I should proof the following for this Markov Chain:
There is a bank account without withdrawal limit. The balance increases by 1000 with probability 0.4, and decreases by 1000 with probability 0.6. Show that "balance=0" is a transient state.
For me this looks like a Random Walk with directional bias, so the state "balance=0" should be transient.
But I have troubles proofing this. My strategy is to establish probabilites for 2n-steps and 2n-1-steps coming back to balance=0 and then sum it up as an infinite sum and show that it converges (meaning that I am not able to come back at all times in infinity). I checked some possibilities, how to get back to balance=0 for 2n-steps (e.g.: $4*(-0.6)+6*0.4$), and for 2n-1 -steps (e.g.: $3*0.4+2*(-0.6)$).
(1) $\mathbb{P}$("back in 2n-steps") = $\binom{2n}{n} p^n (1-p)^n$ = ${2^{2n}\over \sqrt {n\pi}}$ $p^n (1-p)^n$ = ${1\over \sqrt {n\pi}} 4^n p^n (1-p)^n$ = ${1\over \sqrt {n\pi}} (4p(1-p))^n$ = ${1\over \sqrt {n\pi}} (4*0.4*0.6)^n$ = ${1\over \sqrt {n\pi}} 0,96^n$
For the step $\binom{2n}{n}$ = ${2^{2n}\over \sqrt {n\pi}}$ I used the Sterling formula: $n=({n\over e})^n \sqrt {2 \pi n} $. So $\binom{2n}{n}$ = ${2n!\over n!(2n-n)!}$ = ${2n!\over n!n!}$ and we get with the Sterling formula: ${({2n\over e})^{2n} * \sqrt {4\pi n} \over ({n\over e})^{2n} * 2 \pi n}$ = ${2^{2n}\over \sqrt {n \pi}}$
Summing up to infinity: $\Sigma_\inf$ = ${1\over \sqrt {n\pi}} 0,96^n$ brings an infinite sum, that converges (geometric series?)
(2) But for $\mathbb{P}$("back in 2n-1 -steps") I am struggeling because: $\binom{2n-1}{n}$ = ${(2n-1)!\over n!(n-1)!}$. When I insert the Sterling formula I get: ${({(2n-1)\over e})^{2n-1} * \sqrt {2 \pi (2n-1)} \over ({n\over e})^{n} * \sqrt {2 \pi n} * ({(n-1)\over e})^{n-1} * \sqrt {2 \pi (n-1)}}$. And here I get stuck, because I am not able to simplify.
Has anyone a tip how to solve for $\mathbb{P}$("back in 2n-1 -steps") or solve this problem in a different way?
Thanks!
The fortune $S_n$ satisfies $$P(S_{2n}=0)={2n \choose n} \cdot 0.4^n \cdot 0.6^n\le 2^{2n}\cdot 0.4^n \cdot 0.6^n=0.96^n$$ which is summable. No need to use Stirling, we can just sum over all possible sequences of wins and losses. Since $|S_{n+1}-S_n|=1000$, induction shows that $S_n/1000-n$ is an even integer for all $n$, so $$\forall n \quad P(S_{2n-1}=0) =0.$$ Thus this chain is transient.