How to get the Corollary from the Theorem

Question

I would like to understand how this Theorem implies the subsequent Corollary:

Theorem. Let A be an irreducible matrix, related to a continuous time Markov branching process $X(t)$, with dominant eigenvalue $\lambda_1$ and associated eigenvector $v_1$ with $2\Re(\lambda)<\lambda_1$ for $\lambda\not=\lambda_1$. Let $Y(t)=l_1\Re(X(t)\cdot v)+l_2\cdot\Im(X(t)\cdot v)$, where $\cdot$ denotes the inner product, $l_1,l_2\in\mathbb{R}$ are arbitrary, and $v$ is an eigenvector with associated eigenvalue $\lambda$. Then $$\lim_{t\to\infty}\mathbb{P}\bigg(0<x_1\le W\le x_2<\infty,\frac{Y(t)}{\sqrt{X(t)\cdot v}}\le x\bigg)=\mathbb{P}(0<x_1\le W\le x_2<\infty)\Phi\Big(\frac x\sigma\Big),$$ where $\Phi$ denotes the cdf of a normal distribution. The Theorem implies a sort of asymptotic independence of $W$ and $Y(t)/\sqrt{X(t)\cdot v}$. Thus one can deduce from Theorem the following

Corollary. Under the conditions of the Theorem, $$\lim_{t\to\infty}\mathbb{P}\bigg(0<x_1\le W\le x_2<\infty,\frac{Y(t)}{e^{\lambda_1 t/2}}\le x\bigg)=\int_{x_1}^{x_2}\Phi\Big(\frac{x}{\sigma\sqrt{y}}\Big)\text{d}_y\mathbb{P}(W\le y)$$

Probably my "deduction" is not good enough to see how the theorem implies the form of the corollary. What I know is that $$e^{-\lambda_1 t}X(t)\cdot v\xrightarrow{a.s}W\quad\text{and}\quad X(t)\cdot v=o(e^{\lambda_1 t}),$$ I think this would justify the change on the LHS of $\sqrt{X(t)\cdot v}$ with $e^{\lambda_1 t/2}$, but I don't understand how the RHS changes. Any help?

(Edit: the property $X(t)\cdot v=o(e^{\lambda_1 t})$ holds for the eigenvalue assumption of the Theorem, while $e^{-\lambda_1 t}X(t)\cdot v\xrightarrow{a.s}W$ holds in general.)

Source: this is a classical result that can be found e.g. in

• K.B. Athreya - Limit Theorems for Multitype Continuous Time Markov Branching Processes I. The Case of an Eigenvector Linear Functional;
• K.B. Athreya, S. Karlin - Embedding of urn schemes into continuous time Markov branching processes and related limit theorems (without the Corollary);
• K.B. Athreya, P.E. Ney - Branching Processes, chapter V.8.

Answer 1

Here is a non-rigorous demonstration; I will follow your notation as closely as possible, though I think the lettering is a little confusing. Assume that everything in sight admits a density, so that conditioning on $W = y$ (which may be probability 0) corresponds to the usual meaning of integrating against the conditional density. Then the theorem implies that \begin{align*} &\lim_{t \to \infty} \int_{x_1}^{x_2} \mathbb P\left(\frac{Y(t)}{\sqrt{X(t) \cdot v}} \leq x \mid W = y \right) d_y \mathbb P(W \leq y) \\ &= \lim_{t \to \infty} \mathbb P \left( 0 < x_1 \leq W \leq x_2 < \infty, \frac{Y(t)}{\sqrt{X(t) \cdot v}} \leq x\right) \\ &= \mathbb P \left( 0 < x_1 \leq W \leq x_2 < \infty\right) \Phi \left(\frac{x}{\sigma}\right) \\ &= \int_{x_1}^{x_2} \Phi \left(\frac{x}{\sigma}\right) d_y \mathbb P(W \leq y). \end{align*} We may take the limit inside the integral since the integrand is bounded and notice that since $x_1, x_2$ are arbitrary, $$ \lim_{t \to \infty} \mathbb P\left(\frac{Y(t)}{\sqrt{X(t) \cdot v}} \leq x \mid W = y \right) = \Phi \left( \frac{x}{\sigma} \right) $$ as expected from the asymptotic independence. Now, note that since $e^{-\lambda_1 t} X(t) \cdot v \to W$ almost surely, on the event $W = y$, we expect that $$ \lim_{t \to \infty} \mathbb P\left(\frac{Y(t)}{e^{\lambda_1 t / 2}} \leq x \mid W = y \right) = \lim_{t \to \infty} \mathbb P\left(\frac{Y(t)}{\sqrt{X(t) \cdot v}} \leq \frac{x}{\sqrt{y}} \mid W = y \right) = \Phi \left( \frac{x}{\sigma \sqrt{y}} \right). $$ Then we may take \begin{align*} &\lim_{t \to \infty} \mathbb P \left( 0 < x_1 \leq W \leq x_2 < \infty, \frac{Y(t)}{e^{\lambda_1 t / 2}} \leq x\right) \\ &=\lim_{t \to \infty} \int_{x_1}^{x_2} \mathbb P\left(\frac{Y(t)}{e^{\lambda_1 t/ 2}} \leq x \mid W = y \right) d_y \mathbb P(W \leq y) \\ &= \int_{x_1}^{x_2} \Phi \left(\frac{x}{\sigma\sqrt{y}}\right) d_y \mathbb P(W \leq y) \end{align*} similarly to before.