$ is of order $e^{\lambda_i t}$ if $\lambda_i/\lambda_1 > 1/2$ and $P_i(X(t))$ is of order $e^{\lambda_1 t/2}$ if $\lambda_i/\lambda_1 \le 1/2$.

Hence, $\tau_b(z)$ is order $(1/\lambda_1) \log z$, and thus $P_i(X(\tau_b(z)))$ is of order $z(\lambda_i/\lambda_1)$ if $\lambda_i/\lambda_1 > 1/2$ and $P_i(X(\tau_b(z)))$ is of order $z^{1/2}$ if $\lambda_i/\lambda_1 \le 1/2$.

This is essentially said in Theorem 3.1 and Theorem 3.10(i-iii).

Note that in Theorem 3.10, the result isn't stated for individual $P_i$, but it's stated for $P_\lambda$; this is the same if the eigenvalues are distinct, but in general $P_\lambda$ is the sum of all $P_i$ with the same $\lambda_i$. Nonetheless, for these estimates this makes no difference, since we may decompose $P_\lambda(X(t))$ into $P_i(X(t))$.

In Corollary 3.13, it's assumed $\lambda_2/\lambda_1 > 1/2$, and hence, if $\lambda_i < \lambda_2$, what's said above shows that and $P_i(X(\tau_b(z)))$ is of smaller order than $z^{\lambda_2/\lambda_1}$, regardless of whether $\lambda_i/\lambda_1 \le 1/2$ or not.

Convergence in probability or distribution is the right way to formalize this intuition.

user531372
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