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 $$\mathbb P(V_N\leq N^{2/10}\leq W_N)\xrightarrow[N\to +\infty]{} 1$$.

Is that correct in your opinion?