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I have a discrete times birth and death process $\{\Psi_n\}_{n\in \mathbb N}$ with birth probability $p$ and death probability $q$ defined as follows: \begin{align} \Psi_n=\sum_{i=1}^n\eta_i \end{align} where $\{\eta_i\}_{i\in\mathbb N}$ is a family of random variables i.i.d. such that \begin{align} \eta_i= \begin{cases} 1,\; &p\\ -1, \; &q\\ \end{cases} \end{align} and such that $\Psi_0=1$. Call \begin{align} V_N=\min\{n\geq 0: \Psi_n\geq N^{\frac{1}{10}}\} \end{align} and define the process that counts the total number of individuals born in the interval$[0,n]$ as \begin{align} T_n=\sum_{i=1}^n\nu(\eta_i) \end{align} where \begin{align} \nu(\eta_i)= \begin{cases} &1 \;\quad\text{if}\;\quad\eta_i=1\\ &0\; \quad\text{if}\;\quad\eta_i=-1\\ \end{cases} \end{align} Define $$W_N=\min\{n\geq 0: T_n\geq\frac{1}{2}N\}$$ I need to prove that \begin{align}\label{plo} \lim_{N\to +\infty}\mathbb P({V_{N}<{W}_N})=1 \end{align} The idea is to apply the law of large numbers in order to compare the times $V_N$ and $W_N$. From the law of large numbers I know that \begin{align} \frac{\Psi_{N^{\frac{2}{10}}}}{(p-q)N^{\frac{2}{10}}}\xrightarrow[N\to+\infty]{}1 \end{align} almost surely, and \begin{align} \frac{T_{N}}{qN}\xrightarrow[N\to+\infty]{}1 \end{align} almost surely. Then, this allows me to prove that $$\mathbb P(V_N\leq N^{2/10}\leq W_N)\xrightarrow[N\to +\infty]{} 1$$.

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  • $\begingroup$ One stupid question: usually by birth-death process one means something that, at least, terminates when everybody is dead (so you never go below zero). What you have described often goes under the names like "random walk on $\mathbb Z$ with drift", etc. Are you sure that you asked exactly what you intended to ask? (if yes, just say "yes"). $\endgroup$
    – fedja
    Commented May 21, 2018 at 23:30
  • $\begingroup$ Yes, you are right. What I defined is actually a random walk and $T_n$ counts the total number of steps in one direction. I used the term birth and death process because for me this is an auxiliary process and in my original system I have a birth and death process. You are right. Thanks for your comment. $\endgroup$
    – user495333
    Commented May 22, 2018 at 13:20

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