Q) Let $\{X_n\}$ be an asymmetric simple random walk on $\{0,1,2,\cdots\}$ with $P(k,k-1)=q>P(k,k+1)=q$, $p+q=1$ for $k>0$ and $P(0,0)=q$. Show that the stationary distribution is $\pi(k) = (1-p/q)(p/q)^k$
Let $\pi$ be the stationary distribution. For $k>0$,
$$\begin{align} \pi(k) &= \pi(k-1).P(k-1,k)+\pi(k+1).P(k+1,k) \\ &= \pi(k-1).p+\pi(k+1).q \\ \implies \pi(k+1)-\pi(k) &= \frac{p}{q}(\pi(k)-\pi(k-1)) \tag{1} \\ \cdots \\ \implies \pi(2)-\pi(1) &= \frac{p}{q}(\pi(1)-\pi(0)) \tag{2} \end{align} $$
Adding all the equations from $(1)$ to $(2)$, gives:
$$\begin{align} \pi(k+1)-\pi(1) &= \frac{p}{q}(\pi(k)-\pi(0)) \tag{3} \end{align} $$ The boundary equation corresponding to $k=0$ is:
$$\begin{align} \pi(0) &= \pi(1).q+\pi(0).q \\ \pi(1) &= \frac{p}{q}\pi(0) \tag{4} \end{align} $$
Plugging $(4)$ in $(3)$ gives:
$$\begin{align} \pi(k+1)&=\frac{p}{q}\pi(k) \\ &= \left( \frac{p}{q} \right)^k \pi(0) \text{ where } k\geq 1\\ \end{align} $$
Since $\pi$ is a distribution,
$$\begin{align} \sum_{k\geq 1}\left( \frac{p}{q} \right)^k \pi(0) &= 1 \\ \implies \pi(0) &= \frac{1-p/q}{p/q} \\ \implies \pi(k+1) &= \left( \frac{p}{q} \right)^k \pi(0) \\ &= \left( \frac{p}{q} \right)^k \times \frac{1-p/q}{p/q} \\ &= \left( \frac{p}{q} \right)^{k-1}(1-p/q) \end{align} $$
So I have a similar expression but different powers. May I know where I went wrong? Thanks.
$ \sum\limits_{j=0}^{\infty} \pi_j=1$ and not $ \sum\limits_{j=1}^{\infty} \pi_j=1$