birth and death process with changing probability I was given this problem and not very sure where to begin. I drew the transition diagram already but it seems the probability goes on and on ... Could I get some hints?
Consider a birth-and-death Markov chain on $\{0, 1, 2,\ldots\}$ with
transition probabilities given by
$$p(0,1)=1$$
$$p(i,i-1)+p(i,i+1) = 1$$
$$ p(i,i+1) = \frac{(i+1)^2}{i^2} p(i,i-1), i\geq1$$
Assuming $X_0 = 0$, calculate the probability of the event
$\{\forall n\geq 1,X_n \geq 1\}$
ps any one knows how to writing equation in the text box? thanks!
 A: Hint
Let
$$
T_i=\min\big\{n\,\big|\,n\ge1\ \text{ and }\ X_n=i\ \big\}\ ,
$$
be the first time that the chain enters the state $\ i\ $, where $\ \min\emptyset\ $ is taken to be $\ \infty \ $.  Then
\begin{align}
\big\{\forall\,n\ge1,X_n\ge1\,\big\}&=\big\{\forall\,n\ge1,T_n\le T_0\,\big\}\\
&=\bigcap_{n=1}^\infty\big\{T_n
\le T_0\,\big\}\ .
\end{align}
However, since $\ \big\{T_n
\le T_0\,\big\}\subseteq\big\{T_m
\le T_0\,\big\}\ $ whenever $\ n\ge m\ $, it follows that
\begin{align}
P\big(\forall\,n\ge1,X_n\ge1
\big)&=P\Big(\bigcap_{n=1}^\infty\big\{T_n
\le T_0\,\big\}\Big)\\
&=\lim_{n\rightarrow\infty}P\big(T_n
\le T_0\big)\ ,
\end{align}
and $\ P\big(T_n
\le T_0\big)\ $ can be determined by solving a gambler's ruin problem with varying probabilities.
Let
\begin{align}
F_n&=\min\big(T_0,T_n\big)\ \ \text{, and}\\
Y_{nt}&=\cases{X_t&if $\ t\le F_n$\\
X_{F_n}&if $\ t>F_n\ $.}
\end{align}
Then $\ Y_{n1}, Y_{n2},\dots\ $ is a Markov chain on $\ \{0,1,\dots,n\}\ $ with initial state $\ 1\ $, absorbing states $\ 0\ $ and $\ n\ $, and transition probabilities
\begin{align}
P\big(Y_{n\,t+1}=i+1\,\big|
\,Y_{nt}=i\,\big)&=\frac{(i+1)^2}{i^2+(i+1)^2}\\
P\big(Y_{n\,t+1}=i-1\,\big|
\,Y_{nt}=i\,\big)&=\frac{i^2}{i^2+(i+1)^2}
\end{align}
for $\ 1\le i\le n-1\ $. The probability  $\ P\big(T_n
\le T_0\big)\ $ we need to calculate is just the probability that this Markov chain gets absorbed in state $\ n\ $:
$$
P\big(T_n
\le T_0\big)=P\big(Y_{F_n}=n\big)\ .
$$
Calculating the absorption probabilities $\ P\big(Y_{F_n}=n\big)\ $ and $\ P\big(Y_{F_n}=0\big)=$$1-P\big(Y_{F_n}=n\big)\ $ is a gambler's ruin problem which can be solved to get expressions for those absorption probabilities in closed form.  I believe the answer to your question will turn out to be
$$
P\big(\forall\,n\ge1,X_n\ge1\big)=1-\frac{1}{\sum_\limits{j=1}^\infty\frac{1}{j^2}}=1-\frac{6}{\pi^2}
$$
